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 $P$.

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
\[D=\begin{bmatrix}
0 & 0 & 0 \\
0 &\sqrt{2\cos(2x)} &0 \\
0 & 0 & -\sqrt{2\cos(2x)}
\end{bmatrix}.\] Click here if solved 19

The post Diagonalize the Complex Symmetric 3 by 3 Matrix with $\sin x$ and $\cos x$ first appeared on Problems in Mathematics.

Find the inverse matrix of the $3\times 3$ matrix
\[A=\begin{bmatrix}
7 & 2 & -2 \\
-6 &-1 &2 \\
6 & 2 & -1
\end{bmatrix}\] using the Cayley-Hamilton theorem.

Solution.

To apply the Cayley-Hamilton theorem, we first determine the characteristic polynomial $p(t)$ of the matrix $A$.
Let $I$ be the $3\times 3$ identity matrix.
We have
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
7-t & 2 & -2 \\
-6 &-1-t &2 \\
6 & 2 & -1-t
\end{vmatrix}\\[6pt] &=(7-t)\begin{vmatrix}
-1-t & 2\\
2& -1-t
\end{vmatrix}
-2\begin{vmatrix}
-6 & 2\\
6& -1-t
\end{vmatrix}+(-2)\begin{vmatrix}
-6 & -1-t\\
6& 2
\end{vmatrix}\\[6pt] &\text{(by the first row cofactor expansion)}\\[6pt] &=-t^3+5t^2-7t+3.
\end{align*}
(You may also use the rule of Sarrus to compute the $3\times 3$ determinant.)

Thus, we have obtained the characteristic polynomial
\[p(t)=-t^3+5t^2-7t+3\] of the matrix $A$.

The Cayley-Hamilton theorem yields that
\[O=p(A)=-A^3+5A^2-7A+3I,\] where $O$ is the $3\times 3$ zero matrix.
(Here, don’t forget to put the identity matrix $I$.)

Rearranging terms, we have
\begin{align*}
&A^3-5A^2+7A=3I\\[6pt] &\Leftrightarrow A(A^2-5A+7I)=3I\\[6pt] &\Leftrightarrow A\left(\frac{1}{3}(A^2-5A+7I)\right)=I.
\end{align*}
Similarly, we have
\[\left(\frac{1}{3}(A^2-5A+7I)\right)A=I.\] It follows from these two equalities that the matrix
\[\frac{1}{3}(A^2-5A+7I)\] is the inverse matrix of $A$.

Therefore, we have
\begin{align*}
A^{-1}&=\frac{1}{3}(A^2-5A+7I)\\[6pt] &=\frac{1}{3}\left(\, \begin{bmatrix}
25 & 8 & -8 \\
-24 &-7 &8 \\
24 & 8 & -7
\end{bmatrix}-5\begin{bmatrix}
7 & 2 & -2 \\
-6 &-1 &2 \\
6 & 2 & -1
\end{bmatrix}+7\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix} \,\right)\\[6pt] &=\frac{1}{3}\begin{bmatrix}
-3 & -2 & 2 \\
6 &5 &-2 \\
-6 & -2 & 5
\end{bmatrix}.
\end{align*}

In summary, the inverse matrix of $A$ is
\[A^{-1}=\frac{1}{3}\begin{bmatrix}
-3 & -2 & 2 \\
6 &5 &-2 \\
-6 & -2 & 5
\end{bmatrix}.\]

More Exercise

Test whether you understand how to find the inverse matrix using the Cayley-Hamilton theorem by the next problem.

The solution is given in the post “Find the Inverse Matrix Using the Cayley-Hamilton Theorem“.

More Problems about the Cayley-Hamilton Theorem

Problems about the Cayley-Hamilton theorem and their solutions are collected on the page:

The Cayley-Hamilton Theorem

The post How to Use the Cayley-Hamilton Theorem to Find the Inverse Matrix first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & -x & 0 & 0 \\
0 &1 & -x & 0 \\
0 & 0 & 1 & -x \\
0 & 1 & 0 & -1
\end{bmatrix}\] be a $4\times 4$ matrix. Find all values of $x$ so that the matrix $A$ is singular.

Hint.

Use the fact that a matrix is singular if and only if the determinant of the matrix is zero.

To compute the determinant, use a cofactor expansion.

Solution.

We use the fact that a matrix is singular if and only if the determinant of the matrix is zero.

We compute the determinant of $A$ as follows.
\begin{align*}
&\det(A)= \begin{vmatrix}
1 & -x & 0 & 0 \\
0 &1 & -x & 0 \\
0 & 0 & 1 & -x \\
0 & 1 & 0 & -1
\end{vmatrix}\\[6pt] &=\begin{vmatrix}
1 & -x & 0 \\
0 &1 &-x \\
1 & 0 & -1
\end{vmatrix}
-0\begin{vmatrix}
-x & 0 & 0 \\
0 &1 &-x \\
1 & 0 & -1
\end{vmatrix}+0\begin{vmatrix}
-x & 0 & 0 \\
1 &-x &0 \\
1 & 0 & -1
\end{vmatrix}+0\begin{vmatrix}
-x & 0 & 0 \\
1 &-x &0 \\
0 & 1 & -x
\end{vmatrix}
\\[6pt] & \text{ by the first column cofactor expansion}\\[6pt] &=\begin{vmatrix}
1 & -x & 0 \\
0 &1 &-x \\
1 & 0 & -1
\end{vmatrix}\\[6pt] &=\begin{vmatrix}
1 & -x\\
0& -1
\end{vmatrix}-0\begin{vmatrix}
-x & 0\\
0& -1
\end{vmatrix}+\begin{vmatrix}
-x & 0\\
1& -x
\end{vmatrix}\\[6pt] & \text{ by the first column cofactor expansion}\\[6pt] &=-1+x^2.
\end{align*}

Therefore we have $\det(A)=x^2-1$. Thus $\det(A)=0$ if and only if $x=\pm 1$.
We conclude that the matrix $A$ is singular if and only if $x=\pm 1$.

Comment.

You may use the rule of Sarrus to compute the $3\times 3$ determinant instead of the cofactor expansion if you like so.

The post Find All Values of $x$ so that a Matrix is Singular first appeared on Problems in Mathematics.