For a square matrix $M$, its matrix exponential is defined by
\[e^M = \sum_{i=0}^\infty \frac{M^k}{k!}.\]

Suppose that $M$ is a diagonal matrix
\[ M = \begin{bmatrix} m_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_{n n} \end{bmatrix}.\]

Find the matrix exponential $e^M$.

Solution.

First, we find $M^k$ for each integer $k \geq 0$. The first couple powers can be calculated directly,
\[M^0 = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix} , \quad M = \begin{bmatrix} m_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_{n n} \end{bmatrix},\] \[M^2 = \begin{bmatrix} m^2_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^2_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^2_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^2_{n n} \end{bmatrix} , \quad M^3 = \begin{bmatrix} m^3_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^3_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^3_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^3_{n n} \end{bmatrix}.\]

The general pattern can now be seen:
\[M^k = \begin{bmatrix} m^k_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^k_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^k_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^k_{n n} \end{bmatrix}.\]

Now, we can calculate the infinite series $e^M$:
\begin{align*}
e^M &= \sum_{k=0}^{\infty} \frac{ M^k }{k!} \\
&= \sum_{k=0}^\infty \frac{1}{k!} \begin{bmatrix} m^k_{1, 1} & 0 & 0 & \cdots & 0 \\ 0 & m^k_{2, 2} & 0 & \cdots & 0 \\ 0 & 0 & m^k_{3, 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^k_{n, n} \end{bmatrix} \\
&= \begin{bmatrix} \sum_{k=0}^\infty \frac{ m^k_{1 1} }{k!} & 0 & 0 & \cdots & 0 \\ 0 & \sum_{k=0}^\infty \frac{ m^k_{2 2} }{k!} & 0 & \cdots & 0 \\ 0 & 0 & \sum_{k=0}^\infty \frac{ m^k_{3 3} }{k!} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \sum_{k=0}^\infty \frac{ m^k_{n n} }{k!} \end{bmatrix} . \end{align*}

Now, for any real number $c$ we can write $e^c$ as the series
\[e^c = \sum_{k=0}^\infty \frac{ c^k }{k!}.\]

Thus, the matrix exponential $e^M$ is
\[e^M = \begin{bmatrix} e^{ m_{1 1} } & 0 & 0 & \cdots & 0 \\ 0 & e^{ m_{2 2} } & 0 & \cdots & 0 \\ 0 & 0 & e^{ m_{3 3} } & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & e^{ m_{n n} } \end{bmatrix}.\] Click here if solved 31

The post The Matrix Exponential of a Diagonal Matrix first appeared on Problems in Mathematics.

Recall that a matrix $A$ is symmetric if $A^\trans = A$, where $A^\trans$ is the transpose of $A$.

Is it true that if $A$ is a symmetric matrix and in reduced row echelon form, then $A$ is diagonal? If so, prove it.

Otherwise, provide a counterexample.

Proof.

This is true.

If $A$ is in reduced row echelon form, then every term below the diagonal must be 0.
That is, the entry $a_{i j} = 0$ for all $i > j$.

If $A$ is, additionally, symmetric, then for $i < j$ we also have $a_{j i} = a_{i j} = 0$.

These two facts together means that $a_{i j} = 0$ whenever $i \neq j$.

In this case, the only non-zero terms in the matrix $A$ lie on the diagonal, and so $A$ is a diagonal matrix.

The post If a Symmetric Matrix is in Reduced Row Echelon Form, then Is it Diagonal? first appeared on Problems in Mathematics.

Diagonalize the $2\times 2$ matrix $A=\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}$ by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

Solution.

The characteristic polynomial $p(t)$ of the matrix $A$ is
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
2-t & -1\\
-1& 2-1
\end{vmatrix}\\[6pt] &=(2-t)^2-1 =t^2-4t+3\\
&=(t-1)(t-3).
\end{align*}
It follows that the eigenvalues of $A$ are $\lambda=1, 3$ with algebraic multiplicities are both $1$.
Hence, the geometric multiplicities are $1$ and thus any nonzero vector in eahc eigenspace forms a eigenbasis.

Now let us find a eigenbasis for each eigenspace $E_{\lambda}=\calN(A-\lambda I)$.
For the eigenvalue $1$, we have
\[A-I=\begin{bmatrix}
1 & -1\\
-1& 1
\end{bmatrix}\xrightarrow{R_2+R_1} \begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}\] This yields that the eigenvectors corresponding to the eigenvalue $1$ are $x_2\begin{bmatrix}
1 \\
1
\end{bmatrix}$ with $x_2\neq 0$. Hence
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1
\end{bmatrix} \in E_1\] is an eigenbasis for $E_1$.

