Consider the matrix $A=\begin{bmatrix}
a & -b\\
b& a
\end{bmatrix}$, where $a$ and $b$ are real numbers and $b\neq 0$.

(a) Find all eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenspace $E_{\lambda}$.

(c) Diagonalize the matrix $A$ by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

Solution.

(a) Find all eigenvalues of $A$.

The characteristic polynomial $p(t)$ of the matrix $A$ is
\begin{align*}
p(t)&=\det(A-tI) = \begin{vmatrix}
a-t & -b\\
b& a-t
\end{vmatrix}\\[6pt] &=(a-t)^2+b^2.
\end{align*}

The eigenvalues of $A$ are roots of $p(t)$.
So we solve $p(t)=0$. We have
\begin{align*}
& \quad (a-t)^2+b^2=0\\
\Leftrightarrow & \quad (a-t)^2=-b^2\\
\Leftrightarrow &\quad a-t =\pm i b\\
\Leftrightarrow &\quad t= a \pm ib.
\end{align*}
Here $i=\sqrt{-1}$.

Thus, the eigenvalues of $A$ are $a\pm ib$.

(b) For each eigenvalue of $A$, determine the eigenspace $E_{\lambda}$.

We first determine the eigenspace $E_{\lambda}$ for $\lambda = a+ib$.
Recall that by definition $E_{\lambda}=\calN(A-\lambda I)$, the nullspace of $A-\lambda I$.

We compute
\begin{align*}
A-(a+ib)I=\begin{bmatrix}
-ib & -b\\
b& -ib
\end{bmatrix}
\xrightarrow{\frac{i}{b}R_1}
\begin{bmatrix}
1 & -i\\
b& -ib
\end{bmatrix}
\xrightarrow{R_2-bR_1}
\begin{bmatrix}
1 & -i\\
0& 0
\end{bmatrix}.
\end{align*}
Note that in the above row reduction, we needed the assumption $b\neq 0$.

It follows that the general solution of the system is $x_1=i x_2$.
Hence, we have
\[E_{a+ib} =\Span \left(\, \begin{bmatrix}
i \\
1
\end{bmatrix} \,\right).\]

Note that the other eigenvalue $a-ib$ is the complex conjugate of $a+ib$.
It follows that the eigenspace $E_{a-ib}$ is obtained by conjugating the eigenspace $E_{a+ib}$.
Thus,
\[E_{a-ib} =\Span \left(\, \begin{bmatrix}
-i \\
1
\end{bmatrix} \,\right).\]

(c) Diagonalize the matrix $A$

From part (b), we see that
\[\begin{bmatrix}
i \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
-i \\
1
\end{bmatrix}\] form an eigenbasis for $\C^2$.

So, we set
\[S=\begin{bmatrix}
i & -i\\
1& 1
\end{bmatrix} \text{ and } D=\begin{bmatrix}
a+ib & 0\\
0& a-ib
\end{bmatrix},\] and we obtain $S^{-1}AS=D$ by the diagonalization procedure.

The post Find Eigenvalues, Eigenvectors, and Diagonalize the 2 by 2 Matrix first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
a & b\\
-b& a
\end{bmatrix}\] be a $2\times 2$ matrix, where $a, b$ are real numbers.
Suppose that $b\neq 0$.

Prove that the matrix $A$ does not have real eigenvalues.

Proof.

Let $\lambda$ be an arbitrary eigenvalue of $A$.
Then the matrix $A-\lambda I$ is singular, where $I$ is the $2\times 2$ identity matrix.
This is equivalent to having $\det(A-\lambda I)=0$.

We compute the determinant as follows.
We have
\begin{align*}
\det(A-\lambda I)&=\begin{vmatrix}
a-\lambda & b\\
-b& a-\lambda
\end{vmatrix}\\[6pt] &=(a-\lambda)^2-b(-b)\\
&=a^2-2a\lambda+\lambda^2+b^2\\
&=\lambda^2-2a\lambda+a^2+b^2.
\end{align*}

We solve the equation $\lambda^2-2a\lambda+a^2+b^2=0$ by the quadratic formula and obtain
\begin{align*}
\lambda &=\frac{2a\pm\sqrt{4a^2-4(a^2+b^2)}}{2}=\frac{2a\pm\sqrt{-4b^2}}{2}\\[6pt] &=a\pm |b|i.
\end{align*}

Since $b\neq 0$ by assumption, the eigenvalue $\lambda=a\pm|b|i$ is not a real number.
As $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that all eigenvalues of $A$ are not real numbers.

The post An Example of a Real Matrix that Does Not Have Real Eigenvalues 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.

Find all the eigenvalues of the matrix
\[A=\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 &0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0
\end{bmatrix}.\]

(The Ohio State University, Linear Algebra Final Exam Problem)

Solution.

