Submultimi remarcabile ale unui grup. Ecuatia claselor.

In acest articol vom considera unele submultimi remarcabile ale unui grup, vom stabili cateva proprietati ale grupurilor, si vom demonstra ecuatia claselor. In acest scop vom considera cunoscute urmatoarele:

Notiunea de grup. Vom nota un grup cu $\left ( G,\cdot \right )$ .
Notiunea de subgrup al unui grup.
Notiunile de H – comultime stanga si H – comultime dreapta , unde $\left ( H,\cdot \right )$ este un subgrup al unui grup $\left ( G,\cdot \right )$ .
Notatiile $\left ( G/H \right )_{s},\left ( G/H \right )_{d}$ pentru multimea H – comultimilor stanga , respectiv multimea H – comultimilor dreapta .
Teorema lui Lagrange pentru grupurile finite si notiunea de indice al unui subgrup $H$ intr-un grup $G,\; \left [ G:H \right ]$ .

Definitia 1.

Fie $\left ( G,\cdot \right )$ un grup si $\varnothing \neq X\subset G$ . Multimea $Z_{G}(X)=\left \{ g\in G|gx=xg,\forall x\in X \right \}$ se numeste centralizatorul multimii $X$ in $G$ .

Teorema 1.

Daca $\left ( G,\cdot \right )$ este un grup si $\varnothing \neq X\subset G$ , atunci $\left ( Z_{G}(X),\cdot \right )$ este un subgrup al lui $\left ( G,\cdot \right )$ .

Demonstratie.

Fie $g_{1},g_{2}\in Z_{G}\left ( X \right )$ si $x\in X$ . Avem:

$\left ( g_{1}g_{2} \right )x=g_{1}\left ( g_{2}x \right )\overset{g_{2}\in Z_{G}\left ( X \right )}{=}g_{1}\left ( xg_{2} \right )=\left ( g_{1}x \right )g_{2}\overset{g_{1}\in Z_{G}\left ( X \right )}{=}\left ( xg_{1} \right )g_{2}=x\left ( g_{1}g_{2} \right )\Rightarrow$

$g_{1}g_{2}\in Z_{G}\left ( X \right )\;\; \left ( 1 \right )$ .

$g_{1}\in Z_{G}\left ( X \right )\Rightarrow g_{1}x=xg_{1}\Rightarrow xg_{1}^{-1}=g_{1}^{-1}x\Rightarrow g_{1}^{-1}\in Z_{G}\left ( X \right )\; \; \left ( 2 \right )$ .

Din (1) si (2), rezulta $\left ( Z_{G}\left ( X \right ),\cdot \right )$ subgrup al grupului $\left ( G,\cdot \right )$ .

Observatii:

In cazul particular in care $X=\left \{ x \right \}$ , multimea $Z_{G}\left ( X \right )=Z_{G}\left ( \left \{x \right \} \right )\overset{not}{=}Z_{G}\left ( x \right )$ se mai numeste centralizatorul elementului $x\in G$ (elementele sale comuta cu elementul $x$ ).
In cazul particular in care $X=G$ , multimea $Z_{G}\left ( X \right )=Z\left ( G \right )$ se mai numeste centrul grupului $G$ (elementele sale comuta cu orice element din $G$ ) . Grupul $\left ( G,\cdot \right )$ este grup abelian daca si numai daca $Z\left ( G \right )=G$ .
Daca $\varnothing \neq X_{1}\subset X_{2}\subset G$ , se arata usor ca $Z_{G}\left ( X_{2} \right )\subset Z_{G}\left ( X_{1} \right )$ , deci in particular $Z_{G}\left ( G \right )$ este subgrup al oricarui centralizator.
Este valabila egalitatea $Z_{G}\left ( G \right )=\bigcap_{x\in G}^{\, }Z_{G}\left ( x \right )$ .

Definitia 2.

Fie $\left ( G,\cdot \right )$ un grup si $\varnothing \neq X\subset G$ . Normalizatorul multimii $X$ in $G$ este prin definitie multimea $N_{G}\left ( X \right )=\left \{ g\in G|gX=Xg \right \}=\left \{ g\in G|\left \{ gx|x\in X \right \}=\left \{ xg|x\in X \right \} \right \}.$

Teorema 2.

Daca $\left ( G,\cdot \right )$ este un grup si $\varnothing \neq X\subset G$ , atunci $\left ( N_{G}(X),\cdot \right )$ este un subgrup al lui $\left ( G,\cdot \right )$ , iar $\left ( Z_{G}(X),\cdot \right )$ este un subgrup al lui $\left ( N_{G}(X),\cdot \right )$ .

Demonstratie.

Fie $g_{1},g_{2}\in N_{G}\left ( X \right )$ . Avem:

$\left ( g_{1}g_{2} \right )X=g_{1}\left ( g_{2}X \right )\overset{g_{2}\in N_{G}\left ( X \right )}{=}g_{1}\left ( Xg_{2} \right )=$

$\left ( g_{1}X \right )g_{2}\overset{g_{1}\in N_{G}\left ( X \right )}{=}\left ( Xg_{1} \right )g_{2}=X\left ( g_{1}g_{2} \right )\Rightarrow g_{1}g_{2}\in N_{G}\left ( X \right )\; \, \; \left ( 3 \right ).$

$g_{1}\in N_{G}\left ( X \right )\Rightarrow g_{1}X=Xg_{1}\Rightarrow Xg_{1}^{-1}=g_{1}^{-1}X\Rightarrow g_{1}^{-1}\in N_{G}\left ( X \right )\, \; \; \left ( 4 \right ).$

Din (3) si (4), rezulta $\left ( N_{G}\left ( X \right ),\cdot \right )$ subgrup al grupului $\left ( G,\cdot \right )$ . Pentru a arata ca $\left ( Z_{G}(X),\cdot \right )$ este un subgrup al lui $\left ( N_{G}(X),\cdot \right )$ este suficient sa aratam incluziunea $Z_{G}\left ( X \right )\subset N_{G}\left ( X \right )$ . Intr-adevar $g\in Z_{G}\left ( X \right )\Rightarrow gx=xg,\forall x\in X\Rightarrow gX=Xg\Rightarrow g\in N_{G}\left ( X \right )$ , deci rezulta $Z_{G}\left ( X \right )\subset N_{G}\left ( X \right )$ .

Observatie.

In cazul particular in care $X=\left \{ x \right \}$ , multimea $N\left ( X \right )=N\left ( \left \{x \right \} \right )\overset{not}{=}N\left ( x \right )$ ( normalizatorul elementului $x\in G$ ) coincide centralizatorul acestui element .

Exemplu.

Vom considera un grup neabelian cu $8$ elemente, anume grupul diedral $\left ( G,\cdot \right )=\left ( D_{4},\cdot \right )$ , unde $D_{4}=\left \{ e,\sigma ,\sigma ^{2},\sigma ^{3},\tau ,\tau \sigma ,\tau \sigma ^{2},\tau \sigma ^{3} \right \}$ si $ord\left ( \sigma \right )=4,ord\left ( \tau \right )=2,\sigma \tau =\tau \sigma ^{3}$ . Se probeaza imediat ca in acest grup au loc urmatoarele egalitati:

$Z_{G}\left ( e \right )=Z_{G}\left ( \sigma ^{2} \right )=G$ ;
$Z_{G}\left ( \sigma \right )=Z_{G}\left ( \sigma ^{3} \right )=\left \{ e,\sigma ,\sigma ^{2},\sigma ^{3} \right \}$ :
$Z_{G}\left ( \tau \right )=Z_{G}\left ( \tau \sigma ^{2} \right )=\left \{ e,\sigma ^{2},\tau ,\tau \sigma ^{2} \right \}$ ;
$Z_{G}\left ( \tau \sigma \right )=Z_{G}\left ( \tau \sigma ^{3} \right )=\left \{ e,\sigma ^{2},\tau \sigma ,\tau \sigma ^{3} \right \}$ ;
$Z\left ( G \right )=\left \{ e,\sigma ^{2} \right \}$ ;
$N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )=\left \{ e,\sigma ^{2} \right \}$ ;
$N_{G}\left ( \left \{ \sigma ,\sigma ^{3} \right \} \right )=G$ .

Demonstratie.

Tinand cont de faptul ca $\sigma ^{4}=\tau ^{2}=e,\sigma \tau =\tau \sigma ^{3}\;\;\left ( 5 \right )$ , se obtin si egalitatile $\sigma ^{2}\tau =\tau \sigma ^{2},\sigma ^{3}\tau =\tau \sigma,\left ( \tau \sigma \right )^{2}=\left ( \tau \sigma ^{2} \right )^{2}=\left ( \tau \sigma ^{3} \right )^{2}=e\; \; \left ( 6 \right )$ si se poate completa tabla grupului $\left ( D_{4},\cdot \right )$ .

Astfel punctele a), b), c), d) rezulta pe baza simetriilor din tabel fata de diagonala principala si demonstram celelalte puncte.Intra-adevar:

$Z\left ( G \right )=\bigcap_{x\in G}^{\, }Z_{G}\left ( x \right )=Z_{G}\left ( e \right )\cap Z_{G}\left ( \sigma \right )\cap .... \cap Z_{G}\left ( \tau \sigma ^{3} \right )=\left \{ e,\sigma ^{2} \right \}.$

Apoi avem:

$e\left \{ \sigma ,\tau \right \}=\left \{\sigma ,\tau \right \}=\left \{ \sigma ,\tau \right \}e\Rightarrow e\in N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )\; \; \left ( 7 \right )$

$\left.\begin{matrix} \sigma \left \{ \sigma ,\tau \right \}=\left \{ \sigma ^{2},\sigma \tau \right \}=\left \{ \sigma ^{2},\tau \sigma ^{3} \right \}\\ \left \{ \sigma ,\tau \right \}\sigma =\left \{ \sigma ^{2},\tau \sigma \right \}\end{matrix}\right\}\Rightarrow \sigma \left \{ \sigma ,\tau \right \}\neq \left \{ \sigma ,\tau \right \}\sigma\Rightarrow \sigma \notin N_{G}\left ( \left \{ \sigma ,\tau \right \} \right ) \; \; \left ( 8 \right )$

$\left.\begin{matrix} \sigma ^{2}\left \{ \sigma ,\tau \right \}=\left \{ \sigma ^{3},\sigma ^{2}\tau \right \}=\left \{ \sigma ^{3},\tau \sigma ^{2} \right \}\\ \left \{ \sigma ,\tau \right \}\sigma ^{^{2}}=\left \{ \sigma ^{3},\tau \sigma ^{2} \right \}\end{matrix}\right\}\Rightarrow \sigma ^{2}\left \{ \sigma ,\tau \right \}=\left \{ \sigma ,\tau \right \}\sigma ^{2}\Rightarrow \sigma ^{2}\in N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )\; \; \left ( 9 \right )$

$\left.\begin{matrix} \sigma ^{3}\left \{ \sigma ,\tau \right \}=\left \{ \sigma ^{4},\sigma ^{3}\tau \right \}=\left \{ e,\tau\sigma \right \}\\ \left \{ \sigma ,\tau \right \}\sigma ^{3}=\left \{ \sigma ^{4},\tau \sigma ^{3} \right \}=\left \{ e,\tau \sigma ^{3} \right \}\end{matrix}\right\}\Rightarrow \sigma ^{3}\left \{ \sigma ,\tau \right \}\neq \left \{ \sigma ,\tau \right \}\sigma ^{3}\Rightarrow \sigma ^{3}\notin N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )\; \; \left ( 10 \right )$

$\left.\begin{matrix} \tau \left \{ \sigma ,\tau \right \}=\left \{ \tau \sigma ,\tau ^{2} \right \}=\left \{ \tau \sigma ,e \right \}\\ \left \{ \sigma ,\tau \right \}\tau =\left \{ \sigma \tau ,\tau ^{2} \right \}=\left \{ \tau \sigma ^{3},e \right \}\end{matrix}\right\}\Rightarrow \tau \left \{ \sigma ,\tau \right \}\neq \left \{ \sigma ,\tau \right \}\tau\Rightarrow \tau \notin N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )\; \; \left ( 11 \right )$

$\left.\begin{matrix} \tau \sigma \left \{ \sigma ,\tau \right \}=\left \{ \tau \sigma ^{2},\tau (\sigma \tau ) \right \}=\left \{ \tau \sigma ^{2},\tau \left ( \tau \sigma ^{3} \right ) \right \}=\left \{ \tau \sigma ^{2},\sigma ^{3} \right \}\\ \left \{ \sigma ,\tau \right \}\tau \sigma =\left \{ \left ( \sigma \tau \right )\sigma ,\tau ^{2}\sigma \right \}=\left \{ \left ( \tau \sigma ^{3} \right )\sigma ,\sigma \right \}=\left \{ \tau ,\sigma \right \}\end{matrix}\right\}\Rightarrow$

$\tau \sigma \left \{ \sigma ,\tau \right \}\neq \left \{ \sigma ,\tau \right \}\tau \sigma\Rightarrow \tau \sigma \notin N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )\; \; \left (12 \right )$

$\left.\begin{matrix} \tau \sigma ^{2}\left \{ \sigma ,\tau \right \}=\left \{ \tau \sigma ^{3},\tau \left ( \sigma ^{2}\tau \right ) \right \}=\left \{ \tau \sigma ^{3},\tau \left ( \tau \sigma ^{2} \right ) \right \}=\left \{ \tau \sigma ^{3},\sigma ^{2} \right \}\\ \left \{ \sigma ,\tau \right \}\tau \sigma ^{2}=\left \{ \left ( \sigma \tau \right )\sigma ^{2},\tau ^{2}\sigma ^{2} \right \}=\left \{ \left ( \tau \sigma ^{3} \right )\sigma ^{2},\sigma ^{2} \right \}=\left \{ \tau \sigma ,\sigma ^{2} \right \}\end{matrix}\right\}\Rightarrow$

$\tau \sigma ^{2}\left \{ \sigma ,\tau \right \}\neq \left \{ \sigma ,\tau \right \}\tau \sigma ^{2}\Rightarrow \tau \sigma ^{2}\notin N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )\; \; \left ( 13 \right )$

$\left.\begin{matrix} \tau \sigma ^{3}\left \{ \sigma ,\tau \right \}=\left \{ \tau \sigma ^{4},\tau \left ( \sigma ^{3}\tau \right ) \right \}=\left \{ \tau ,\tau \left ( \tau \sigma \right ) \right \}=\left \{ \tau ,\sigma \right \}\\ \left \{ \sigma ,\tau \right \}\tau \sigma ^{3}=\left \{ \left ( \sigma \tau \right )\sigma ^{3},\tau ^{2}\sigma ^{3} \right \}=\left \{ \left ( \tau \sigma ^{3} \right )\sigma ^{3},\sigma ^{3} \right \}=\left \{ \tau \sigma ^{2},\sigma ^{3} \right \}\end{matrix}\right\}\Rightarrow$

$\tau \sigma ^{3}\left \{ \sigma ,\tau \right \}\neq \left \{ \sigma ,\tau \right \}\tau \sigma ^{3}\; \; \Rightarrow \tau \sigma ^{3}\notin N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )\; \; \left ( 14 \right )\;.$

Din (7),(8),….,(14) rezulta ca normalizatorul multimii $\left \{ \sigma ,\tau \right \}$ in $G$ este multimea $N_{G}\left ( \left \{ \sigma ,\tau \right \} \right )=\left \{ e,\sigma ^{2} \right \}$ si prin aceeasi metoda se arata ca $N_{G}\left ( \left \{ \sigma ,\sigma ^{3} \right \} \right )=G.$

Observatie.