Similarly, as we have
\[A-3I=\begin{bmatrix}
-1 & -1\\
-1& -1
\end{bmatrix} \xrightarrow{-R_1}\begin{bmatrix}
1 & 1\\
-1& -1
\end{bmatrix} \xrightarrow{R_2+R_1} \begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix},\] we see that
\[\mathbf{v}_2=\begin{bmatrix}
-1 \\
1
\end{bmatrix} \in E_3\] is an eigenbasis for $E_3$.

Let
\[S=\begin{bmatrix}
\mathbf{v}_1 & \mathbf{v}_2
\end{bmatrix}=
\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix} \text{ and } D=\begin{bmatrix}
1 & 0\\
0& 3
\end{bmatrix}.\]

Then the diagonalization procedure yields that $S$ is nonsingular and $S^{-1}AS= D$.

The post Diagonalize a 2 by 2 Symmetric Matrix first appeared on Problems in Mathematics.

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

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.

The post Diagonalize the \times 2$ Hermitian Matrix by a Unitary Matrix first appeared on Problems in Mathematics.

Prove that the matrix
\[A=\begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}\] is diagonalizable.
Prove, however, that $A$ cannot be diagonalized by a real nonsingular matrix.
That is, there is no real nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix.

Proof.

We first find the eigenvalues of $A$ by computing its characteristic polynomial $p(t)$.
We have
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
-t & 1\\
-1& -t
\end{vmatrix}=t^2+1.
\end{align*}
Solving $p(t)=t^2+1=0$, we obtain two distinct eigenvalues $\pm i$ of $A$.
Hence the matrix $A$ is diagonalizable.

To prove the second statement, assume, on the contrary, that $A$ is diagonalizable by a real nonsingular matrix $S$.
Then we have
\[S^{-1}AS=\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}\] by diagonalization.
As the matrices $A, S$ are real, the left-hand side is a real matrix.
Taking the complex conjugate of both sides, we obtain
\[\begin{bmatrix}
-i & 0\\
0& i
\end{bmatrix}=\overline{\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}}=\overline{S^{-1}AS}=S^{-1}AS=\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}.\] This equality is clearly impossible.
Hence the matrix $A$ cannot be diagonalized by a real nonsingular matrix.

The post A Diagonalizable Matrix which is Not Diagonalized by a Real Nonsingular Matrix first appeared on Problems in Mathematics.

Consider the $2\times 2$ complex matrix
\[A=\begin{bmatrix}
a & b-a\\
0& b
\end{bmatrix}.\]

(a) Find the eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenvectors.

(c) Diagonalize the matrix $A$.

(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$.

Solution.

(a) Find the eigenvalues of $A$.

Since $A$ is an upper triangular matrix, eigenvalues are diagonal entries.
Hence $a, b$ are eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenvectors.

Suppose now that $a\neq b$.
Let us find eigenvectors corresponding to the eigenvalue $a$.
We have
\begin{align*}
A-aI=\begin{bmatrix}
0 & b-a\\
0& b-a
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
0 & b-a\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{b-a}R_1}
\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors corresponding to $a$ are
\[x_1\begin{bmatrix}
1 \\
0
\end{bmatrix},\] where $x_1$ is any nonzero complex number.

Next, we find the eigenvectors corresponding to the eigenvalue $b$.
We have
\begin{align*}
A-bI=\begin{bmatrix}
a-b & b-a\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{a-b}R_1}
\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}.
\end{align*}
Hence the eigenvectors corresponding to $b$ are
\[x_1\begin{bmatrix}
1 \\
1
\end{bmatrix},\] where $x_1$ is any nonzero complex number.

(c) Diagonalize the matrix $A$.

When $a=b$, then $A$ is already diagonal matrix. So let us consider the case $a\neq b$.
In the previous parts, we obtained the eigenvalues $a, b$, and corresponding eigenvectors
\[\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } \begin{bmatrix}
1 \\
1
\end{bmatrix}.\] Let $S=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}$ be a matrix whose column vectors are the eigenvectors.
Then $S$ is invertible and we have
\[S^{-1}AS=\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}\] by the diagonalization process.

Remark that this formula is also true even when $a=b$.

(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$.

Using the result of the diagonalization in part (c), we have
\[A=S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}S^{-1}.\] For each positive integer $k$, we have
\begin{align*}
A^k&=\left(\, S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}S^{-1} \,\right)^k\\[6pt] &=S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}^k S^{-1}=S\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}S^{-1}\\[6pt] &=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}
\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}
\begin{bmatrix}
1 & -1\\
0& 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
a^k & b^k-a^k\\
0& b^k
\end{bmatrix}.
\end{align*}

In summary, we have the formula
\[A^k=\begin{bmatrix}
a^k & b^k-a^k\\
0& b^k
\end{bmatrix}.\] Click here if solved 62

The post Diagonalize the Upper Triangular Matrix and Find the Power of the Matrix first appeared on Problems in Mathematics.

