# Powers of a Diagonal Matrix

## Problem 7

Let $A=\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}$.
Show that

(1) $A^n=\begin{bmatrix} a^n & 0\\ 0& b^n \end{bmatrix}$ for any $n \in \N$.

(2) Let $B=S^{-1}AS$, where $S$ be an invertible $2 \times 2$ matrix.
Show that $B^n=S^{-1}A^n S$ for any $n \in \N$

Contents

## Hint.

Use mathematical induction.

## Proof.

(1) We prove $A^n=\begin{bmatrix} a^n & 0\\ 0& b^n \end{bmatrix}$ by induction on $n$.
The base case $n=1$ is true by definition.

Suppose that $A^k=\begin{bmatrix} a^k & 0\\ 0& b^k \end{bmatrix}$. Then we have
$A^{k+1}=AA^k =\begin{bmatrix} a & 0\\ 0& b \end{bmatrix} \begin{bmatrix} a^k & 0\\ 0& b^k \end{bmatrix} =\begin{bmatrix} a^{k+1} & 0\\ 0& b^{k+1} \end{bmatrix}.$ Here we used the induction hypothesis in the second equality.
Hence the inductive step holds. This completes the proof.

(2) We show that $B^n=S^{-1}A^n S$ by induction on $n$.
When $n=1$, this is just the definition of $B$.

For induction step, assume that $B^k=S^{-1} A^k S$.
Then we have
$B^{k+1}=B B^k=(S^{-1} A S) (S^{-1} A^k S)=S^{-1}A A^k S=S^{-1} A^{k+1} S,$ where we used the induction hypothesis in the second equality and the third equality follows by canceling $S S^{-1}=I_2$ in the middle.

Thus the inductive step holds, and this competes the proof.

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