# Diagonalize the $2\times 2$ Hermitian Matrix by a Unitary Matrix

## Problem 585

Consider the Hermitian matrix
$A=\begin{bmatrix} 1 & i\\ -i& 1 \end{bmatrix}.$

(a) Find the eigenvalues of $A$.

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

(c) Diagonalize the Hermitian matrix $A$ by a unitary matrix. Namely, find a diagonal matrix $D$ and a unitary matrix $U$ such that $U^{-1}AU=D$.

## Solution.

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

To find the eigenvalues of the matrix $A$, we first compute the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
1-t & i\\
-i& 1-t
\end{vmatrix}\6pt] &=(1-t)(1-t)-(i)(-i)=t^2-2t+1-1\\ &=t(t-2). \end{align*} Solving p(t)=0, we obtain the eigenvalues 0 and 2. ### (b) For each eigenvalue of A, find the eigenvectors. Let us first find the eigenvectors corresponding to the eigenvalue 0. We solve the equation (A-0I)\mathbf{x}=\mathbf{0}. We have \begin{align*} A\xrightarrow{R_2+iR_1} \begin{bmatrix} 1 & i\\ 0& 0 \end{bmatrix}. \end{align*} Hence the eigenvalues are \[c\begin{bmatrix} 1 \\ i \end{bmatrix} for any nonzero complex number $c$.

Next, we find the eigenvectors corresponding to the eigenvalue $2$. We solve the equation $(A-2I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A-2I&=\begin{bmatrix}
-1 & i\\
-i& -1
\end{bmatrix}\xrightarrow{-R_1}\6pt] &\begin{bmatrix} 1 & -i\\ -i& -1 \end{bmatrix}\xrightarrow{R_2+iR_1} \begin{bmatrix} 1 & -i\\ 0& 0 \end{bmatrix}. \end{align*} Hence the eigenvalues are \[c\begin{bmatrix} 1 \\ -i \end{bmatrix} for any nonzero complex number $c$.

### (c) Diagonalize the Hermitian matrix $A$ by a unitary matrix.

To diagonalize the Hermitian matrix $A$ by a unitary matrix $U$, we find an orthonormal basis for each eigenspace of $A$.
As each eigenspace of $A$ is $1$-dimensional by part (b), we just need to normalize any eigenvector for each eigenvalue.

By part (b), we know that $\mathbf{v}_1:=\begin{bmatrix} 1 \\ i \end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $0$.
The length of this vector is
$\|\mathbf{v}_1\|=\sqrt{1\cdot 1 +(i)(-i)}=\sqrt{2}.$ Hence the vector
$\mathbf{u}_1=\frac{\mathbf{v}_1}{\|\mathbf{v}_1\|}=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ i \end{bmatrix}$ is a unit eigenvector.

Similarly, from (b) we see that $\begin{bmatrix} 1 \\ -i \end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $2$. The length of the vector is $\sqrt{2}$.
Hence
$\mathbf{u}_2=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -i \end{bmatrix}$ is a unit eigenvector.

It follows that the matrix
$U=\begin{bmatrix} \mathbf{u}_1 & \mathbf{u}_2 \end{bmatrix}=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ i& -i \end{bmatrix}$ is unitary and
$U^{-1}AU=\begin{bmatrix} 0 & 0\\ 0& 2 \end{bmatrix}$ by diagonalization process.

### 1 Response

1. 10/15/2017

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