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High powers of a square matrix.

Let $D= \text{diag}(\alpha_1, \alpha_2, \dots , \alpha_r)$. By induction, we prove that $\forall n \in \bbb{N}, D^n =\text{diag}(\alpha_1^n, \alpha_2^n, \dots , \alpha_r^n)$.

Let A be a square matrix; we wish to compute An for some positive integer n. If n is a large number, this is a tedious task. Suppose now that A is diagonalizable, i.e. there exists an invertible matrix P such that D=P-1AP is a diagonal matrix. Then A= PDP-1 and we have:


\begin{displaymath}A^n= PD \underbrace{P^{-1}P}_{=I}D \underbrace{P^{-1}P}_{=I}D \underbrace{P^{-1} \quad }_{=I} \dots PDP^{-1}
=P D^n P^{-1}.
\end{displaymath}

Example 11.7.1   The sequences (un) and (vn) are defined by their first terms u0 and v0 and the relations $\begin{cases}u_{n+1}= \frac{1}{12} (5 u_n + v_n) \\ v_{n+1}= \frac{1}{12} (u_n + 5 v_n) \end{cases}$. We wish to know whether these sequences converge , and eventually towards which limits.

Write the given relations in matricial form: $\begin{pmatrix}u_{n+1} \\ v_{n+1} \end{pmatrix}
= \underbrace{\begin{pmatrix}\...
...d{pmatrix}}
_{\underset{def}{=} A}
\begin{pmatrix}u_{n} \\ v_{n} \end{pmatrix}$.

By induction, we show that $\begin{pmatrix}u_n \\ v_n \end{pmatrix} = A^n \begin{pmatrix}u_0 \\ v_0 \end{pmatrix}$.

The characteristic polynomial of A is $P_A(\lambda)=\frac 16 (6\lambda^2 -5 \lambda + 1) = \left( m- \frac 13
\right) \left( m- \frac 12 \right)$. the matrix A has two eigenvalues, namely $\frac 12$ and $\frac 13$, therefore A is diagonalizable and is similar to $D= \begin{pmatrix}\frac 12 & 0 \\ 0 & \frac 13 \end{pmatrix}$. We denote by P the change of basis matrix.

Then we have: An= PDnP-1 and $\underset{n \rightarrow \infty}{\lim} A^n= P \underset{n \rightarrow \infty}{\lim} D^n P^{-1}$.

As $D^n=\begin{pmatrix}\left( \frac 12 \right)^n & 0 \\ 0 & \left( \frac 13 \right)^n \end{pmatrix}$, we have: $\underset{n \rightarrow \infty}{\lim} D^n =0$, therefore $\underset{n \rightarrow \infty}{\lim} A^n =0$. As a consequence, we have that the sequences (un) and (vn) are convergent and have 0 as their common limit.


next up previous contents
Next: The exponential of a Up: Eigenvalues and eigenvectors. Previous: Diagonalization of a square
Noah Dana-Picard
2001-02-26