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:

  1. Coloanele matricei produs C  sunt combinatii liniare ale coloanelor matricei A;
  2. 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 ) .
  1. Daca A  este nesingulara, atunci det(X)=det(A)\cdot det\left ( D-CA^{-1}B \right );
  2. Daca 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 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 Aik este egal cu numarul de linii al submatricei Bkj. 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 Xn+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 ) .
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