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

Diagonalization Problems and Solutions in Linear Algebra

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

 
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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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1 Response

  1. 10/15/2017

    […] The solution is given in the post ↴ Diagonalize the $2times 2$ Hermitian Matrix by a Unitary Matrix […]

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