# Diagonalize the Complex Symmetric 3 by 3 Matrix with $\sin x$ and $\cos x$

## Problem 533

Consider the complex matrix

\[A=\begin{bmatrix}

\sqrt{2}\cos x & i \sin x & 0 \\

i \sin x &0 &-i \sin x \\

0 & -i \sin x & -\sqrt{2} \cos x

\end{bmatrix},\]
where $x$ is a real number between $0$ and $2\pi$.

Determine for which values of $x$ the matrix $A$ is diagonalizable.

When $A$ is diagonalizable, find a diagonal matrix $D$ so that $P^{-1}AP=D$ for some nonsingular matrix $P$.

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## Solution.

Let us first find the eigenvalues of the matrix $A$.

To do so, we compute the characteristic polynomial $p(t)=\det(A-tI)$ of $A$ as follows.

Using Sarrus’s rule to compute the $3\times 3$ determinant, we have

\begin{align*}

&p(t)=\det(A-tI)\\[6pt]
&=\begin{bmatrix}

\sqrt{2}\cos x -t & i \sin x & 0 \\

i \sin x & -t &-i \sin x \\

0 & -i \sin x & -\sqrt{2} \cos x-t

\end{bmatrix}\\[6pt]
&=-t(\sqrt{2}\cos x-t)(-\sqrt{2}\cos x -t)

-\left(\, -(\sin^2 x) (-\sqrt{2}\cos x-t)-(\sin^2 x) (\sqrt{2}\cos x -t) \,\right)\\

&=-t^3+2(\cos^2 x-\sin ^2 x)t\\

&=-t^3+2\cos(2x) t.

\end{align*}

The eigenvalues of $A$ are the roots of

\[p(t)=-t^3+2\cos(2x) t=-t(t^2-2\cos(2x)).\]
Hence the eigenvalues are

\[t=0, \quad\pm \sqrt{2\cos(2x)}.\]

Note that if $\sqrt{2\cos(2x)}=-\sqrt{2\cos(2x)}$ then we have $\cos(2x)=0$ and hence $x=\pi/4, 3\pi/4$.

It follows that if $x=\pi/4, 3\pi/4$, then the matrix $A$ has only one eigenvalue $0$ with algebraic multiplicity $3$.

Since $A$ is not the zero matrix, the rank of $A$ is greater than or equal to $1$.

Hence the nullity of $A$ is less than or equal to $2$ by the rank-nullity theorem.

It follows that the geometric multiplicity (=nullity) of the eigenvalue $0$ is strictly less than the algebraic multiplicity of $0$ and $A$ is not diagonalizable in this case.

Now suppose that $x\neq \pi/4, 3\pi/4$.

In this case, the matrix $A$ has three distinct eigenvalues $0, \pm \sqrt{2\cos(2x)}$.

This implies that $A$ is diagonalizable.

Let $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ be eigenvectors corresponding to eigenvalues $0, \pm \sqrt{2\cos(2x)}$, respectively.

Define the $3\times 3$ matrix $P$ by $P=\begin{bmatrix}

\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 \\

\end{bmatrix}$.

It follows from the general procedure of the diagonalization that $P$ is a nonsingular matrix and

\[P^{-1}AP=D,\]
where $D$ is a diagonal matrix

\[D=\begin{bmatrix}

0 & 0 & 0 \\

0 &\sqrt{2\cos(2x)} &0 \\

0 & 0 & -\sqrt{2\cos(2x)}

\end{bmatrix}.\]

### Summary

In summary, when $x=\pi/4, 3\pi/4$ the matrix $A$ is not diagonalizable.

When $x \neq \pi/4, 3\pi/4$, the matrix $A$ is diagonalizable and we can take the diagonal matrix $D$ as

\[D=\begin{bmatrix}

0 & 0 & 0 \\

0 &\sqrt{2\cos(2x)} &0 \\

0 & 0 & -\sqrt{2\cos(2x)}

\end{bmatrix}.\]

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