We compute the characteristic polynomial $p(t)$ of the matrix $A$ as follows.
We have
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
-t & 1 & 0 & 0 \\
0 &-t & 1 & 0 \\
0 & 0 & -t & 1 \\
1 & 0 & 0 & -t
\end{vmatrix}\\[6pt] &=-t\begin{vmatrix}
-t & 1 & 0 \\
0 &-t &1 \\
0 & 0 & -t
\end{vmatrix}
-\begin{vmatrix}
0 & 1 & 0 \\
0 &-t &1 \\
1 & 0 & -t
\end{vmatrix} \tag{*}
\end{align*}
by the first row cofactor expansion.
The left determinant of the $3\times 3$ matrix in (*) is $(-t)^3$ since it is a diagonal matrix.

We apply the first column cofactor expansion to the right determinant in (*) and obtain
\begin{align*}
\begin{vmatrix}
0 & 1 & 0 \\
0 &-t &1 \\
1 & 0 & -t
\end{vmatrix} =\begin{vmatrix}
1 & 0\\
-t& 1
\end{vmatrix}=1.
\end{align*}

It follows from (*) that
\begin{align*}
p(t)&=(-t)(-t)^3-1=t^4-1.
\end{align*}

The eigenvalues of $A$ are the roots of the characteristic polynomial $p(t)$.
Solving $t^4-1=0$, we obtain the eigenvalues
\[\pm 1, \pm i,\] where $i=\sqrt{-1}$.
Note that $t^4-1=(t-1)(t+1)(t-i)(t+i)$.

Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

1. Find All the Eigenvalues of 4 by 4 Matrix (This page)
2. Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue
3. Diagonalize a 2 by 2 Matrix if Diagonalizable
4. Find an Orthonormal Basis of the Range of a Linear Transformation
5. The Product of Two Nonsingular Matrices is Nonsingular
6. Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not
7. Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials
8. Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable
9. Idempotent Matrix and its Eigenvalues
10. Diagonalize the 3 by 3 Matrix Whose Entries are All One
11. Given the Characteristic Polynomial, Find the Rank of the Matrix
12. Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$
13. Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$

The post Find All the Eigenvalues of 4 by 4 Matrix first appeared on Problems in Mathematics.

Let $A$ be a $3\times 3$ real orthogonal matrix with $\det(A)=1$.

(a) If $\frac{-1+\sqrt{3}i}{2}$ is one of the eigenvalues of $A$, then find the all the eigenvalues of $A$.

(b) Let
\[A^{100}=aA^2+bA+cI,\] where $I$ is the $3\times 3$ identity matrix.
Using the Cayley-Hamilton theorem, determine $a, b, c$.

(Kyushu University, Linear Algebra Exam Problem)
 

Solution.

(a) Find the all the eigenvalues of $A$.

Since $A$ is a real matrix and $\frac{-1+\sqrt{3}i}{2}$ is a complex eigenvalue, its conjugate $\frac{-1-\sqrt{3}i}{2}$ is also an eigenvalue of $A$.
As $A$ is a $3\times 3$ matrix, it has one more eigenvalue $\lambda$.

Note that the product of all eigenvalues of $A$ is the determinant of $A$.
Thus, we have
\[\frac{-1+\sqrt{3}i}{2} \cdot \frac{-1-\sqrt{3}i}{2}\cdot \lambda =\det(A)=1.\] Solving this, we obtain $\lambda=1$.
Therefore, the eigenvalues of $A$ are
\[\frac{-1+\sqrt{3}i}{2}, \frac{-1-\sqrt{3}i}{2}, 1.\]

(a) Using the Cayley-Hamilton theorem, determine $a, b, c$.

To use the Cayley-Hamilton theorem, we first need to determine the characteristic polynomial $p(t)=\det(A-tI)$ of $A$.
Since we found all the eigenvalues of $A$ in part (a) and the roots of characteristic polynomials are the eigenvalues, we know that
\begin{align*}
p(t)&=-\left(\, t-\frac{-1+\sqrt{3}i}{2} \,\right)\left(\, t-\frac{-1-\sqrt{3}i}{2} \,\right)(t-1) \tag{*}\\
&=-(t^2+t+1)(t-1)\\
&=-t^3+1.
\end{align*}
(Remark that if your definition of the characteristic polynomial is $\det(tI-A)$, then the first negative sign in (*) should be omitted.)

Then the Cayley-Hamilton theorem yields that
\[P(A)=-A^3+I=O,\] where $O$ is the $3\times 3$ zero matrix.

Hence we have $A^3=I$.
We compute
\begin{align*}
A^{100}=(A^3)^{33}A=I^{33}A=IA=A.
\end{align*}

Thus, we conclude that $a=0, b=1, c=0$.

Comment.

Observe that we did not use the assumption that $A$ is orthogonal.

The post Use the Cayley-Hamilton Theorem to Compute the Power $A^{100}$ 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.

The post Diagonalize the 3 by 3 Matrix if it is Diagonalizable first appeared on Problems in Mathematics.