# Diagonalize the Upper Triangular Matrix and Find the Power of the Matrix

## Problem 583

Consider the $2\times 2$ complex matrix
$A=\begin{bmatrix} a & b-a\\ 0& b \end{bmatrix}.$

(a) Find the eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenvectors.

(c) Diagonalize the matrix $A$.

(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$.

## Solution.

### (a) Find the eigenvalues of $A$.

Since $A$ is an upper triangular matrix, eigenvalues are diagonal entries.
Hence $a, b$ are eigenvalues of $A$.

### (b) For each eigenvalue of $A$, determine the eigenvectors.

Suppose now that $a\neq b$.
Let us find eigenvectors corresponding to the eigenvalue $a$.
We have
\begin{align*}
A-aI=\begin{bmatrix}
0 & b-a\\
0& b-a
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
0 & b-a\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{b-a}R_1}
\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors corresponding to $a$ are
$x_1\begin{bmatrix} 1 \\ 0 \end{bmatrix},$ where $x_1$ is any nonzero complex number.

Next, we find the eigenvectors corresponding to the eigenvalue $b$.
We have
\begin{align*}
A-bI=\begin{bmatrix}
a-b & b-a\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{a-b}R_1}
\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}.
\end{align*}
Hence the eigenvectors corresponding to $b$ are
$x_1\begin{bmatrix} 1 \\ 1 \end{bmatrix},$ where $x_1$ is any nonzero complex number.

### (c) Diagonalize the matrix $A$.

When $a=b$, then $A$ is already diagonal matrix. So let us consider the case $a\neq b$.
In the previous parts, we obtained the eigenvalues $a, b$, and corresponding eigenvectors
$\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{ and } \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$ Let $S=\begin{bmatrix} 1 & 1\\ 0& 1 \end{bmatrix}$ be a matrix whose column vectors are the eigenvectors.
Then $S$ is invertible and we have
$S^{-1}AS=\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}$ by the diagonalization process.

Remark that this formula is also true even when $a=b$.

### (d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$.

Using the result of the diagonalization in part (c), we have
$A=S\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}S^{-1}.$ For each positive integer $k$, we have
\begin{align*}
A^k&=\left(\, S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}S^{-1} \,\right)^k\6pt] &=S\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}^k S^{-1}=S\begin{bmatrix} a^k & 0\\ 0& b^k \end{bmatrix}S^{-1}\\[6pt] &=\begin{bmatrix} 1 & 1\\ 0& 1 \end{bmatrix} \begin{bmatrix} a^k & 0\\ 0& b^k \end{bmatrix} \begin{bmatrix} 1 & -1\\ 0& 1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} a^k & b^k-a^k\\ 0& b^k \end{bmatrix}. \end{align*} In summary, we have the formula \[A^k=\begin{bmatrix} a^k & b^k-a^k\\ 0& b^k \end{bmatrix}.

### 1 Response

1. 10/11/2017

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