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.

Prove that if $A$ is a diagonalizable nilpotent matrix, then $A$ is the zero matrix $O$.

Definition (Nilpotent Matrix)

A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k=O$.

Proof.

Main Part

Since $A$ is diagonalizable, there is a nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix whose diagonal entries are eigenvalues of $A$.

As we show below, the only eigenvalue of any nilpotent matrix is $0$.
Thus, $S^{-1}AS$ is the zero matrix.
Hence $A=SOS^{-1}=O$.

The only eigenvalue of each nilpotent matrix is $0$

It remains to show that the fact we used above: the only eigenvalue of the nilpotent matrix $A$ is $0$.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$.
That is,
\[A\mathbf{v}=\lambda \mathbf{v}, \tag{*}\]

Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$.

Then we use the relation (*) inductively and obtain
\begin{align*}
A^k\mathbf{v}&=A^{k-1}A\mathbf{v}\\
&=\lambda A^{k-1}\mathbf{v} && \text{by (*)}\\
&=\lambda A^{k-2}A\mathbf{v}\\
&=\lambda^2 A^{k-2}\mathbf{v} && \text{by (*)}\\
&=\dots =\lambda^k \mathbf{v}.
\end{align*}

Hence we have
\[\mathbf{0}=O\mathbf{v}=A^k\mathbf{v}=\lambda^k \mathbf{v}.\]

Note that the eigenvector $\mathbf{v}$ is a nonzero vector by definition.
Thus, we must have $\lambda^k=0$, hence $\lambda=0$.
This proves that the only eigenvalue of the nilpotent matrix $A$ is $0$, and this completes the proof.

Another Proof of the Fact

Even though the fact proved above is true regardless of diagonalizability of $A$, we can make use that $A$ is diagonalizable to prove the fact as follows.

Let $\lambda_1, \dots, \lambda_n$ be the eigenvalues of the $n\times n$ nilpotent matrix $A$.
Then we have
\[S^{-1}AS=D,\] where
\[D:=\begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 &\lambda_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n
\end{bmatrix}.\]

Then we have
\(\require{cancel}\)
\begin{align*}
D^k&=(S^{-1}AS)^k\\
&=(S^{-1}A\cancel{S})(\cancel{S}^{-1}A\cancel{S})\cdots (\cancel{S}^{-1}AS)\\
&=S^{-1}A^kS\\
&=S^{-1}OS=O.
\end{align*}

Since
\[D^k=\begin{bmatrix}
\lambda_1^k & 0 & \dots & 0 \\
0 &\lambda_2^k & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n^k
\end{bmatrix},\] it yields that $\lambda_i^=0$, and hence $\lambda_i=0$ for $i=1, \dots, n$.

Related Question.

The converse of the above fact is also true: if the only eigenvalue of $A$ is $0$, then $A$ is a nilpotent matrix.

See the post ↴
Nilpotent Matrix and Eigenvalues of the Matrix
for a proof of this fact.

The post Every Diagonalizable Nilpotent Matrix is the Zero Matrix first appeared on Problems in Mathematics.

Let
\[D=\begin{bmatrix}
d_1 & 0 & \dots & 0 \\
0 &d_2 & \dots & 0 \\
\vdots & & \ddots & \vdots \\
0 & 0 & \dots & d_n
\end{bmatrix}\] be a diagonal matrix with distinct diagonal entries: $d_i\neq d_j$ if $i\neq j$.
Let $A=(a_{ij})$ be an $n\times n$ matrix such that $A$ commutes with $D$, that is,
\[AD=DA.\] Then prove that $A$ is a diagonal matrix.

Proof.

We prove that the $(i,j)$-entry of $A$ is $a_{ij}=0$ for $i\neq j$.

We compare the $(i,j)$-entries of both sides of $AD=DA$.
Let $D=(d_{ij})$. That is, $d_{ii}=d_i$ and $d_{ij}=0$ if $i\neq j$.
The $(i,j)$-entry of $AD$ is
\begin{align*}
(AD)_{ij}=\sum_{k=1}^n a_{ik}d_{kj}=a_{ij}d_j.
\end{align*}

The $(i,j)$-entry of $DA$ is
\[(DA)_{ij}=\sum_{k=1}^n d_{ik}a_{kj}=d_ia_{ij}.\]

Hence we have
\[a_{ij}d_j=a_{ij}d_i.\] Or equivalently we have
\[a_{ij}(d_j-d_i)=0.\]

Since $d_i\neq d_j$, we have $d_j-d_i\neq 0$.
Thus, we must have $a_{ij}=0$ for $i\neq j$.

Hence $A=(a_{ij})$ is a diagonal matrix.

