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.

Let $I$ be the $2\times 2$ identity matrix.
Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$.

Proof.

Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies
\[ABA^{-1}=-B. \tag{*}\] Taking the trace, we have
\[-\tr(B)=\tr(-B)=\tr(ABA^{-1})=tr(BAA^{-1})=\tr(B),\] hence the trace $\tr(B)=0$.
Thus, the characteristic polynomial of $B$ is
\[x^2-\tr(B)x+\det(B)=x^2+1.\] Hence the eigenvalues of $B$ are $\pm i$.

Note that the matrix $\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}$ has also eigenvalues $\pm i$.
Thus this matrix is similar to the matrix $B$ as both matrices are similar to the diagonal matrix $\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}$.
Let $P$ be a nonsingular matrix such that
\[B’:=P^{-1}BP=\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}.\] Let $A’=P^{-1}AP$.

The relation (*) is equivalent to $AB=-BA$.
Using this we have
\begin{align*}
A’B’&=(P^{-1}AP)(P^{-1}BP)=P^{-1}(AB)P\\
&=P^{-1}(-BA)P=-(P^{-1}BP)(P^{-1}AP)=-B’A’.
\end{align*}

Let $A’=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$.
Then $A’B’=-B’A’$ gives
\begin{align*}
\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}
\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}
=-\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}\\[6pt] \Leftrightarrow
\begin{bmatrix}
b & -a\\
d& -c
\end{bmatrix}=\begin{bmatrix}
c & d\\
-a& -b
\end{bmatrix}.
\end{align*}
Hence we obtain $d=-a$ and $c=b$.

Then
\begin{align*}
1=\det(A)=\det(PA’P^{-1})=\det(A’)=\begin{vmatrix}
a & b\\
b& -a
\end{vmatrix}=-a^2-b^2,
\end{align*}
which is impossible.
Therefore, the matrix $-I$ cannot be written as a commutator $[A, B]$ for any $2\times 2$ matrices $A, B$ with determinant $1$.

The post An Example of a Matrix that Cannot Be a Commutator 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.

Diagonalize the matrix
\[A=\begin{bmatrix}
1 & 1 & 1 \\
1 &1 &1 \\
1 & 1 & 1
\end{bmatrix}.\] Namely, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

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

Hint.

To diagonalize the matrix $A$, we need to find eigenvalues $A$ and bases of eigenspaces.

For a procedure of the diagonalization, see the post “How to Diagonalize a Matrix. Step by Step Explanation.“.

Below, we will find eigenvalues and eigenvectors without using the characteristic polynomial although you may use it.

Solution.

We use an indirect way to find eigenvalues and eigenvectors.
(We will not use the characteristic polynomial.)

Applying the elementary row operations, we have
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 1 \\
1 &1 &1 \\
1 & 1 & 1
\end{bmatrix}\xrightarrow{\substack{R_2-R_1\\ R_3-R_1}}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
Hence the solutions $\mathbf{x}$ of $A\mathbf{x}=\mathbf{0}$ satisfy
\[x_1=-x_2-x_3.\] Thus, every vector in the null space is of the form
\[\mathbf{x}=\begin{bmatrix}
-x_2-x_3 \\
x_2 \\
x_3
\end{bmatrix}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\] for some scalars $x_2, x_3$.

It follows that
\[\left\{\, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right\}\] is a basis of the null space $\calN(A)$.

Hence $0$ is an eigenvalue and the geometric multiplicity corresponding to $0$, which is the nullity of $A$, is $2$.
It follows that the algebraic multiplicity of the eigenvalue $0$ is either $2$ or $3$.
We see that it is $2$ shortly.

Note that by inspection we have
\[A\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}=3\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}.\] This yields that $3$ is an eigenvalue of $A$ and $\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}$ is a corresponding eigenvector.

The sum of algebraic multiplicities of all eigenvalues of $A$ is $3$.
Hence the algebraic multiplicity of $0$ must be $2$, and that of $3$ must be $1$.
In particular, the vector $\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}$ forms a basis of the eigenspace $E_3$.

In summary so far, we have eigenvalues $0, 3$ and basis vectors of eigenspaces are
\[\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix},\] respectively.

Thus, we put
\[S=\begin{bmatrix}
-1 & -1 & 1 \\
1 &0 &1 \\
0 & 1 & 1
\end{bmatrix}\] and obtain
\[S^{-1}AS=D,\] where
\[D=\begin{bmatrix}
0 & 0 & 0 \\
0 &0 &0 \\
0 & 0 & 3
\end{bmatrix}.\]

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
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 (This page)
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 Diagonalize the 3 by 3 Matrix Whose Entries are All One first appeared on Problems in Mathematics.

For which values of constants $a, b$ and $c$ is the matrix
\[A=\begin{bmatrix}
7 & a & b \\
0 &2 &c \\
0 & 0 & 3
\end{bmatrix}\] diagonalizable?

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

Solution.

Note that the matrix $A$ is an upper triangular matrix.
Hence the eigenvalues of $A$ are diagonal entries $7, 2, 3$.

So the $3\times 3$ matrix $A$ has three distinct eigenvalues.
This implies that $A$ is diagonalizable.

Hence, regardless of the values of $a, b, c$, the matrix $A$ is always diagonalizable.
Thus, $a, b, c$ can take arbitrary values.

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
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 Values of $a, b, c$ such that the Given Matrix is Diagonalizable first appeared on Problems in Mathematics.

Let $A$ be a real symmetric $n\times n$ matrix with $0$ as a simple eigenvalue (that is, the algebraic multiplicity of the eigenvalue $0$ is $1$), and let us fix a vector $\mathbf{v}\in \R^n$.