Centralizatoarele $Z_{G}\left ( \sigma \right ),Z_{G}\left ( \sigma ^{3} \right )$ sunt subgrupuri ciclice de ordinul $4$ ale lui $G=D_{4}$ , iar centralizatoarele $Z_{G}\left ( \tau \right ),Z_{G}\left ( \tau \sigma \right ),Z_{G}\left ( \tau \sigma ^{2} \right ),Z_{G}\left ( \tau \sigma ^{3} \right )$ sunt subgrupuri ale lui $D_{4}$ izomorfe cu grupul lui Klein .

Definitia 3.

Fie $\left ( G,\cdot \right )$ un grup si $\varnothing \neq X\subset G$ . Clasa de conjugare a multimii $X$ in $G$ este prin definitie multimea de multimi $\overline{C_{G}}\left ( X \right )=\left \{ gXg^{-1} |g\in G\right \}=\left \{ \left \{ gxg^{-1}|x\in X \right \}|g\in G \right \}.$

Teorema 3.

Demonstratie.

Fie $\left ( G,\cdot \right )$ este un grup si $\varnothing \neq X\subset G$ . Atunci multimile $\overline{C_{G}}\left ( X \right ) si \left ( G/N_{G}\left ( X \right ) \right )_{s}$ sunt echipotente. Daca $G$ este grup finit, atunci cardinalul multimii $\overline{C_{G}}\left ( X \right )$ este egal cu indicele lui $N_{G}\left ( X \right )$ in $G$ .

Fie functia $F:\overline{C_{G}}\left ( X \right )\rightarrow \left (G/N_{G}\left ( X \right )\right )_{s},F\left ( gXg^{-1} \right )=gN_{G}\left ( X \right )$ . Evident $F$ este functie surjectiva si aratam ca este si injectiva. Intr-adevar daca $g_{1},g_{2}\in G$ , astfel incat $F\left ( g_{1} \right )=F\left ( g_{2} \right )$ , avem:

$g_{1}N_{G}\left ( X \right )=g_{2}N_{G}\left ( X \right )\Rightarrow g_{2}\in g_{1}N_{G}\left ( X \right )\Rightarrow \exists n\in N_{G}\left ( X \right )\: a.i.\: g_{2}=g_{1}n\Rightarrow$

$g_{2}Xg_{2}^{-1}=\left ( g_{1}n \right )X\left ( g_{1}n \right )^{-1}=g_{1}\left ( nXn^{-1} \right )g_{1}^{-1}\overset{n\in N_{G}\left ( X \right )}{=}g_{1}Xg_{1}^{-1},$

de unde rezulta $F$ injectiva. Deci multimile $\overline{C_{G}}\left ( X \right ) si \left ( G/N_{G}\left ( X \right ) \right )_{s}$ sunt echipotente iar in cazul in care grupul este finit, cele doua multimi avand acelasi cardinal, se obtine $card\left ( C_{G}\left ( X \right ) \right )=\left [ G:N_{G}\left ( X \right ) \right ]$ .

Teorema 4.

Fie $\left ( G,\cdot \right )$ un grup si $\varnothing \neq X,Y\subset G$ . Atunci:

$\overline{C_{G}}\left ( X \right )= \overline{C_{G}}\left ( Y \right )$ daca si numai daca $Y\in \overline{C_{G}}\left ( X \right )$ .
$\overline{C_{G}}\left ( X \right )\cap \overline{C_{G}}\left ( Y \right )=\varnothing$ sau $\overline{C_{G}}\left ( X \right )= \overline{C_{G}}\left ( Y \right )$ .

Demonstratie.

a)Implicatia directa este evidenta si demonstram implicatia inversa prin dubla incluziune. Avem:

$Y\in \overline{C_{G}}\left ( X \right )\Rightarrow\exists g_{1}\in G\: a.i.\: Y=g_{1}Xg_{1}^{-1}\Rightarrow$

$gYg^{-1}=\left ( gg_{1} \right )X\left ( gg_{1} \right )^{-1 }\in \overline{C_{G}}\left ( X \right ),\forall g\in G\Rightarrow \overline{C_{G}}\left ( Y \right )\subset \overline{C_{G}}\left ( X \right )$

Apoi din $Y=g_{1}Xg_{1}^{-1}$ rezulta $X=g_{1}^{-1}Y\left ( g_{1}^{-1} \right )^{-1}$ ( deci $X\in \overline{C_{G}}\left ( Y \right )$ ) si analog se arata ca $\overline{C_{G}}\left ( X \right )\subset \overline{C_{G}}\left ( Y \right )$ .

b)Fie $\overline{C_{G}}\left ( X \right )\cap \overline{C_{G}}\left ( Y \right )\neq \varnothing$ , deci exista $g_{1},g_{2}\in G$ astfel incat $g_{1}Xg_{1}^{-1}=g_{2}Yg_{2}^{-1}$ , de unde rezulta $Y=\left ( g_{2}^{-1}g_{1} \right )X\left ( g_{2}^{-1}g_{1} \right )^{-1}\in \overline{C_{G}}\left ( X \right )$ si conform punctului (a) $\overline{C_{G}}\left ( X \right )=\overline{C_{G}}\left ( Y \right )$ .

Observatii:

In cazul particular in care $X=\left \{ x \right \}$ , multimea $\overline{C_{G}}\left ( X \right )=\overline{C_{G}}\left ( \left \{x \right \} \right )\overset{not}{=}\overline{C_{G}}\left ( x \right )$ se numeste clasa de conjugare a elementului $x\in G$ .
Multimile $\overline{C_{G}}\left ( x \right )\: si\: \left ( G/Z_{G}\left ( x \right ) \right )_{s}$ sunt echipotente ( normalizatorul $N_{G}\left ( x \right )$ coincide cu centralizatorul $Z_{G}\left ( x \right )$ ).
Daca $x\in Z\left ( G \right )$ avem $\left \{ gxg^{-1}|g\in G \right \}=\left \{ x \right \}$ , deci $\overline{C_{G}}\left ( x \right )=\left \{ x \right \}$ .
Daca $G$ este grup finit, atunci $card\left ( \overline{C_{G}}\left ( X \right ) \right )|ord\left ( G \right )$ pentru orice submultime nevida $X\subset G$ si $card\left ( \overline{C_{G}}\left ( x \right ) \right )=\left [ G:Z_{G}\left ( x \right ) \right ]$ pentru orice element $x\in G$ .

Teorema 5 (ecuatia claselor)

Daca $\left ( G,\cdot \right )$ este un grup finit, atunci exista exista $x_{1},x_{2},....,x_{p}\in G - Z\left ( G \right )$ astfel incat

$G=Z\left ( G \right )\cup \overline{C_{G}}\left ( x_{1} \right )\cup ....\cup \overline{C_{G}}\left ( x_{p} \right )$ si multimile $Z\left ( G \right ), \overline{C_{G}}\left ( x_{1} \right ), ...., \overline{C_{G}}\left ( x_{p} \right )$ sunt disjuncte doua cate doua .
$card\left ( G \right )=card\left ( Z\left ( G \right ) \right )+\left [ G:Z_{G}\left ( x_{1} \right ) \right ]+....+\left [ G:Z_{G}\left ( x_{p} \right ) \right ]$ .

Demonstratie.

a)Fie $a\in Z\left ( G \right )\;\;si\;\;x\in G-Z\left ( G \right )$ . Avem $\overline{C_{G}}\left ( a \right )=\left \{ a \right \}$ si cum $x\notin \overline{C_{G}}\left ( a \right )\overset{teorema\: 4}{\Rightarrow }$ $\overline{C_{G}}\left ( x \right )\cap \overline{C_{G}}\left ( a \right )=\varnothing$ , deci $\overline{C_{G}}\left ( x \right )\cap Z\left ( G \right )=\varnothing \; \; \left ( 15 \right )$ .

Alegem $x_{1}\in G-Z\left ( G \right ),x_{2}\in G-\left [ Z\left ( G \right )\cup \overline{C_{G}}\left ( x_{1} \right ) \right ],x_{3}\in G-\left [ Z\left ( G \right )\cup \overline{C_{G}}\left ( x_{1} \right )\cup \overline{C_{G}}\left ( x_{2} \right ) \right ],....$ si tinand cont de teorema 4, relatia (15) si de faptul ca grupul este finit, rezulta concluzia.

b)Din punctul (a) si si egalitatea $card\left ( \overline{C_{G}}\left ( x \right ) \right )=\left [ G:Z_{G}\left ( x \right ) \right ]$ (teorema 3).

Exemplu.

In grupul diedral $\left ( G,\cdot \right )$ (din exemplul anterior) sunt adevarate urmatoarele egalitati:

$\overline{C_{G}}\left ( e \right )=\left \{ e \right \}$ .
$\overline{C_{G}}\left ( \sigma \right )=\overline{C_{G}}\left ( \sigma^{3} \right )=\left \{ \sigma ,\sigma ^{3} \right \}$ ;
$\overline{C_{G}}\left ( \sigma ^{2} \right )=\left \{ \sigma ^{2} \right \}$ ;
$\overline{C_{G}}\left ( \tau \right )=\overline{C_{G}}\left ( \tau \sigma ^{2} \right )=\left \{ \tau ,\tau \sigma ^{2} \right \}$ ;
$\overline{C_{G}}\left ( \tau \sigma \right )=\overline{C_{G}}\left ( \tau \sigma ^{3} \right )=\left \{ \tau \sigma ,\tau \sigma ^{3} \right \}$ .

Demonstratie.

Tinand cont din nou de egalitatile $\left ( 5 \right )\; si\: \left ( 6 \right )$ obtinem:

$a)\overline{C_{G}}\left ( e \right )=\left \{ geg^{-1} |g\in G\right \}\overset{e\in Z\left ( G \right )}{=}\left \{ e \right \}.$

$b)\overline{C_{G}}\left ( \sigma \right )=\left \{ g\sigma g^{-1}|g\in G \right \}=$

$\left \{ e\sigma e^{-1},\sigma \sigma \sigma ^{-1},\sigma ^{2}\sigma \sigma ^{-2},\sigma ^{3}\sigma \sigma ^{-3},\tau \sigma \tau ^{-1},\left ( \tau \sigma \right )\sigma \left ( \tau \sigma \right )^{-1},\left ( \tau \sigma ^{2} \right )\sigma \left ( \tau \sigma ^{2} \right )^{-1},\left ( \tau \sigma ^{3} \right )\sigma \left ( \tau \sigma ^{3} \right )^{-1} \right \}=$

$\left \{ \sigma,\tau \sigma \tau \right \}=\left \{ \sigma ,\tau \left ( \sigma \tau \right ) \right \}=\left \{ \sigma, \tau \left ( \tau \sigma ^{3} \right ) \right \}=\left \{ \sigma ,\sigma ^{3} \right \}\overset{analog}{=}\overline{C_{G}}\left ( \sigma ^{3} \right ).$

$c)C_{G}\left ( \sigma ^{2} \right )=\left \{ g\sigma ^{2}g^{-1}|g\in G \right \}\overset{\sigma ^{2}\in Z\left ( G \right )}{=}\left \{ \sigma ^{2} \right \}.$

$d)C_{G}\left ( \tau \right )=\left \{ g\tau g^{-1}|g\in G \right \}=$

$\left \{ e\tau e^{-1},\sigma \tau \sigma ^{-1},\sigma ^{2}\tau \sigma ^{-2},\sigma ^{3}\tau \sigma ^{-3},\tau \tau \tau ^{-1},\left ( \tau \sigma \right )\tau \left (\tau \sigma \right )^{-1},\left ( \tau \sigma ^{2} \right )\tau \left ( \tau \sigma ^{2} \right )^{-1},\left ( \tau \sigma ^{3} \right )\tau \left ( \tau \sigma ^{3} \right )^{-1} \right \}=$

$\left \{ \tau,\sigma \tau \sigma ^{3},\sigma ^{2}\tau \sigma ^{2},\sigma ^{3}\tau \sigma ,\tau ,\left ( \tau \sigma \right )\tau \left ( \tau \sigma \right ),\left ( \tau \sigma ^{2} \right )\tau \left ( \tau \sigma ^{2} \right ),\left ( \tau \sigma ^{3} \right )\tau \left ( \tau \sigma ^{3} \right ) \right \}=$

$\left \{ \tau ,\tau \sigma ^{6},\tau \sigma ^{4},\tau \sigma ^{2},\tau ,\tau \sigma ^{2},\tau \sigma ^{4},\tau \sigma ^{6} \right \}=\left \{ \tau ,\tau \sigma ^{2} \right \}\overset{analog}{=}\overline{C_{G}}\left ( \tau \sigma ^{2} \right ).$

Punctul ( e ) se demonstreaza asemanator si se constata verificarea ecuatiei claselor:

$card\left ( G \right )=card\left ( Z\left ( G \right ) \right )+card\left ( \overline{C_{G}}\left (\sigma \right ) \right )+card\left ( \overline{C_{G}}\left (\tau \right ) \right )+card\left ( \overline{C_{G}}\left (\tau \sigma \right ) \right )=$

$card\left ( Z\left ( G \right ) \right )+\left [ G:Z_{G}\left ( \sigma \right ) \right ]+\left [ G:Z_{G}\left ( \tau \right ) \right ]+\left [ G:Z_{G}\left ( \tau \sigma \right ) \right ].$

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Teorema lui Lagrange

In continuare vom considera cunoscute urmatoarele notiuni si proprietati:

Multimea numerelor intregi impreuna cu adunarea formeaza o structura algebrica de grup abelian, pe care il vom nota $\left ( \mathbb{Z},+ \right )$ ;
Subgrupurile grupului $\left ( \mathbb{Z},+ \right )$ sunt multimile $n\mathbb{Z}=\left \{ kn|k\in \mathbb{Z} \right \}$ , unde $n\in \mathbb{Z}^{*}$ ;
Daca $n\in \mathbb{Z}^{*}$ si $a\in \mathbb{Z}$ , atunci $clasa\; lui\; a\; modulo\; n$ este $\widehat{a}=\left \{ a+kn|k\in \mathbb{Z} \right \}$ care mai poate fi scrisa ca $\widehat{a}=a+n\mathbb{Z}$ si privita ca fiind o translatie a subgrupului $n\mathbb{Z}$ in grupul $\left ( \mathbb{Z},+ \right )$ ;
Ideile anterioare pot fi generalizate pentru orice grup cu conditia ca, in cazul grupurilor neabeliene, sa facem distinctie intre translatia la stanga si translatia la dreapta .

