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

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 .

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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