# 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 $k$.

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## 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}.\]

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