The post A Matrix Commuting With a Diagonal Matrix with Distinct Entries is Diagonal first appeared on Problems in Mathematics.

Determine whether the matrix
\[A=\begin{bmatrix}
0 & 1 & 0 \\
-1 &0 &0 \\
0 & 0 & 2
\end{bmatrix}\] is diagonalizable.

If it is diagonalizable, then find the invertible matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

How to diagonalize matrices.

For a general procedure of the diagonalization of a matrix, please read the post “How to Diagonalize a Matrix. Step by Step Explanation“.

Solution.

We first determine the eigenvalues of the matrix $A$.
To do so, we compute the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
&p(t)=\det(A-tI)\\
&=\begin{vmatrix}
-t & 1 & 0 \\
-1 &-t &0 \\
0 & 0 & 2-t
\end{vmatrix}\\[6pt] &=(-1)^{3+3}(2-t)\begin{vmatrix}
-t & 1\\
-1& -t
\end{vmatrix} && \text{by the third row cofactor expansion}\\
&=(2-t)(t^2+1).
\end{align*}
Thus the eigenvalues of $A$ are $2, \pm i$.
Since the $3\times 3$ matrix $A$ has three distinct eigenvalues, it is diagonalizable.

To diagonalize $A$, we now find eigenvectors.
For the eigenvalue $2$, we compute
\begin{align*}
&A-2I=\begin{bmatrix}
-2 & 1 & 0 \\
-1 &-2 &0 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{-R_2}
\begin{bmatrix}
-2 & 1 & 0 \\
1 &2 &0 \\
0 & 0 & 0
\end{bmatrix}\\[6pt] &\xrightarrow{R_1 \leftrightarrow R_2}\begin{bmatrix}
1 & 2 & 0 \\
-2 &1 &0 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_2+2R_1}\begin{bmatrix}
1 & 2 & 0 \\
0 &5 &0 \\
0 & 0 & 0
\end{bmatrix}\\[6pt] &\xrightarrow{\frac{1}{5}R_2}\begin{bmatrix}
1 & 2 & 0 \\
0 &1 &0 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_1-2R_2}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
Thus, the solutions $\mathbf{x}$ of $(A-2I)=\mathbf{0}$ satisfy $x=y=0$.
Hence the eigenspace is
\[E_2=\calN(A-2I)=\Span\left\{\, \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \,\right\}.\]

For the eigenvalue $i$, we compute
\begin{align*}
A-iI=\begin{bmatrix}
-i & 1 & 0 \\
-1 &-i &0 \\
0 & 0 & 2-i
\end{bmatrix}
\xrightarrow{iR_1}
\begin{bmatrix}
1 & i & 0 \\
-1 &-i &0 \\
0 & 0 & 2-i
\end{bmatrix}\\[6pt] \xrightarrow{\substack{R_2+R_1\\ \frac{1}{2-i}R_3}}
\begin{bmatrix}
1 & i & 0 \\
0 &0 &0 \\
0 & 0 & 1
\end{bmatrix}
\xrightarrow{R_2 \leftrightarrow R_3}
\begin{bmatrix}
1 & i & 0 \\
0 & 0 & 1\\
0 &0 &0
\end{bmatrix}.
\end{align*}
So the solutions $\mathbf{x}$ of $(A-iI)\mathbf{x}=\mathbf{0}$ satisfy
\[x=-iy \text{ and } z=0.\] Thus, the eigenspace is
\[E_i=\calN(A-iI)=\Span\left\{\, \begin{bmatrix}
1 \\
i \\
0
\end{bmatrix} \,\right\}.\]

Since $i$ and $-i$ are complex conjugate, their eigenspaces are also complex conjugate.
Hence the eigenspace for $-i$ is
\[E_{-i}=\Span\left\{\, \begin{bmatrix}
1 \\
-i \\
0
\end{bmatrix} \,\right\}.\]

From these computations, we have obtained eigenvalues $2, i, -i$ and eigenvector corresponding to these are
\[\mathbf{v}_{2}=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}, \mathbf{v}_i=\begin{bmatrix}
1 \\
i \\
0
\end{bmatrix}, \mathbf{v}_{-i}=\begin{bmatrix}
1 \\
-i \\
0
\end{bmatrix}.\]

Let
\[S=\begin{bmatrix}
\mathbf{v}_2 & \mathbf{v}_i & \mathbf{v}_{-i} \\
\end{bmatrix}=\begin{bmatrix}
0 & 1 & 1 \\
0 &i &-i \\
1 & 0 & 0
\end{bmatrix}\] and
\[D=\begin{bmatrix}
2 & 0 & 0 \\
0 &i &0 \\
0 & 0 & -i
\end{bmatrix}.\] Then $S$ is invertible and we have $S^{-1}AS=D$ by the diagonalization process.

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