Definitia 1.

Fie $H$ un subgrup al grupului $\left ( G,\cdot \right )$ . Pentru $a\in G$ , definim urmatoarele multimi:

$aH=\left \{ ah|h\in H \right \}$ (H – comultimea stanga , prin elementul $a$ ) ;
$Ha=\left \{ ha|h\in H \right \}$ (H – comultimea dreapta , prin elementul $a$ ) .

Observatii.

O H – comultime este de fapt o translatie (stanga, respectiv dreapta) a subgrupului $H$ ;
$H$ este o H – comultime prin elementul neutru $e$ ;
Deoarece $a=ae=ea$ ( $e$ este elementul neutru din grup), rezulta $a\in aH\; si\; a\in Ha$ ;
In general $aH\neq Ha$ . Daca $\left ( G,\cdot \right )$ este grup abelian , atunci $aH=Ha$ ;
Daca $a\in H$ , atunci $aH=Ha=H$ si reciproc daca $aH=H\, \; sau\; Ha=H$ , atunci $a\in H$ ;
Daca operatia din grup este notata aditiv, atunci H- comultimea va fi notata $a+H$ , respectiv $H+a$ .

Lema 1.

Fie $H$ un subgrup al grupului $\left ( G,\cdot \right )$ si $a,b\in G$ . Atunci:

Multimile $H,aH\; si\; Ha$ sunt echivalente;
Multimile $aH,bH,Ha\; si\; Hb$ sunt echivalente.

Demonstratie.

Este suficient sa demonstram ca functiile $f_{a}:H\rightarrow aH,\: f_{a}\left ( x \right )=ax$ , respectiv $g_{a}:H\rightarrow Ha,\: g_{a}\left ( x \right )=xa$ sunt bijective. Intr-adevar cele doua functii sunt corect definite si din $f_{a}\left ( x \right )=f_{a}\left ( y \right )\Leftrightarrow ax=ay\Rightarrow x=y$ , deci $f_{a}$ este injectiva. Apoi daca $y=ax\in aH,\: x\in H$ avem $y=f_{a}\left ( x \right )$ , de unde se obtine ca functia $f_{a}$ este si surjectiva.
Analog se arata ca functia $g_{a}$ este bijectiva.
Din punctul anterior multimile $aH,Ha,bH,Hb$ sunt echivalente cu multimea $H$ , deci sunt echivalente intre ele.

Lema 2.

Fie $H$ un subgrup al grupului $\left ( G,\cdot \right )$ si $a,b\in G$ . Atunci:

$aH=bH$ ( $Ha=Hb$ ) daca si numai daca $b\in aH$ ( $b\in Ha$ );
$aH\cap bH=\O$ ( $Ha\cap Hb=\O$ ) daca si numai daca $b\notin aH$ ( $b\notin Ha$ ).

Demonstratie.

$"\Rightarrow":aH=bH\; si\; b\in bH\Rightarrow b\in aH$ , respectiv $Ha=Hb\; si\; b\in Hb\Rightarrow b\in Ha$ .

$"\Leftarrow":$ Fie $b=ah,x=ah_{1}\in aH,y=bh_{2}\in bH$ . Avem $x=\left ( bh^{-1} \right )h_{1}=b\left ( h^{-1}h_{1} \right )\in bH$ , deci $aH\subset bH$ si $y=\left ( ah \right )h_{2}=a\left ( hh_{2} \right )\in aH$ , deci $bH\subset aH$ . Rezulta $aH=bH$ si demonstratie asemanatoare pentru H – comultimile dreapta ;

$"\Rightarrow ":$ Daca prin reducere la absurd $b\in aH\, \; \left ( b\in Ha \right )$ rezulta din punctul anterior $aH=bH\: \left ( Ha=Hb \right )\Rightarrow aH\cap bH\neq \O \: \left ( Ha\cap Hb\neq \O \right )-Contradictie.$

$"\Leftarrow":$ Fie prin reducere la absurd $x=ah_{1}=bh_{2}\in aH\cap bH,\: h_{1},h_{2}\in H$ . Rezulta $b=xh_{2}^{-1}=$ $=\left ( ah_{1} \right )h_{2}^{-1}=a\left ( h_{1}h_{2}^{-1} \right )\in aH-Contradictie$ si demonstratie asemanatoare pentru H – comultimile dreapta .

Teorema 1(Lagrange).

Intr-un grup finit ordinul oricarui subgrup divide ordinul grupului.

Demonstratie.

Fie un grup finit $\left ( G,\cdot \right )$ avand ordinul $n$ si $H$ un subgrup al lui $G$ avand ordinul $k$ ( $0<k\leqslant n$ ). Tinand cont de faptul ca multimea $G$ este finita, aceasta se poate scrie ca o reuniune finita de H – comultimi stanga , adica $G=a_{1}H\cup a_{2}H\cup ....\cup a_{m}H$ .

Daca $a_{2}\notin a_{1}H,\: a_{3}\notin a_{1}H\cup a_{2}H,\: a_{4}\notin a_{1}H\cup a{_{2}}H\cup a_{3}H,.........$ , atunci, conform lemei 2, H – comultimile $a_{1}H,a_{2}H,...,a_{m}H$ vor fi distincte doua cate doua si, cum fiecare dintre ele contine $k$ elemente (lema 1), vom avea $n=km$ , de unde rezulta $ord\left ( H \right )|ord\left ( G \right )$ .

Corolar 1. Intr-un grup finit ordinul oricarui element divide ordinul grupului.

Demonstratie.

Fie un grup finit $\left ( G,\cdot \right )$ avand ordinul $n$ si $a\in G$ . Avem $ord\left ( a \right )=ord\left ( H \right )$ , unde $H$ este subgrupul generat de elementul $a$ , deci conform teoremei lui Lagrange $ord\left ( a \right )|ord\left ( G \right )$ .

Corolar 2. Fie un grup finit $\left ( G,\cdot \right )$ , de element neutru $e$ , avand ordinul $n$ si $a\in G$ . Atunci $a^{n}=e$ .

Demonstratie.

Fie $ord(a)=k$ . Din corolarul 1 avem $n=k\cdot l,\: l\in \mathbb{N}^{*}$ , de unde rezulta $a^{n}=\left ( a^{k} \right )^{l}=e^{l}=e$ .

Corolar 3. Orice grup finit, de ordin un numar prim , este grup ciclic.

Demonstratie.

Fie un grup finit $\left ( G,\cdot \right )$ avand ordinul $n$ si $a\in G-\left \{ e \right \}$ ( $e$ este elementul neutru ). Din $n$ numar prim si din corolarul 1 rezulta $ord(a)=n$ , deci $G$ este grupul generat de elemenul $a$ , in concluzie este grup ciclic.

Teorema 2.

Fie $H$ un subgrup al grupului $\left ( G,\cdot \right )$ . Atunci multimile $\left ( G/H \right )_{s}=\left \{ xH|x\in G \right \}$ si $\left ( G/H \right )_{d}=\left \{ Hx|x\in G \right \}$ sunt echivalente.

Demonstratie.

Fie functia $\Phi :\left ( G/H \right )_{s}\rightarrow \left ( G/H \right )_{d},\: \Phi (xH)=Hx^{-1}$ . Avem $Hx^{-1}=Hy^{-1}\overset{Lema\: 2}{\Rightarrow }y^{-1}\in Hx^{-1}$ $\Rightarrow y\in xH\overset{Lema\: 2}{\Rightarrow }xH=yH$ , deci functia $\Phi$ este injectiva. Pe de alta parte pentru $Hy\in \left ( G/H \right )_{d}$ avem $Hy=\Phi \left ( y^{-1}H \right )$ , ceea ce inseamna ca functia $\Phi$ este si surjectiva si in concluzie multimile $\left ( G/H \right )_{s}=\left \{ xH|x\in G \right \}$ si $\left ( G/H \right )_{d}=\left \{ Hx|x\in G \right \}$ sunt echivalente.

Observatie.

Teorema anterioara ne arata ca numarul de H – comultimi stanga este egal cu numarul de H – comultimi dreapta.

Definitie 2.

Fie $H$ un subgrup al grupului $\left ( G,\cdot \right )$ . Numim indicele lui $H$ in $G$ numarul de H – comultimi stanga ( numarul de H – comultimi dreapta ).

Observatii:

Indicele subgrupului $H$ in grupul $\left ( G,\cdot \right )$ se noteaza cu $\left [ G:H \right ]$ si poate fi un numar natural nenul sau infinit. El reprezinta cardinalul multimii $\left ( G/H \right )_{s}$ sau cardinalul multimii $\left ( G/H \right )_{d}$ (conform teoremei 2 cele doua multimi sunt echivalente) ;
Indicele unui subgrup intr-un grup ne spune de fapt de cate ori trebuie sa translatam subgrupul ( la stanga de exemplu ) pentru a acoperii intregul grup;
Daca $H$ este un subgrup al unui grup finit, teorema lui Lagrange stabileste egalitatea:

$ord\left ( G \right )=ord\left ( H \right )\cdot \left [ G:H \right ]$

Exemple:

In cazul grupului $\left ( \mathbb{Z},+ \right )$ , pentru $n\in \mathbb{N}^{*}$ avem $\left [\mathbb{Z}:n\mathbb{Z} \right ]=n$ si $\left ( \mathbb{Z}/n\mathbb{Z} \right )_{s}=\left ( \mathbb{Z}/n\mathbb{Z} \right )_{d}=\mathbb{Z}_{n}$ ;
In cazul grupului $\left ( S_{3},\circ \right )$ si subgrupului $H=\left \{ e,\tau \right \}$ , unde $\tau =\left ( 12 \right )=\left ( \begin{matrix} 1 &2 &3 \\ 2 &1 &3 \end{matrix} \right )$ , pentru $\sigma =\left ( \begin{matrix} 1 &2 &3 \\ 2 &3 &1 \end{matrix} \right )$ avem:

$\left [ S_{3}:H \right ]=3,\; \left ( S_{3}/H \right )_{s}=\left \{ H,\sigma H,\sigma ^{2}H \right \},\; \left ( S_{3}/H \right )_{d}=\left \{ H,H\sigma,H\sigma ^{2} \right \}$ , unde

$H=\left \{ e,\tau \right \}=\left \{ \left ( \begin{matrix} 1 &2 &3 \\ 1 &2 &3 \end{matrix} \right ),\left ( \begin{matrix} 1 &2 &3 \\ 2 &1 &3 \end{matrix} \right ) \right \},\: \sigma H=\left \{ \sigma ,\sigma \tau \right \}=\left \{ \left ( \begin{matrix} 1 &2 &3 \\ 2 &3 &1 \end{matrix} \right ),\left ( \begin{matrix} 1 &2 &3 \\ 3 &2 &1 \end{matrix} \right ) \right \}$

$\sigma ^{2}H=\left \{ \sigma ^{2},\sigma ^{2}\tau \right \}=\left \{ \left ( \begin{matrix} 1 &2 &3 \\ 3 &1 &2 \end{matrix} \right ),\left ( \begin{matrix} 1 &2 &3 \\ 1 &3 &2 \end{matrix} \right ) \right \}$ ,

$H\sigma =\left \{ \sigma ,\tau \sigma \right \}=\left \{ \left ( \begin{matrix} 1 &2 &3 \\ 2 &3 &1 \end{matrix} \right ),\left ( \begin{matrix} 1 &2 &3 \\ 1 &3 &2 \end{matrix} \right ) \right \},H\sigma ^{2}=\left \{ \sigma ^{2},\tau \sigma ^{2} \right \}=\left \{ \left ( \begin{matrix} 1 &2 &3 \\ 3 &1 &2 \end{matrix} \right ),\left ( \begin{matrix} 1 &2 &3 \\ 3 &2 &1 \end{matrix} \right ) \right \}$ .

Aplicatie.

Determinati indicele subgrupului $15\mathbb{Z}$ in grupul $3\mathbb{Z}$ .

Solutie.

$3\mathbb{Z}$ se poate scrie ca o reuniune de $15\mathbb{Z}$ – comultimi disjuncte doua cate doua, astfel:

$3\mathbb{Z}=15\mathbb{Z}\cup \left ( 3+15\mathbb{Z} \right )\cup \left ( 6+15\mathbb{Z} \right )\cup \left ( 9+15\mathbb{Z} \right )\cup \left ( 12+15\mathbb{Z} \right )\, \; \; \left ( 1 \right )$

Din (1) rezulta $\left [ 3\mathbb{Z}:15\mathbb{Z} \right ]=5$ .

