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

Diagonalization Problems and Solutions in Linear Algebra

Problem 533

Consider the complex matrix
\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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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
&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.

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 \\

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


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

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