# 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 $U^{-1}AU=D$.

Contents

## 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.

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