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Siruri de matrice

In cele ce urmeaza consideram:

$b_{1},b_{2},....,b_{r}$ , r numere complexe fixate;
un sir de matrice $\left ( U_{n} \right )_{n\geqslant 0}$ , unde $U_{n}\in \mathit{M}_{p,q}\left ( \mathbb{C} \right ) , \forall n\geq 0$ . Sirul $\left ( U_{n} \right )_{n\geqslant 0}$ este definit astfel:

$U_{0},U_{1},....,U_{r-1}\in \mathit{M}_{p,q}\left ( \mathbb{C} \right )$ si $U_{n}=b_{1}U_{n-1}+b_{2}U_{n-2}+....+b_{r}U_{n-r},\forall n\geq r\;\; \left ( 1 \right )$ .

Vom spune ca sirul $\;\; \left ( 1 \right )$ este definit printr-o recurenta liniara de ordinul r .

ecuatia algebrica de gradul r , $x^{r}-b_{1}x^{r-1}-b_{2}x^{r-2}-....-b_{r-1}x-b_{r}=0\;\; \left ( 2 \right )$ , pe care o vom numi ecuatia caracteristica atasata recurentei $\left ( 1 \right )$ .
$\lambda _{1},\lambda_{2},....,\lambda _{r}\in \mathbb{C}$ , cele r radacini ale ecuatiei $\;\; \left ( 2 \right )$ . Sa observam ca:

$\lambda _{k}^{n}=b_{1}\lambda _{k}^{n-1}+b_{2}\lambda _{k}^{n-2}+.....+b_{r}\lambda _{k}^{n-r},\forall n\geq r,\forall k=\overline{1,r}\;\; \left ( 3 \right )$ .

Teorema 1.

In cazul in care radacinile ecuatiei caracteristice $(2)$ , $\lambda _{1},\lambda_{2},....,\lambda _{r}$ , sunt distincte doua cate doua , exista matricile $\Gamma _{1},\Gamma _{2},...,\Gamma _{r}\in \mathit{M}_{p,q}\left ( \mathbb{C} \right )$ astfel incat

$U_{n}=\lambda _{1}^{n}\Gamma _{1}+\lambda _{2}^{n}\Gamma _{2}+....+\lambda _{r}^{n}\Gamma _{r},\forall n\geq 0$ .

Demonstratie.

Sistemul matricial $\left\{\begin{matrix} \: X_{1}+ &\: X_{2}+ &.... &+\; X_{r}=U_{0} \\ \lambda _{1}X_{1}+ &\lambda _{2}X_{2}+ &.... &+\lambda _{r}X_{r}=U_{1} \\ .... &.... &.... &.... \\ \lambda _{1}^{r-1}X_{1}+ &\lambda _{2}^{r-1}X_{2}+ &.... &+\lambda _{r}^{r-1}X_{r}=U_{r-1} \end{matrix}\right.,X_{1},X_{2},....,X_{r}\in \mathit{M_{p,q}\left ( \mathbb{C} \right )}\; \; (4)$ are solutie unica, deoarece rezolvarea sa se reduce la rezolvarea in multimea numerelor complexe a $p\cdot q$ sisteme de ecuatii de tip Cramer de forma $\left\{\begin{matrix} \: x_{1}+ &\: x_{2}+ &.... &+\; x_{r}=u_{0} \\ \lambda _{1}x_{1}+ &\lambda _{2}x_{2}+ &.... &+\lambda _{r}x_{r}=u_{1} \\ .... &.... &.... &.... \\ \lambda _{1}^{r-1}x_{1}+ &\lambda _{2}^{r-1}x_{2}+ &.... &+\lambda _{r}^{r-1}x_{r}=u_{r-1} \end{matrix}\right. \; \; (5)$ , deci cu solutie unica (determinantul sistemului $(5)$ este de tip Vendermonde si cum $\lambda _{1},\lambda_{2},....,\lambda _{r}$ sunt distincte, va fi nenul).

Fie $\Gamma _{1},\Gamma _{2},...,\Gamma _{r}\in \mathit{M}_{p,q}\left ( \mathbb{C} \right )$ solutia unica a sistemului $(4)$ . Demonstram prin inductie propozitia $P\left ( n \right ):U_{n}=\lambda _{1}^{n}\Gamma _{1}+\lambda _{2}^{n}\Gamma _{2}+....+\lambda _{r}^{n}\Gamma _{r},\forall n\geq 0$ .

Intr-adevar pentru $n\in \left \{ 0,1,....,r-1 \right \}$ propozitia este adevarata prin modul de alegere a matricilor $\Gamma _{1},\Gamma _{2},...,\Gamma _{r}\in \mathit{M}_{p,q}\left ( \mathbb{C} \right )$ . Apoi, pentru n ≥ r, daca $P\left ( n-1 \right ),P\left ( n-2 \right ),....,P\left ( n-r \right )$ sunt adevarate, avem:

$U_{n}=b_{1}U_{n-1}+b_{2}U_{n-2}+....+b_{r}U_{n-r}=$

$b_{1}\sum_{k=1}^{r}\lambda _{k}^{n-1}\Gamma_{k}+b_{2}\sum_{k=1}^{r}\lambda _{k}^{n-2}\Gamma_{k}+....+b_{r}\sum_{k=1}^{r}\lambda _{k}^{n-r}\Gamma_{k}=$

$\sum_{k=1}^{r}\left ( b_{1}\lambda _{k}^{n-1} +b_{2}\lambda _{k}^{n-2}+....+b_{r}\lambda _{k}^{n-r}\right )\Gamma _{k}=\sum_{k=1}^{r}\lambda _{k}^{n}\Gamma _{k}$ ,

deci propozitia $P\left ( n \right )$ este adevarata, ceea ce incheie demonstratia.

Corolar.

Daca $A\in M_{r}\left ( \mathbb{C} \right )$ este o matrice avand valorile proprii $\lambda _{1},\lambda _{2},....,\lambda _{r}$ distincte doua cate doua, atunci exista matricile $\Gamma _{1},\Gamma _{2},...,\Gamma _{r}\in \mathit{M}_{r}\left ( \mathbb{C} \right )$ astfel incat

$A^{n}=\lambda _{1}^{n}\Gamma _{1}+\lambda _{2}^{n}\Gamma _{2}+....+\lambda _{r}^{n}\Gamma _{r},\forall n\geq 0$

Demonstratie.

Daca $det\left ( xI_{r}-A \right )=x^{r}-b_{1}x^{r-1}-b_{2}x^{r-2}-...-b_{r-1}x-b_{0}$ , atunci conform teoremei Hamilton-Cayley (pe care o consideram cunoscuta ), matricea A satisface relatia:

$A^{r}=b_{1}A^{r-1}+b_{2}A^{r-2}+....+b_{r-1}A+b_{r}I_{r}\; \; \left ( 6 \right )$ .

Inmultind in (6) cu $A^{n-r}$ si considerand sirul de matrice $U_{n}=A^{n},n\geq 0$ , obtinem pentru sirul $\left ( U_{n} \right )_{n\geqslant 0}$ , recurenta liniara de ordinul r

$U_{0},U_{1},....,U_{r-1}\in \mathit{M}_{r}\left ( \mathbb{C} \right )$ si $U_{n}=b_{1}U_{n-1}+b_{2}U_{n-2}+....+b_{r}U_{n-r},\forall n\geq r\;\; \left ( 7 \right )$ .

Ecuatia caracteristica a recurentei (7), este chiar ecuatia caracteristica a matricei A, adica

$det\left ( xI_{r}-A \right )=0 \Leftrightarrow x^{r}-b_{1}x^{r-1}-b_{2}x^{r-2}-...-b_{r-1}x-b_{0}=0\;\; \left ( 8 \right )$ .

Radacinile ecuatiei (8) fiind chiar valorile proprii ale matricei A, demonstratia este incheiata.

Aplicatie.

Daca $n\in \mathbb{N}$ si $A=\begin{pmatrix} 1 &5 &0 \\ 1 &1 &-1 \\ 0 &1 &1 \end{pmatrix}\in M_{3}\left ( \mathbb{R} \right )$ , sa se determine $A^{n}$ .

Solutie.

Avem $det(xI_{3}-A)=0\Leftrightarrow \left ( x-1 \right )\left ( x+1 \right )\left ( x-3 \right )=0$ , deci conform corolarului exista matricile $M,N,P\in M_{3}\left ( \mathbb{R} \right )$ astfel incat $A^{n}=\left ( -1 \right )^{n}M+1^{n}N+3^{n}P$ .

Rezolvand sistemul $\left\{\begin{matrix} M+N+P=I_{3}\\ -M+N+3P=A\\ M+N+9P=A^{2}\end{matrix}\right.$ , obtinem $\left\{\begin{matrix} M=\frac{1}{8}\left ( A^{2}-4A+3I_{3} \right )\\ N=\frac{1}{4}\left ( -A^{2}+2A+3I_{3} \right )\\ P=\frac{1}{8}\left ( A^{2}-I_{3} \right )\end{matrix}\right.$ si apoi

$A=\frac{1}{8}\left \{ \left [ 3^{n}+\left ( -1 ^{n}-2\right ) \right ]A^{2}+4\left [ 1-\left ( -1 \right )^{n} \right ]A+\left [ 3\left ( -1 \right )^{n}+6-3^{n} \right ]I_{3} \right \}=$

$=\frac{1}{8}\begin{pmatrix} 5\cdot 3^{n}+5\left ( -1 \right )^{n}-2 &10\cdot 3^{n}-10\left ( -1 \right )^{n} &-5\cdot 3^{n}-5\left ( -1 \right )^{n}+10 \\ 2\cdot 3^{n}-2\left ( -1 \right )^{n} &4\cdot 3^{n}+4\left ( -1 \right )^{n} &-2\cdot 3^{n}+2\left ( -1 \right )^{n} \\ 3^{n}+\left ( -1 \right )^{n}-2 &2\cdot 3^{n}-2\left ( -1 \right )^{n} &-3^{n}-\left ( -1 \right )^{n}+10 \end{pmatrix}$ .

In cazul in care ecuatia caracteristica are radacini multiple admitem fara demonstratie teorema si corolarul urmator.

Teorema 2.

Daca ecuatia caracteristica $(2)$ are radacinile $\lambda _{1},\lambda _{2},....,\lambda _{s}$ , avand respectiv ordinele de multiplicitate $\l _{1},\l _{2},....,\l _{s}\in \mathbb{N}^{*}$ , unde $\lambda _{1},\lambda _{2},....,\lambda _{s}$ sunt distincte doua cate doua si $\l _{1}+\l _{2}+....+\l _{s}=r$ , atunci exista matricile $\Gamma _{0}^{k},\Gamma _{1}^{k},....,\Gamma _{l_{k}-1}^{k}\in M_{p,q}\left ( \mathbb{C} \right ),k=\overline{1,s}$ , astfel incat

$U_{n}=\lambda _{1}^{n}\left ( \Gamma _{0}^{1}+n\Gamma _{1}^{1}+..+n^{l_{1}-1}\Gamma _{l_{1}-1}^{1} \right )+...+\lambda _{s}^{n}\left ( \Gamma _{0}^{s}+n\Gamma _{1}^{s}+..+n^{l_{s}-1}\Gamma _{l_{s}-1}^{s} \right ) ,$ $\forall n\geq 0$ .

Corolar.

Daca $A\in M_{r}\left ( \mathbb{C} \right )$ este o matrice avand valorile proprii $\lambda _{1},\lambda _{2},....,\lambda _{s}$ cu ordinele de multiplicitate $\l _{1},\l _{2},....,\l _{s}\in \mathbb{N}^{*}$ , unde $\lambda _{1},\lambda _{2},....,\lambda _{s}$ sunt distincte doua cate doua si $\l _{1}+\l _{2}+....+\l _{s}=r$ , atunci exista matricile $\Gamma _{0}^{k},\Gamma _{1}^{k},....,\Gamma _{l_{k}-1}^{k}\in M_{r}\left ( \mathbb{C} \right ),k=\overline{1,s}$ , astfel incat

Aplicatie.

Daca $n\in \mathbb{N}$ si $A=\begin{pmatrix} 4 &-2 &1 \\ 2 &0 &1 \\ -2 &2 &1 \end{pmatrix}\in M_{3}\left ( \mathbb{R} \right )$ , sa se determine $A^{n}$ .

Solutie.

Avem $det(xI_{3}-A)=0\Leftrightarrow \left ( x-1 \right )\left ( x-2 \right )^{2}=0$ , deci conform corolarului exista matricile $M,N,P\in M_{3}\left ( \mathbb{R} \right )$ astfel incat $A^{n}=1^{n}M+\left ( N+nP \right )2^{n}$ .

Rezolvand sistemul $\left\{\begin{matrix} M+N=I_{3}\\ M+2N+2P=A\\ M+4N+8P=A^{2}\end{matrix}\right.$ , obtinem $\left\{\begin{matrix} M=A^{2}-4A+4I_{3}\\ N=-A^{2}+4A-3I_{3}\\ P=\frac{1}{2}\left ( A^{2}-3A+2I_{3} \right )\end{matrix}\right.$ si apoi

$A^{n}=\frac{1}{2}\left [ \left ( n\cdot 2^{n}-2\cdot 2^{n}+2 \right )A^{2}+\left ( -3n\cdot 2^{n}+8\cdot 2^{n}-8 \right )A+\left ( 2n\cdot 2^{n}-6\cdot 2^{n}+8 \right )I_{3} \right ]=$

$=\begin{pmatrix} 3\cdot 2^{n}-2 &2-2^{n+1} &2^{n}-1 \\ 2^{n+1}-2 &2-2^{n} &2^{n}-1 \\ 2-2^{n+1} &2^{n+1}-2 & 1 \end{pmatrix}$ .

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Polinoame ciclotomice

Pentru $n\in \mathbb{N}^{*}$ vom considera numarul complex $\zeta =cos\frac{2\pi }{n}+isin\frac{2\pi }{n}$ si vom nota cu $U_{n}=\left \{ x\in \mathbb{C}|x^{n}=1 \right \}=\left \{ \zeta ^{k}|k=\overline{0,n-1} \right \}$ (multimea radacinilor de ordinul n ale unitatii ). Vom presupune cunoscute urmatoarele proprietati:

$\left ( U_{n},\cdot \right )$ este un subgrup ciclic al grupului abelian $\left ( \mathbb{C}^{*},\cdot \right )$ ;
orice generator al grupului ciclic $\left ( U_{n},\cdot \right )$ se numeste radacina primitiva de ordinul n a unitatii. Vom nota multimea radacinilor primitive de ordinul n ale unitatii cu $P_{n}$ ;
$\zeta =cos\frac{2\pi }{n}+isin\frac{2\pi }{n}$ este una dintre radacile primitive de ordinul n ale unitatii, iar multimea acestora este $P_{n}=\left \{ \zeta ^{k}|k\in \left \{1,...,n-1 \right \},\left ( k,n \right )=1 \right \}$ , iar pentru n numar prim, $P_{n}=\left \{ \zeta ,\zeta ^{2},....,\zeta ^{n-1} \right \}$ ;
cardinalul multimii $P_{n}$ este $\varphi \left ( n \right )$ ( indicatorul lui Euler ), iar pentru n numar prim, cardinalul multimii $P_{n}$ este $\varphi \left ( n \right )=n-1$ ;
daca $\varepsilon \in P_{n}$ , atunci $P_{n}=\left \{ \varepsilon ,\varepsilon ^{2},....,\varepsilon ^{\varphi \left ( n \right )} \right \}=\left \{ \zeta ,\zeta ^{2},....,\zeta ^{\varphi \left ( n \right )} \right \}$ ;

Definitie. Pentru $n\in \mathbb{N}^{*}$ , polinomul $\Phi _{n}\left ( X \right )=\prod_{\zeta \in P_{n}}^{}\left ( X-\zeta \right )$ se numeste al n – lea polinom ciclotomic.

Pentru a stabili anumite proprietati ale polinoamelor ciclotomice, demonstram mai intai urmatoarea lema:

Lema. Familia de multimi $\left ( P_{d} \right )$ , unde d parcurge multimea divizorilor naturali ai lui n , este o partitie a multimii $U_{n}$ , adica:

$\bigcup_{d\in \mathbb{N}^{*},d|n}^{}P_{d}=U_{n}$ ;
$d_{1},d_{2}\in \mathbb{N}^{*},d_{1}|n,d_{2}|n\, \; si\: d_{1}\neq d_{2}\Rightarrow P_{d_{1}}\bigcap P_{d_{2}}=\O$ ;

Demonstratie.

Avem $x\in P_{d}\Rightarrow x\in U_{d}\Rightarrow x^{d}=1\overset{d|n }{\Rightarrow}x^{n}= \left ( x^{d} \right )^{\frac{n}{d}}=1$ $\Rightarrow x\in U_{n}$ , deci $\bigcup_{d\in \mathbb{N}^{*},d|n}^{}P_{d}\subset U_{n}$ . Pentru a demonstra incluziunea inversa fie $x=cos\frac{2k\pi }{n}+isin\frac{2k\pi }{n}\in U_{n}$ si $\left ( k,n \right )=d,k=d\cdot k^{'}$ , $n=d\cdot n^{'}$ . Atunci $x=cos\frac{2k^{'}\pi}{n^{'}}+isin\frac{2k^{'}\pi}{n^{'}}$ si cum $\left ( k^{'},n^{'} \right )=1$ , avem $x\in P_{n^{'}}$ , deci este adevarata si incluziunea inversa $U_{n}\subset \bigcup_{n^{'}\in \mathbb{N}^{*},n^{'}|n}P_{n^{'}}$ .
Fie prin reducere la absurd $x=cos\frac{2k_{1}\pi}{d_{1}}+isin\frac{2k_{1}\pi}{d_{1}}=cos\frac{2k_{2}\pi}{d_{2}}+isin\frac{2k_{2}\pi}{d_{2}}\in P_{d_{1}}\bigcap P_{d_{2}}$ , unde $\left ( k_{1},d_{1} \right )=1\, \: si\: \left ( k_{2},d_{2} \right )=1$ (1) . Rezulta $\frac{2k_{1}\pi }{d_{1}}-\frac{2k_{2}\pi }{d_{2}}=2p\pi ,p\in \mathbb{N}\Leftrightarrow k_{1}d_{2}-k_{2}d_{1}=pd_{1}d_{2}$ si tinand cont de (1) obtinem $d_{1}|d_{2}\: si\: d_{2}|d_{1}$ , deci $d_{1}=d_{2}$ – contradictie .

Teorema. ( Dedekind ) Pentru $n\in \mathbb{N}^{*}$ are loc egalitatea $X^{n}-1=\prod_{d\in \mathbb{N}^{*},d|n}^{}\Phi _{d}\left ( X \right )$ .

Demonstratie.

Tinand cont de lema de mai sus obtinem:

$X^{n}-1=\prod_{\zeta \in U_{n}}\left ( X-\zeta \right )=\prod_{d\in \mathbb{N}^{*},d|n}\prod_{\zeta \in P_{d}}\left ( X-\zeta \right )=\prod_{d\in \mathbb{N}^{*},d|n}\Phi _{d}\left ( X \right )$ .

Observatii:

egaland, in teorema de mai sus, gradele polinoamelor din cei doi membrii se regaseste o egalitate interesanta din teoria numerelor, anume $n=\sum_{d\in \mathbb{N}^{*},d|n}\varphi \left ( d \right )$ ;
Teorema de mai sus permite determinarea pe cale recursiva a polinoamelor ciclotomice. Astfel avem:
1. $\Phi _{1}\left ( X \right )=\left ( X-1 \right )$ ;
2. $X^{2}-1=\Phi _{1}\left ( X \right )\Phi _{2}\left ( X \right )\Rightarrow \Phi _{2}\left ( X \right )=X+1$ ;
3. $X^{3}-1=\Phi _{1}\left ( X \right )\Phi _{3}\left ( X \right )\Rightarrow \Phi _{3}\left ( X \right )=X^{2}+X+1$ ;
4. $X^{4}-1=\Phi _{1}\left ( X \right )\Phi _{2}\left ( X \right )\Phi _{4}\left ( X \right )\Rightarrow \Phi _{4}\left ( X \right )=X^{2}+1$ ;
5. $X^{5}-1=\Phi _{1}\left ( X \right )\Phi _{5}\left ( X \right )\Rightarrow \Phi _{5}\left ( X \right )=X^{4}+X^{3}+X^{2}+X+1$ ;
6. $X^{6}-1=\Phi _{1}\left ( X \right )\Phi _{2}\left ( X \right )\Phi _{3}\left ( X \right )\Phi _{6}\left ( X \right )\Rightarrow \Phi _{6}\left ( X \right )=X^{2}-X+1$ ;
7. $X^{7}-1=\Phi _{1}\left ( X \right )\Phi _{7}\left ( X \right )\Rightarrow \Phi _{7}\left ( X \right )=X^{6}+X^{5}+X^{4}+X^{3}+X^{2}+X+1$ ;
8. $X^{8}-1=\Phi _{1}\left ( X \right )\Phi _{2}\left ( X \right )\Phi _{4}\left ( X \right )\Phi _{8}\left ( X \right )\Rightarrow \Phi _{8}\left ( X \right )=X^{4}+1$ ;
9. $X^{9}-1=\Phi _{1}\left ( X \right )\Phi _{3}\left ( X \right )\Phi _{9}\left ( X \right )\Rightarrow \Phi _{9}\left ( X \right )=X^{6}+X^{3}+1$ ;
10. $X^{10}-1=\Phi _{1}\left ( X \right )\Phi _{2}\left ( X \right )\Phi _{5}\left ( X \right )\Phi _{10}\left ( X \right )\Rightarrow \Phi _{10}\left ( X \right )=X^{4}-X^{3}+X^{2}-X+1$ ;
11. $X^{p}-1=\Phi _{1}\left ( X \right )\cdot \Phi _{p}\left ( X \right )\Rightarrow \Phi _{p}\left ( X \right )=X^{p-1}+X^{p-2}+....+X+1$ pentru orice numar natural prim p ;

Corolar.Polinoamele ciclotomice sunt polinoame monice cu coeficienti intregi ( $\Phi _{n}\in \mathbb{Z}\left [ X \right ]$ pentru orice numar natural nenul n ).

Se poate demonstra prin inductie . Intr-adevar avem $\Phi _{1}\left ( X \right )\in \mathbb{Z}\left [ X \right ]$ monic si daca presupunem $\Phi _{k}\left ( X \right )\in \mathbb{Z}\left [ X \right ]$ si monice pentru orice 1 < k < n, cum $X^{n}-1=\Phi _{n}\left ( X \right )\cdot \prod_{d\in \mathbb{N}^{*},d|n,d<n}\Phi _{d}\left ( X \right )$ , rezulta ca si $\Phi _{n}\left ( X \right )$ va fi un polinom monic cu coeficienti intregi.

Alte proprietati ale polinoamelor ciclotomice:

$\Phi _{n}\left ( X \right )$ este ireductibil in inelul $\mathbb{Z}\left [ X \right ]$ , pentru orice $n\in \mathbb{N}^{*}$ ;
$\Phi _{n}\left ( X \right )$ este un polinom reciproc pentru orice $n\in \mathbb{N}^{*}$ ;
$\Phi _{2n}\left ( X \right )=\Phi _{n}\left ( -X \right )$ pentru orice $n\in \mathbb{N},n>1$ si n impar;
$\Phi _{n}\left ( X \right )=\Phi _{p_{1}\cdot ....\cdot p_{k}}\left ( X^{p_{1}^{r_{1}-1}\cdot ....\cdot p_{k}^{r_{k}-1}} \right )$ , unde $n=p_{1}^{r_{1}}\cdot ....\cdot p_{k}^{r_{k}}$ este descompunerea in factori primi a lui n .

Aplicatie. Sa se arate ca polinomul $f=X^{8}-X^{7}+X^{5}-X^{4}+X^{3}-X+1\in \mathbb{Z}\left [ X \right ]$ este ireductibil peste $\mathbb{Z}$ .

Solutie. Conform teoremei de mai sus avem egalitatea:

$X^{15}-1=\Phi _{1}\left ( X \right )\cdot \Phi _{3}\left ( X \right )\cdot \Phi _{5}\left ( X \right )\cdot \Phi _{15}\left ( X \right )\Leftrightarrow$

$X^{15}-1=\left ( X-1 \right )\left ( X^{2}+X+1 \right )\left ( X^{4}+X^{3}+X^{2}+X+1 \right )\Phi _{15}\left ( X \right )$

de unde rezulta $\Phi _{15}\left ( X \right )=f$ , deci $f$ este ireductibil peste $\mathbb{Z}$ .

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Clase de polinome

Consideram cunoscut faptul ca multimea numerelor intregi si multimea polinoamelor cu coeficienti intr-un corp comutativ au o structura algebrica de inel comutativ. Asadar vom considera $\left (\mathbb{K},+,\cdot \right )$ un corp comutativ si inelele comutative $\left ( \mathbb{Z},+,\cdot \right ),\left ( \mathbb{K}[X],+,\cdot \right )$ . In continuare vom considera un numar natural $n\geqslant 2$ , un polinom $\pi\in \mathbb{K}[X],\: grad(\pi)=n$ si vom face mai multe analogii:

Folosim notatiile:
- $\textit{R}=\left \{ 0,1,....,n-1 \right \}$ multimea resturilor care pot aparea prin impartirea unui numar intreg la numarul n ;
- prin analogie, $\textit{R}^{'}=\left \{ r\in \mathbb{K}[X]|grad(r)\leq n-1 \right \}$ , multimea resturilor care pot aparea prin impartirea unui polinom din $\left (\mathbb{K},+,\cdot \right )$ la polinomul $\pi$ de gradul n ;
Pentru $r\in \mathit{R}$ , respectiv $r^{'}\in \mathit{R}^{'}$ , folosim notatiile:
- $\hat{r}=\left \{ nk+r|k\in \mathbb{Z} \right \}$ multimea numerelor intregi care impartite la n dau restul $r\in \mathit{R}$ ;
- prin analogie, $\hat{r^{'}}=\left \{ \pi\cdot f+r^{'}|f\in \mathbb{K}[X] \right \}$ , multimea polinomelor din $\left (\mathbb{K},+,\cdot \right )$ care prin impartire la polinomul $\pi$ dau restul $\hat{r^{'}}\in \mathit{R^{'}}$ .
Consideram multimile:
- $\mathbb{Z}/(n)=\left \{ \hat{0},\hat{1},.....,\widehat{n-1}\right \}=\mathbb{Z}_{n}$ multimea claselor de resturi modulo n ;
- prin analogie $\mathbb{K}[X]/(\pi)=\left \{ \hat{r^{'}}|r^{'}\in \mathit{R^{'}} \right \}$ multimea claselor de polinoame modulo $\pi$ ;
Alegerea reprezentantilor:
- daca $\hat{r}\in \mathbb{Z}/(n)$ si $a\in \hat{r}$ , notam $\hat{a}=\hat{r}$ si spunem ca $a$ este un reprezentant al clasei $\hat{r}$ ;
- prin analogie, daca $\hat{r^{'}}\in \mathit{R^{'}}$ si $f\in \mathbb{K}[X]/(\pi)$ , notam $\hat{f}=\hat{r^{'}}$ si spunem ca $f$ este un reprezentant al clasei $\hat{r^{'}}$ .
Introducerea operatiilor:
- pe multimea $\mathbb{Z}/(n)$ se introduc operatiile de adunare si inmultire modulo n astfel: $\hat{a}\oplus \hat{b}=\widehat{a+b},\: \hat{a}\otimes \hat{b}=\widehat{ab}$ . Se demonstreaza ca cele doua operatii sunt corect definite (nu depind de alegerea reprezentantilor) si determina pe multimea $\mathbb{Z}/(n)$ o structura de inel comutativ ;
- prin analogie, pe multimea $\mathbb{K}[X]/(\pi)$ se introduc operatiile de adunare si inmultire modulo $f$ astfel: $\hat{f}\oplus \hat{g}=\widehat{f+g},\, \: \hat{f}\otimes \hat{g}=\widehat{fg}$ .Se demonstreaza ca cele doua operatii sunt corect definite (nu depind de alegerea reprezentantilor) si determina pe multimea $\mathbb{K}[X]/(\pi)$ o structura de inel comutativ .
Unitatile inelelor:
- se demonstreaza ca $\hat{a}\in \mathbb{Z}/(n)$ este unitate a inelului $\left ( \mathbb{Z}/(n),\oplus ,\otimes \right )$ daca si numai daca $a$ este prim cu $n$ ;
- prin analogie, se demonstreaza ca $\hat{f}\in \mathbb{K}[X]/(\pi)$ este unitate a inelului $\left ( \mathbb{K}[X]/(\pi),\oplus ,\otimes \right )$ daca si numai daca polinoamele $f$ si $\pi$ sunt relativ prime.
Extinderea corpului :
- se demonstreaza ca inelul $\left ( \mathbb{K}[X]/(\pi),\oplus ,\otimes \right )$ devine corp daca si numai daca polinomul $\pi\in \mathbb{K}[X]$ este ireductibil peste $\mathbb{K}$ ;
- in cazul in care polinomul $\pi\in \mathbb{K}[X]$ este ireductibil peste $\mathbb{K}$ , corpul $\left ( \mathbb{K}[X]/(\pi),\oplus ,\otimes \right )$ poate fi privit ca o extindere a corpului $\left ( \mathbb{K},+,\cdot \right )$ , iar polinomul $\pi$ (ireductibil peste $\mathbb{K}$ ) este reductibil peste $\mathbb{K}[X]/(\pi)$ , avand in corpul $\left ( \mathbb{K}[X]/(\pi),\oplus ,\otimes \right )$ radacina $\widehat{X}$ .

Exemplu:

Daca $\left ( \mathbb{K},+,\cdot \right )=\left ( \mathbb{R},+,\cdot \right )$ si $\pi=X^{2}+1$ , atunci corpul $\left ( \mathbb{K}[X]/(\pi),\oplus ,\otimes \right )$ devine izomorf cu corpul numerelor complexe.

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Corpuri finite

In continuare vom considera $\left ( \mathbb{K} ,+,\cdot \right )$ un corp si vom presupune cunoscute urmatoarele notiuni si proprietati:

caracteristica corpului (notata cu $char\left ( \mathbb{K} \right )$ ) este cel mai mic numar natural nenul k cu proprietatea $\underset{k\: ori}{\underbrace{1+1+....+1}}=0$ , in cazul in care exista un asemenea numar, si vom spune ca caracteristica corpului este 0 in caz contrar;
daca caracteristica unui corp este nenula, atunci caracteristica corpului este un numar prim;
cel mai mic subcorp propriu al corpului $\left ( \mathbb{K} ,+,\cdot \right )$ (fata de relatia de incluziune) se numeste subcorpul prim al corpului $\left ( \mathbb{K} ,+,\cdot \right )$ si il vom nota cu $P_{1}$ ;
orice corp finit este un corp abelian (teorema lui Wedderburn) si caracteristica sa este un numar prim ;
daca $char\left ( \mathbb{K} \right )=0$ , atunci subcorpul prim al lui $\mathbb{K}$ este infinit, $P_{1}=\left \{ 0,1,1+1,1+1+1,.... \right \}$ si este izomorf cu corpul $\left ( \mathbb{Q},+,\cdot \right )$ ;
daca $char\left ( \mathbb{K} \right )=p\neq 0$ , atunci subcorpul prim al lui $\mathbb{K}$ este finit, $P_{1}=\left \{ 0,1,1+1,....,\underset{\underbrace{p-1\: ori}}{1+1+....+1} \right \}$ si este izomorf cu corpul $\left ( \mathbb{Z}_{p},+,\cdot \right )$

Vom demonstra doua teoreme legate de corpurile finite.

Teorema 1.

Daca $\left ( \mathbb{K} ,+,\cdot \right )$ este un corp finit, atunci exista numerele naturale nenule p , n astfel incat p este numar prim si $card\left ( \mathbb{K} \right )=p^{n}$ .

Demonstratie.

Fie p caracteristica corpului si $P_{1}=\left \{ 0,1,1+1,....,\underset{\underbrace{p-1\: ori}}{1+1+....+1} \right \}$ subcorpul prim al lui $\mathbb{K}$ ;
Daca $P_{1}=\mathbb{K}$ , atunci $card\left ( \mathbb{K} \right )=card\left ( P_{1} \right )=p$ , altfel alegem $\omega _{2}\in \mathbb{K}-P_{1}$ si consideram multimea $P_{2}=\left \{ a+b\cdot \omega _{2}|a,b\in P_{1} \right \}$ ;
Daca $P_{2}=\mathbb{K}$ , atunci $card\left ( \mathbb{K} \right )=card\left ( P_{2} \right )=p \cdot p=p^{2}$ , altfel alegem $\omega _{3}\in \mathbb{K}-P_{2}$ si consideram multimea $P_{3}=\left \{ a+b\cdot \omega _{3}|a,b\in P_{2} \right \}$ ;
Daca $P_{3}=\mathbb{K}$ , atunci $card\left ( \mathbb{K} \right )=card\left ( P_{3} \right )=p^{2} \cdot p=p^{3}$ , altfel alegem $\omega _{3}\in \mathbb{K}-P_{3}$ si consideram multimea $P_{4}=\left \{ a+b\cdot \omega _{4}|a,b\in P_{3} \right \}$ ;
……………………………………………………………………………………………………………

Continuand rationamentul, cum $\mathbb{K}$ este multime finita, va exista numarul natural n astfel incat $P_{n}=\mathbb{K}$ , deci $card\left ( \mathbb{K} \right )=card\left ( P_{n} \right )=p^{n-1} \cdot p=p^{n}$ .

Teorema 2.

(Teorema elementului primitiv ).Daca $\left ( \mathbb{K} ,+,\cdot \right )$ este un corp finit, atunci grupul $\left ( \mathbb{K^{*},\cdot } \right )$ este grup ciclic.

Demonstratie.

Consideram $card\left ( \mathbb{K}^{*} \right )=h=p_{1}^{\alpha _{1}}\cdot .....\cdot p_{r}^{\alpha _{r}}$ , $p_{i}$ numar prim, $f_{i}=X^{\frac{h}{p_{i}}}-1\in \mathbb{K}\left [ X \right ]$ , $i=\overline{1,r}$ ;
Deoarece un polinom din $\mathbb{K}\left [ X \right ]$ nu poate avea mai multe radacini in $\mathbb{K}$ decat gradul sau, pentru fiecare $i\in \left \{ 1,2,...,r \right \}$ exista $a_{i}\in \mathbb{K}$ astfel incat $f_{i}\left ( a_{i} \right )\neq 0$ ;
$\textrm{Pentru} \; b_{i}=a_{i}^{\frac{h}{p_{i}^{\alpha _{i}}}}\; \textrm{avem}\; b_{i}^{p_{i}^{\alpha _{i}}}=a_{i}^{h}=1\: \; \textrm{si\; }\; b_{i}^{p_{i}^{\alpha _{i}-1}}=a_{i}^{\frac{h}{p_{i}}}\neq 1\: \textrm{, de unde rezulta ca ordinul elementului}$ $b_{i}$ in grupul $\left ( \mathbb{K}^{*},\cdot \right )$ este egal cu $p_{i}^{\alpha _{i}}$ , pentru fiecare $i\in \left \{ 1,2,....,r \right \}$ ;
Pentru $b=b_{1}b_{2}....b_{r}$ avem $b^{h}=b_{1}^{h}b_{2}^{h}....b_{r}^{h}=1$ , deci ordinul elementului $b$ in grupul $\left ( \mathbb{K}^{*},\cdot \right )$ este un divizor al lui $h$ .
Daca prin absurd $ord\left ( b \right )=d<h$ , atunci d este un divizor propriu al lui h , deci $d=p_{1}^{\beta _{1}}p_{2}^{\beta _{2}}....p_{r}^{\beta {_{r}}}$ cu $\beta _{1}<\alpha _{1},\beta _{2}\leq \alpha _{2},....,\beta _{r}\leq \alpha _{r}$ (dupa o eventuala renumerotare a indicilor).
In cazul de mai sus $d|p_{1}^{\alpha _{1}-1}p_{2}^{\alpha _{2}}....p_{r}^{\alpha _{r}}$ si cum $b^{d}=1$ , se obtine

$b^{p_{1}^{\alpha _{1}-1}\cdot p_{2}^{\alpha {_{2}}}\cdot .....\cdot p_{r}^{\alpha {_{r}}}}=1\Leftrightarrow \left ( b_{1} \cdot b_{2}\cdot .....\cdot b_{r}\right )^{p_{1}^{\alpha _{1}-1}\cdot p_{2}^{\alpha {_{2}}}\cdot .....\cdot p_{r}^{\alpha {_{r}}}}=1\Leftrightarrow b_{1}^{p_{1}^{\alpha _{1}-1}}=1$

in contradictie cu faptul ca $ord\left ( b_{1} \right )=p_{1}^{\alpha _{1}}$ .

In concluzie $ord(b)=h$ si grupul $\left ( \mathbb{K^{*},\cdot } \right )$ este ciclic ( b este un generator sau element primitiv al grupului).

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Matrici partitionate de ordinul doi

In continuare vom stabili o formula de calcul a determinantului unei matrice partitionate de ordinul doi cand una din submatricile patratice componente este nesingulara .

Vom considera, pentru inceput, urmatoarea lema:

Lema.

Daca $A=\left ( a_{ij} \right )_{1\leq i\leq m,1\leq j\leq n},B=\left ( b_{ij} \right )_{1\leq i\leq n,1\leq j\leq p}$ , iar $C=A\cdot B=\left ( c_{ij} \right )_{1\leq i\leq m,1\leq j\leq p}$ , atunci:

Coloanele matricei produs C sunt combinatii liniare ale coloanelor matricei A;
Liniile matricei produs C sunt combinatii liniare ale liniilor matricei B.

Demonstratie.
1. Pentru orice $j\in \left \{ 1,2,....,p \right \}$ avem:

$\begin{pmatrix} c_{1j}\\ c_{2j}\\ .....\\ c_{mj}\end{pmatrix}=\begin{pmatrix} \sum_{k=1}^{n}a_{1k} b_{kj}\\ \sum_{k=1}^{n}a_{2k} b_{kj}\\ ....\\ \sum_{k=1}^{n}a_{mk} b_{kj}\end{pmatrix}=\sum_{k=1}^{n}\begin{pmatrix} a_{1k}b_{kj}\\ a_{2k}b_{kj}\\ .....\\ a_{mk}b_{kj}\end{pmatrix}=\sum_{k=1}^{n}b_{kj}\begin{pmatrix} a_{1k}\\ a_{2k}\\ ....\\ a_{mk}\end{pmatrix}$ ,

deci coloanele matricei C sunt combinatii liniare ale coloanelor matricei A.

2. Pentru orice $i\in \left \{ 1,2,....,m \right \}$ avem:

$\begin{pmatrix} c_{i1} &c_{i2} &.... &c_{ip} \end{pmatrix}=\begin{pmatrix} \sum_{k=1}^{n}a_{ik}b_{k1} &\sum_{k=1}^{n}a_{ik}b_{k2} &..... &\sum_{k=1}^{n}a_{ik}b_{kp} \end{pmatrix}=$

$=\sum_{k=1}^{n}\begin{pmatrix} a_{ik}b_{k1} &a_{ik}b_{k2} &..... &a_{ik}b_{kp} \end{pmatrix}=\sum_{k=1}^{n}a_{ik}\begin{pmatrix} b_{k1} &b_{k2} &.... &b_{kp} \end{pmatrix}$ ,

deci liniile matricei C sunt combinatii liniare ale liniilor matricei B.

Teorema.

Fie matricea bloc $X=\begin{pmatrix}A &B \\ C &D \end{pmatrix}$ , unde $A\in M_{p}\left ( \mathbb{C} \right ),B\in M_{p,q}\left ( \mathbb{C} \right ),C\in M_{q,p}\left ( \mathbb{C} \right ),D\in M_{q}\left ( \mathbb{C} \right )$ .

Daca A este nesingulara, atunci $det(X)=det(A)\cdot det\left ( D-CA^{-1}B \right )$ ;
Daca D este nesingulara, atunci $det(X)=det(D)\cdot det\left ( A-BD^{-1}C \right )$ .

Demonstratie.

1.Adaugand la ultimele q coloane combinatii liniare de primele p colane (tinand cont de lema) si dezvoltand in final dupa primele p linii, folosind regula lui Laplace obtinem:

$det\left ( X \right )=\begin{vmatrix} A &B \\ C &D \end{vmatrix}=\begin{vmatrix} A &B-A\left ( A^{-1}B \right ) \\ C &D-C\left ( A^{-1}B \right ) \end{vmatrix}=$

$=\begin{vmatrix} A &O_{p,q} \\ C &D-CA^{-1}B \end{vmatrix}=det\left ( A \right )\cdot det\left ( D-CA^{-1}B \right )$ .

2.Adaugand la primele p coloane combinatii liniare de ultimele q colane (tinand cont de lema) si dezvoltand in final dupa primele p linii, folosind regula lui Laplace obtinem:

$det\left ( X \right )=\begin{vmatrix} A &B \\ C &D \end{vmatrix}=\begin{vmatrix} A-\left (B D^{-1} \right )C &B-\left ( BD^{-1} \right )D \\ C &D \end{vmatrix}=$

$=\begin{vmatrix} A- BD^{-1}C &O_{p,q} \\ C &D \end{vmatrix}=det\left ( D \right ) \cdot det\left ( A-BD^{-1}C \right )$ .

Aplicatie.

Daca a, b, c, d sunt numere reale, sa se demonstreze egalitatea:

$\begin{vmatrix} a &b &c &d \\ -b &a &-d &c \\ -c &d &a &-b \\ -d &-c &b &a \end{vmatrix}=\left ( a^{2}+b^{2}+c^{2}+d^{2} \right )^{2}$ .

Solutie.

Vom considera cazul $a^{2}+b^{2}\neq 0$ (daca a = b = 0 egalitatea se obtine folosind regula lui Laplace ). Facand notatiile $A=\begin{pmatrix} a &b \\ -b &a \end{pmatrix},B=\begin{pmatrix} c &d \\-d &c \end{pmatrix},C=\begin{pmatrix} -c &d \\ -d &-c \end{pmatrix},D=\begin{pmatrix} a &-b \\ b &a \end{pmatrix}$ si folosind teorema de mai sus obtinem:

$\begin{vmatrix} a &b &c &d \\ -b &a &-d &c \\ -c &d &a &-b \\ -d &-c &b &a \end{vmatrix}=det\left ( A \right )\cdot det\left ( D-CA^{-1}B \right )=$

$det\left [ \begin{pmatrix} a &b \\ -b &a\ \end{pmatrix} \right ]\cdot det\left [ \begin{pmatrix} a &-b \\ b &a \end{pmatrix} -\begin{pmatrix} -c &d \\ -d &-c \end{pmatrix}\begin{pmatrix} a &b \\- b &a \end{pmatrix}^{-1}\begin{pmatrix} c &d \\ -d &c \end{pmatrix}\right ]=$

$\left ( a^{2}+b^{2} \right )\cdot det\left [ \begin{pmatrix} a &-b \\ b &a \end{pmatrix}-\frac{1}{a^{2}+b^{2}} \begin{pmatrix} -c &d \\ -d &-c \end{pmatrix}\begin{pmatrix} a &-b \\ b &a \end{pmatrix}\begin{pmatrix} c &d \\ -d &c \end{pmatrix}\right ]=$

$\frac{1}{a^{2}+b^{2}}\cdot det\left [ \begin{pmatrix} a\left ( a^{2}+b^{2} \right ) &-b\left ( a^{2}+b^{2} \right ) \\ b\left ( a^{2}+b^{2} \right ) &a\left ( a^{2}+b^{2} \right ) \end{pmatrix}-\begin{pmatrix} bd-ac &ad+bc \\ -ad-bc &bd-ac \end{pmatrix} \begin{pmatrix} c &d \\ -d &c \end{pmatrix}\right ]=$

$\frac{1}{a^{2}+b^{2}}\begin{vmatrix} a\left ( a^{2}+b^{2}+c^{2}+d^{2} \right ) &-b\left ( a^{2}+b^{2}+c^{2}+d^{2} \right ) \\ b\left ( a^{2}+b^{2}+c^{2}+d^{2} \right ) &a\left ( a^{2}+b^{2}+c^{2}+d^{2} \right ) \end{vmatrix}=\left ( a^{2}+b^{2}+c^{2}+d^{2} \right )^{2}$ .

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Partitionarea unei matrice (matrice bloc)

In continuare urmeaza mai multe definitii si exemple legate de matricile bloc, precum si unele proprietati ale acestora.

Definitie 1.

Daca $A\in M_{m,p}(\mathbb{C})$ , $B\in M_{m,q}(\mathbb{C})$ definim matricea bloc $X=\left ( A|B \right )=\left (A\; B \right )\in M_{m,p+q}(\mathbb{C})$ , matricea obtinuta prin adaugarea coloanelor matricei B la drepta coloanelor matricei A .
Daca $A\in M_{p,n}(\mathbb{C})$ , $B\in M_{q,n}(\mathbb{C})$ definim matricea bloc $Y=\begin{pmatrix} A\\ \overline{\; \; }\\ B\end{pmatrix}=\begin{pmatrix} A\\ \; \\ B\end{pmatrix}\in M_{p+q,n}\left ( \mathbb{C} \right )$ , matricea obtinuta prin adaugarea randurilor matricei B sub randurile matricei A .

Exemple:

$A=\begin{pmatrix} 1 &2 \\ 3 &4 \end{pmatrix},B=\begin{pmatrix} 1 &2 &1 \\ 0 &-1 &-2 \end{pmatrix}\Rightarrow X=\left (A|B \right )=\left ( A\; B \right )=\begin{pmatrix} 1 &2 &1 &2 &1 \\ 3 &4 &0 &-1 &-2 \end{pmatrix}$ ;

$A=\begin{pmatrix} 1 &2 \\ 0 &3 \end{pmatrix},B=\begin{pmatrix} -1 &2 \\ 0 &1 \\ -3 &4 \end{pmatrix}\Rightarrow Y=\begin{pmatrix} A\\ \overline{\; \; }\\ B\end{pmatrix}=\begin{pmatrix} A\\ \; \\ B\end{pmatrix}=\begin{pmatrix} 1 &2 \\ 0 &3 \\ -1 &2 \\ 0 &1 \\ -3 &4 \end{pmatrix}$ .

Definitie 2.

Daca $A_{1}\in M_{m,p_{1}}\left ( \mathbb{C} \right ),A_{2}\in M_{m,p_{2}}\left ( \mathbb{C} \right ),....,A_{k}\in M_{m,p_{k}}\left ( \mathbb{C} \right )$ ,…., definim recursiv matricile bloc $X_{1}\in M_{m,p_{1}}\left ( \mathbb{C} \right ),X_{2}\in M_{m,p_{1}+p_{2}}\left ( \mathbb{C} \right ),....,X_{k}\in M_{m,p_{1}+p_{2}+....+p_{k}}\left ( \mathbb{C} \right )$ ,…. astfel:

$X_{k}=\left \{ \begin{matrix} A_{1},\: k=1 & \\ \left ( X_{k-1}|A_{k} \right ),\: k\geqslant 2 & \end{matrix} \right.$ .

Daca $B_{1}\in M_{p_{1},n}\left ( \mathbb{C} \right ),B_{2}\in M_{p_{2},n}\left ( \mathbb{C} \right ),....,B_{k}\in M_{p_{k},n}\left ( \mathbb{C} \right )$ ,…., definim recursiv matricile bloc $Y_{1}\in M_{p_{1},n}\left ( \mathbb{C} \right ),Y_{2}\in M_{p_{1}+p_{2},n}\left ( \mathbb{C} \right ),....,Y_{k}\in M_{p_{1}+p_{2}+....+p_{k},n}\left ( \mathbb{C} \right )$ ,….. astfel:

$Y_{k}=\left\{\begin{matrix} B_{1},\; k=1\\ \begin{pmatrix} Y_{k-1}\\ \overline{\; \, \; }\\ B_{k}\end{pmatrix},\; k\geqslant 2\end{matrix}\right.$ .

Pentru matricile bloc $X_{k},Y_{k}$ se vor folosi si notatiile $X_{k}=\left ( A_{1}|A_{2}|....|A_{k} \right )$ sau $X_{k}=\left ( A_{1}\; A_{2}\; ....\; A_{k} \right )$ , respectiv $Y_{k}=\begin{pmatrix} B_{1}\\ \overline{B_{2}}\\ .....\\ \overline{B_{k}}\end{pmatrix}$ sau $Y_{k}=\begin{pmatrix} B_{1}\\ .....\\ B_{n}\end{pmatrix}$ ca in exemplele urmatoare:

Exemple:

$\begin{matrix} A_{1}=\begin{pmatrix} 1\\ -1\end{pmatrix},A_{2}=\begin{pmatrix} 2 &3 &-2 \\ 0 &-1 &1 \end{pmatrix},A_{3}=\begin{pmatrix} -2 &0 \\ 1 &4 \end{pmatrix}\Rightarrow \\ X_{3}=\begin{pmatrix} A_{1}|A_{2}|A_{3} \end{pmatrix}=\begin{pmatrix} A_{1} &A_{2} &A_{3} \end{pmatrix}=\begin{pmatrix} 1 &2 &3 &-2 &-2 &0 \\ -1 &0 &-1 &1 &1 &4 \end{pmatrix} \end{matrix}$ ;

$\begin{matrix} B_{1}=\begin{pmatrix} 1 &2 \\ 3 &-2 \end{pmatrix},B_{2}=\begin{pmatrix} 1 &1 \\ -1 &-1 \\ 0 &2 \end{pmatrix},B_{3}=\begin{pmatrix} 4 &5 \end{pmatrix}\Rightarrow \\ Y_{3}=\begin{pmatrix} B_{1}\\ \overline{B_{2}}\\ \overline{B_{3}}\end{pmatrix}=\begin{pmatrix} B_{1}\\ \; \\ B_{2}\\ \; \\ B_{3}\end{pmatrix}=\begin{pmatrix} 1 &2 \\ 3 &-2 \\ 1 &1 \\ -1 &-1 \\ 0 &2 \\ 4 &5 \end{pmatrix}\end{matrix}$ .

Definitie 3.

Daca $A_{ij}\in M_{m_{i},n_{j}}\left ( \mathbb{C} \right ),i=\overline {1,p},j=\overline{1,q}$ si pentru orice $i\in \left \{1,2,....,p \right \}$ , $X_{i}$ este matricea bloc $\begin{pmatrix} A_{i1}|A_{i2}|....|A_{iq} \end{pmatrix}$ , definim matricea bloc A astfel:

$A=\begin{pmatrix} \underline{X_{1}}\\ \underline{X_{2}}\\ ....\\ \overline{X_{p}}\end{pmatrix}$ $=\begin{pmatrix} X_{1}\\ X_{2}\\ ....\\ X_{p}\end{pmatrix}$ .

Se poate observa ca notand, pentru $j\in \left \{ 1,2,....,q \right \}$ , cu $Y_{j}$ matricea bloc $\begin{pmatrix} A_{1j}\\ \overline{A_{2j}}\\ .....\\ \overline{A_{pj}}\end{pmatrix}$ , atunci matricea A se poate scrie si sub forma $A=\begin{pmatrix} Y_{1}| &Y_{2}| &.... &|Y_{q} \end{pmatrix}=\begin{pmatrix} Y_{1}\, \; &Y_{2}\; &.... &Y_{q} \end{pmatrix}$ . De asemenea, se vor folosi pentru matricea bloc A notatiile:

$A=\begin{pmatrix} A_{11}\; \; \; | &A_{12}\; \; \; | &.... &|A_{1q}\; \; \; \; \\ \overline{A_{21}\; \; \; }| &\overline{A_{22}\; \; \; \; }| &.... &|\overline{A_{2q}\; \; \; \; } \\ .... &..... &.... &.... \\ \overline{A_{p-11}}| &\overline{A_{p-12}}| &.... &|\overline{A_{p-1q}} \\ \overline{A_{p1}\: \: \; \: }| &\overline{A_{p2}\: \; \; \: }| &.... &|\: \, \overline{ A_{pq}\; \; } \end{pmatrix}$ $=\begin{pmatrix} A_{11}\: &A_{12}\: &....\: &A_{1q} \\ A_{21}\: &A_{22}\: &....\: &A_{2q} \\ .... &.... &.... &.... \\ A_{p-11}\: &A_{p-12}\: &....\: &A_{p-1q} \\ A_{p1}\: &A_{p2}\: &....\: &A_{pq} \end{pmatrix}=\left ( A_{ij} \right )_{\begin{matrix} 1\leq i\leq p\\ 1\leq j\leq q \end{matrix}}$

Observatii.

Fie $\left ( A_{ij} \right )_{1\leq i\leq p;\: 1\leq j\leq q}$ o matrice bloc. Atunci:

Matricile $A_{ij},i=\overline{1,p},j=\overline{1,q}$ se mai numesc blocuri sau submatrice ale matricei A ;
Pentru orice $i\in \left \{ 1,2,....,p \right \}$ submatricile $A_{i1},A_{i2},....,A_{iq}$ au acelasi numar de linii si pentru orice $j\in \left \{ 1,2,....,q \right \}$ submatricile $A_{1j},A_{2j},....,A_{pj}$ au acelasi numar de coloane;
Daca p=1, q=n si pentru orice $j\in \left \{ 1,2,...,n \right \}$ $A_{1j}\in M_{m,1}\left ( \mathbb{C} \right )$ , atunci blocurile devin coloanele matricei A si facand notatia $A_{1j}=C_{j}$ se poate scrie $A=\begin{pmatrix} C_{1}\: &C_{2}\: &....\: &C_{n} \end{pmatrix}$ ;
Daca p=m, q=1 si pentru orice $i\in \left \{ 1,2,...,m \right \}$ $A_{i1}\in M_{1,n}\left ( \mathbb{C} \right )$ , atunci blocurile devin liniile matricei A si facand notatia $A_{i1}=L_{i}$ se poate scrie $A=\begin{pmatrix} L_{1}\\ L_{2} \\ .... \\L_{m} \end{pmatrix}$ ;

Privitor la adunarea si inmultirea matricilor bloc, se pot demonstrea urmatoarele teoreme:

Teorema 1.

Fie $A=\left ( A_{ij} \right )_{1\leq i\leq p;1\leq j\leq q},\: B=\left ( B_{ij} \right )_{1\leq i\leq p;1\leq j\leq q}$ doua matrice bloc la fel partitionate (adica pentru orice $i\in \left \{ 1,2,....,p \right \}$ si pentru orice $j\in \left \{ 1,2,....,q \right \}$ matricile bloc $A_{ij}\: si\: B_{ij}$ sunt de acelasi tip). Atunci matricile A si B se pot aduna pe blocuri dupa regula:

$A+B=\left ( A_{ij}+B_{ij} \right )_{1\leq i\leq p,1\leq j\leq q}$ .

Teorema 2.

Fie $A=\left ( A_{ij} \right )_{1\leq i\leq p;1\leq j\leq q},\: B=\left ( B_{ij} \right )_{1\leq i\leq q;1\leq j\leq r}$ doua matrice bloc cu proprietatea ca pentru orice $i\in \left \{ 1,2,...,p \right \},k\in \left \{1,2,...,q \right \},j\in \left \{ 1,2,....,q \right \}$ se poate efectua produsul $A_{ik}\cdot B_{kj}$ (adica numarul de coloane al submatricei A_ik este egal cu numarul de linii al submatricei B_kj. Atunci matricile A si B se pot inmulti pe blocuri dupa regula:

$A\cdot B=\left ( \sum_{k=1}^{q}A_{ik}\cdot B_{kj} \right )_{1\leq i\leq p,1\leq j\leq r}$ .

Exemplu:

$\begin{pmatrix} -1 &0 &2 &1 &3 \\ 0 &-1 &1 &-2 &1 \\ 0 &0 &0 &1 &1 \\ 1 &0 &0 &0 &1 \\ 0 &1 &1 &0 &2 \end{pmatrix}\cdot \begin{pmatrix} 1 &2 &0 \\ -1 &0 &1 \\ 2 &1 &0 \\ 3 &0 &-1 \\ 1 &1 &1 \end{pmatrix}=$

$\begin{pmatrix} \begin{pmatrix} -1 &0 \\ 0 &-1 \end{pmatrix}\cdot \begin{pmatrix} 1 &2 &0 \\ -1 &0 &1 \end{pmatrix}+\begin{pmatrix} 2 &1 &3 \\ 1 &-2 &1 \end{pmatrix}\cdot \begin{pmatrix} 2 &1 &0 \\ 3 &0 &-1 \\ 1 &1 &1 \end{pmatrix}\\ \begin{pmatrix} 0 &0 \\ 1 &0 \\ 0 &1 \end{pmatrix}\cdot \begin{pmatrix} 1 &2 &0 \\ -1 &0 &1 \end{pmatrix}+\begin{pmatrix} 0 &1 &1 \\ 0 &0 &1 \\ 1 &0 &2 \end{pmatrix}\cdot \begin{pmatrix} 2 &1 &0 \\ 3 &0 &-1 \\ 1 &1 &1 \end{pmatrix} \end{pmatrix}=$

$\begin{pmatrix} \begin{pmatrix} -1 &-2 &0 \\ 1 &0 &-1 \end{pmatrix}+\begin{pmatrix} 10 &5 &2 \\ -3 &2 &3 \end{pmatrix}\\ \begin{pmatrix} 0 &0 &0 \\ 1 &2 &0 \\ -1 &0 &1 \end{pmatrix}+\begin{pmatrix} 4 &1 &0 \\ 1 &1 &1 \\ 4 &3 &2 \end{pmatrix} \end{pmatrix}=\begin{pmatrix} 9 &3 &2 \\ -2 &2 &2 \\ 4 &1 &0 \\ 2 &3 &1 \\ 3 &3 &3 \end{pmatrix}$ .

Aplicatie (formula lui Schur):

Daca $X_{n}=\begin{pmatrix} A_{11} &A_{12} &..... &A_{1n} \\ 0 &A_{22} &..... &A_{2n} \\ .... &.... &.... &.... \\ 0 &0 &.... &A_{nn} \end{pmatrix}$ este o matrice bloc triunghiulara, astfel incat matricile $A_{11},A_{22},....,A_{nn}$ sunt patratice, atunci $det(X_{n})=det\left ( A_{11} \right )\cdot det\left ( A_{22} \right )\cdot ....\cdot det\left ( A_{nn} \right )$ .

Pentru demonstratie folosim metoda inductie matematice.
Intr-adevar: formula se verifica pentru n=1, iar pentru calculul determinantului matricei X_n+1 , folosind formula lui Laplace si tinand cont de ipoteza inductiva obtinem:

$det\left ( X_{n+1} \right )=det\left ( X_{n} \right )\cdot det\left ( A_{n+1n+1} \right )=$ .

$det\left ( A_{11} \right )\cdot det\left ( A_{22} \right )\cdot ....\cdot det\left ( A_{nn} \right ).\cdot det\left ( A_{n+1n+1} \right )$ .

Academia de matematica aici:http://robeauty.ro

Formula Cauchy-Binet

Vom considera numerele naturale nenule m si n , m ≤ n si matricile $A=\left ( a_{ij} \right )\in M_{m,n}\left ( \mathbb{C} \right )$ , $B=\left ( b_{ij} \right )\in M_{n,m}\left ( \mathbb{C} \right )$ . Pentru $1\leqslant j_{1}\leqslant j_{2}\leqslant ....\leqslant j_{m}\leqslant n$ facem notatiile:

$A^{j_{1}j_{2}....j_{m}}=\left | \begin{matrix} a_{1j_{1}} &a_{1j_{2}} &.... &a_{1j_{m}} \\ a_{2j_{1}} &a_{2j_{2}} &.... &a_{2j_{m}} \\ .... &.... &.... &.... \\ a_{mj_{1}} &a_{mj_{2}} &.... &a_{mj_{m}} \end{matrix} \right |$ , $B_{j_{1}j_{2}....j_{m}}=\left | \begin{matrix} b_{j_{1}1} &b_{j_{1}2} &.... &b_{j_{1}m} \\ b_{j_{2}1} &b_{j_{2}2} &.... &b_{j_{2}m} \\ .... &.... &.... &.... \\ b_{j_{m}1} &b_{j_{m}2} &.... &b_{j_{m}m} \end{matrix} \right |$

si demonstram urmatoarea egalitate cunoscuta sub numele de formula Cauchy-Binet:

$det(A\cdot B)=\sum_{1\leqslant j_{1}<j_{2}<....< j_{m}\leqslant n}^{ }A^{j_{1}j_{2}....j_{m}}\cdot B_{j_{1}j_{2}....j_{m}}$ .

Pentru demonstratia acestei formule, vom nota $A\cdot B=C=\left ( c_{ij} \right )\in M_{m}\left ( \mathbb{C} \right )$ si avem:

$det(A\cdot B)=\sum_{\sigma \in S_{m}}^{ }\varepsilon\left ( \sigma \right )c_{1\sigma \left ( 1 \right )}c_{2\sigma \left ( 2 \right )}....c_{m\sigma \left ( m \right )}=$

$=\sum_{\sigma \in S_{m}}^{ }\varepsilon \left ( \sigma \right )\left [ \sum_{k_{1}=1}^{n}a_{1k_{1}}b_{k_{1}\sigma \left ( 1 \right )} \right ]\left [ \sum_{k_{2}=1}^{n}a_{2k_{2}}b_{k_{2}\sigma \left ( 2 \right )} \right ]....\left [ \sum_{k_{m}=1}^{n}a_{mk_{m}}b_{k_{m}\sigma \left ( m \right )} \right ]=$

$=\sum_{\sigma \in S_{m}}^{ }\varepsilon \left ( \sigma \right )\sum_{k_{1},k_{2},....,k_{m}=1}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}b_{k_{1\sigma \left ( 1 \right )}}b_{k_{2}\sigma \left ( 2 \right )}....b_{k_{m}\sigma \left ( m \right )}=$

$=\sum_{k_{1},k_{2},....,k_{m}=1}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}\sum_{\sigma \in S_{m}}^{ }\varepsilon \left ( \sigma \right )b_{k_{1}\sigma \left ( 1 \right )}b_{k_{2}\sigma \left ( 2 \right )}....b_{k_{m}\sigma \left ( m \right )}=$

$=\sum_{k_{1},k_{2},....,k_{m}=1}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}B_{k_{1}k_{2}....k_{m}}=$

$=\sum_{\begin{matrix} k_{1},k_{2},....,k_{m}=1\\ B_{k_{1},k_{2},....,k_{m}}\neq 0 \end{matrix}}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}B_{k_{1}k_{2}....k_{m}}=$

$=\sum_{\begin{matrix} k_{1},k_{2},....,k_{m}=1\\ k_{1},k_{2},....,k_{m}\: distincte \end{matrix}}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}B_{k_{1}k_{2}....k_{m}}=$

$=\sum_{1\leqslant j_{1}<j_{2}<....<j_{m}\leqslant n}^{ }\sum_{\tau \in S_{m}}^{ }a_{1\tau \left ( j_{1} \right )}a_{2\tau \left ( j_{2} \right )}....a_{m\tau \left ( j_{m} \right )}B_{\tau \left ( j_{1} \right )\tau \left ( j_{2} \right )....\tau \left ( j_{m} \right )}=$

$=\sum_{1\leqslant j_{1}<j_{2}<....<j_{m}\leqslant n}^{ }\sum_{\tau \in S_{m}}^{ }\varepsilon \left ( \tau \right )a_{1\tau \left ( j_{1} \right )}a_{2\tau \left ( j_{2} \right )}....a_{m\tau \left ( j_{m} \right )}B_{ j_{1} j_{2} .... j_{m} }=$

$=\sum_{1\leqslant j_{1}<j_{2}<....<j_{m}\leqslant n}^{ }B_{ j_{1} j_{2} .... j_{m} }\sum_{\tau \in S_{m}}^{ }\varepsilon \left ( \tau \right )a_{1\tau \left ( j_{1} \right )}a_{2\tau \left ( j_{2} \right )}....a_{m\tau \left ( j_{m} \right )}=$

$\sum_{1\leqslant j_{1}<j_{2}<....< j_{m}\leqslant n}^{ }A^{j_{1}j_{2}....j_{m}}\cdot B_{j_{1}j_{2}....j_{m}}$ .

Observatie. In cazul particular n = m, formula Cauchy-Binet ne conduce la egalitatea cunoscuta $det\left ( A\cdot B \right )=det\left ( A \right )\cdot det\left ( B \right )$ .

Aplicatie:Daca n ≥3 si $a_{i},b_{i},c_{i}\in \mathbb{R},i=\overline{1,n}$ , atunci:

$\left | \begin{matrix} a_{1}^{2}+....+a_{n}^{2} &a_{1}b_{1}+....+a_{n}b_{n} &a_{1}c_{1}+....+a_{n}c_{n} \\ a_{1}b_{1}+....+a_{n}b_{n} &b_{1}^{2}+....+b_{n}^{2} &b_{1}c_{1}+....+b_{n}c_{n} \\ a_{1}c_{1}+....+a_{n}c_{n} &b_{1}c_{1}+....+b_{n}c_{n} & c_{1}^{2}+....+c_{n}^{2} \end{matrix} \right |\geqslant 0$ .

Solutie.Considerand matricile $A=\left ( \begin{matrix} a_{1} &.... &a_{n} \\ b_{1} &.... &b_{n} \\ c_{1} &.... &c_{n} \end{matrix} \right ),B=\left ( \begin{matrix} a_{1} &b_{1} &c_{1} \\ .... &.... &.... \\ a_{n} &b_{n} &c_{n} \end{matrix} \right )$ , notand cu Δ determinantul din enunt si tinand cont de formula Cauchy-Binet obtinem:

$\bigtriangleup =det\left ( A\cdot B \right )=\sum_{1\leqslant i<j<k\leqslant n}^{ }A^{ijk}\cdot B_{ijk}=$

$=\sum_{1\leqslant i<j<k\leqslant n}^{ } \left | \begin{matrix} a_{i} &a_{j} &a_{k} \\ b_{i} &b_{j} &b_{k} \\ c_{i} &c_{j} &c_{k} \end{matrix} \right |\cdot \left | \begin{matrix} a_{i} &b_{i} &c_{i} \\ a_{j} &b_{j} &c_{j} \\ a_{k} &b_{k} &c_{k} \end{matrix} \right |=\sum_{1\leqslant i<j<k\leqslant n}^{ } \left | \begin{matrix} a_{i} &a_{j} &a_{k} \\ b_{i} &b_{j} &b_{k} \\ c_{i} &c_{j} &c_{k} \end{matrix} \right |^{^{2}}\geqslant 0$ .

Academia de matematica aici:http://robeauty.ro

Determinantul matricei adjuncte

Consideram matricea $A=\begin{pmatrix} a_{11} &.... &a_{1j} &.... &a_{1n} \\ .... &.... &.... &.... &.... \\ a_{i1} &.... &a_{ij} &.... &a_{in} \\ .... &.... &.... &.... &.... \\ a_{n1} &.... &a_{nj} &.... &a_{nn} \end{pmatrix}$ cu elemente numere complexe si definim:

$\Delta {_{ij}}$ determinantul matricei obtinuta din matricea A prin eliminarea liniei „i ” si coloanei „j ” , $i,j\epsilon \left \{ 1,2,....,n \right \}$ ;
$A {_{ij}}=(-1)^{i+j}\Delta {_{ij}}$ complementul algebric al elementului $a {_{ij}}$ , $i,j\epsilon \left \{ 1,2,....,n \right \}$ ;
$A^{*}=\begin{pmatrix} A_{11} &.... &A_{j1} &.... &A_{n1} \\ .... &.... &.... &.... &.... \\ A_{1i} &.... &A_{ji} &.... &A_{ni} \\ .... &.... &.... &.... &.... \\ A_{1n} &.... &A_{jn} &.... &A_{nn} \end{pmatrix}$ adjuncta matricii A .

In conditiile notatiilor de mai sus demonstram egalitatea:

$det\left ( A^{*} \right )=\left [ det(A) \right ]^{n-1}$ (1).

Pentru demonstratie consideram urmatoarele cazuri:

$\mathit{Cazul\: 1:}det(A)\neq 0$ . In acest caz matricea A este inversabila si inversa sa este $A^{-1}=\frac{1}{det(A)}A^{*}$ . Se obtine

$1=det\left ( I_{n} \right )=det(A)\cdot det\left ( A^{-1} \right )=det(A)\cdot \left (\frac{1}{det(A)} \right )^{n}\cdot det\left ( A^{*} \right )$ ,

de unde rezulta egalitatea (1).

$\mathit{Cazul\: 2:}A=O_{n}$ . Avem evident $A^{*}=O_{n}$ , deci $det\left ( A^{*} \right )=0$ .

$\mathit{Cazul\: 3:}A\neq O_{n}\: si\: det(A)=0$ . Pentru a demonstra egalitatea (1) vom considera sistemul omogen:

$^{t}\left ( A^{*} \right )X=O_{n,1}\Leftrightarrow \begin{pmatrix} A_{11} &.... &A_{1j} &.... &A_{1n} \\ .... &.... &.... &.... &.... \\ A_{i1} &.... &A_{ij} &.... &A_{in} \\ .... &.... &.... &.... &... \\ A_{n1} &.... &A_{nj} &.... &A_{nn} \end{pmatrix}\cdot \begin{pmatrix} x_{1}\\ ....\\ x_{i}\\ ....\\ x_{n}\end{pmatrix}=\begin{pmatrix} 0\\ ....\\ 0\\ ....\\ 0\end{pmatrix}\; \;(S)$

Cum $A\neq O_{n}$ , exista i,j∈{1,2,….,n } astfel incat $a_{ij}\neq 0$ si atunci sistemul omogen (S ) va avea solutia nebanala $x_{1}=a_{i1},....,x_{j}=a_{ij},....,x_{n}=a_{in}$ , ceea ce conduce la $det\left ( A^{*} \right )=0$ .

Academia de matematica aici:http://robeauty.ro

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