(a) Prove that for sufficiently small positive real $\epsilon$, the equation
\[A\mathbf{x}+\epsilon\mathbf{x}=\mathbf{v}\] has a unique solution $\mathbf{x}=\mathbf{x}(\epsilon) \in \R^n$.

(b) Evaluate
\[\lim_{\epsilon \to 0^+} \epsilon \mathbf{x}(\epsilon)\] in terms of $\mathbf{v}$, the eigenvectors of $A$, and the inner product $\langle\, ,\,\rangle$ on $\R^n$.

 
(University of California, Berkeley, Linear Algebra Qualifying Exam)

Proof.

(a) Prove that $A\mathbf{x}+\epsilon\mathbf{x}=\mathbf{v}$has a unique solution $\mathbf{x}=\mathbf{x}(\epsilon) \in \R^n$.

Recall that the eigenvalues of a real symmetric matrices are all real numbers and it is diagonalizable by an orthogonal matrix.

Note that the equation $A\mathbf{x}+\epsilon\mathbf{x}=\mathbf{v}$ can be written as
\[(A+\epsilon I)\mathbf{x}=\mathbf{v}, \tag{*}\] where $I$ is the $n\times n$ identity matrix. Thus to show that the equation (*) has a unique solution, it suffices to show that the matrix $A+\epsilon I$ is invertible.

Since $A$ is diagonalizable, there exists an invertible matrix $S$ such that
\[S^{-1}AS=\begin{bmatrix}
\lambda_1 & 0 & \cdots & 0 \\
0 & \lambda_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n
\end{bmatrix},\] where $\lambda_i$ are eigenvalues of $A$.
Since the algebraic multiplicity of $0$ is $1$, without loss of generality, we may assume that $\lambda_1=0$ and $\lambda_i, i > 1$ are nonzero.

Then we have
\begin{align*}
S^{-1}(A+\epsilon I)S&=S^{-1}AS+\epsilon I=\begin{bmatrix}
\epsilon & 0 & \cdots & 0 \\
0 & \epsilon+\lambda_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \epsilon+\lambda_n
\end{bmatrix}.
\end{align*}

If $\epsilon > 0$ is smaller than the lengths of $|\lambda_i|, i > 1$, then none of the diagonal entries $\epsilon+ \lambda_i$ are zero.

Hence we have
\begin{align*}
\det(A+\epsilon I)&=\det(S)^{-1}\det(A+\epsilon I)\det(S)\\
&=\det\left(\, S^{-1}(A+\epsilon I) S \,\right)\\
&=\epsilon(\epsilon+\lambda_2)\cdots (\epsilon+\lambda_n)\neq 0.
\end{align*}
Since $\det(A+\epsilon I)\neq 0$, it yields that $A$ is invertible, hence the equation (*) has a unique solution
\[\mathbf{x}(\epsilon)=(A+\epsilon I)^{-1}\mathbf{v}.\]

Remark

This result is in general true for any square matrix.
Instead of using the diagonalization, we can use the triangulation of a matrix.

(b) Evaluate $\lim_{\epsilon \to 0^+} \epsilon \mathbf{x}(\epsilon)$

As noted earlier that a real symmetric matrix can be diagonalizable by an orthogonal matrix.
This means that there is an eigenvector $\mathbf{v}_i$ corresponding to the eigenvalue $\lambda_i$ for each $i$ such that the eigenvectors $\mathbf{v}_i$ form an orthonormal basis of $\R^n$.
That is,
\begin{align*}
A\mathbf{v}_i=\lambda_i \mathbf{v}_i \\
\langle \mathbf{v}_i,\mathbf{v}_j \rangle=\delta_{i,j},
\end{align*}
where $\delta_{i,j}$ is the Kronecker delta symbol, where $\delta_{i,i}=1, \delta_{i,j}=0$ if $i\neq j$.
From this, we deduce that
\begin{align*}
(A+\epsilon I)\mathbf{v}_i=(\lambda_i+\epsilon)\mathbf{v}_i\\
(A+\epsilon I)^{-1}\mathbf{v}_i=\frac{1}{\lambda_i+\epsilon}\mathbf{v}_i. \tag{**}
\end{align*}
Using the basis $\{\mathbf{v}_i\}$, we write
\[\mathbf{v}=\sum_{i=1}^nc_i \mathbf{v}_i\] for some $c_i\in \R$.

Then we compute
\begin{align*}
A\mathbf{x}(\epsilon)&=A(A+\epsilon I)^{-1}\mathbf{v} && \text{by part (a)}\\
&=A(A+\epsilon I)^{-1}\left(\, \sum_{i=1}^nc_i \mathbf{v}_i \,\right)\\
&=\sum_{i=1}^n c_iA(A+\epsilon I)^{-1}\mathbf{v}_i\\
&=\sum _{i=1}^n c_iA\left(\, \frac{1}{\lambda_i+\epsilon}\mathbf{v}_i \,\right) && \text{by (**)}\\
&=\sum_{i=1}^n c_i\frac{\lambda_i}{\lambda_i+\epsilon}\mathbf{v}_i && \text{since $A\mathbf{v}_i=\lambda_i\mathbf{v}_i$}\\
&=\sum_{i=2}^n c_i\frac{\lambda_i}{\lambda_i+\epsilon}\mathbf{v}_i && \text{since $\lambda_1=0$}.
\end{align*}

Therefore we have
\begin{align*}
\lim_{\epsilon \to 0^+} \epsilon \mathbf{x}(\epsilon)&=\lim_{\epsilon \to 0^+}\left(\, \mathbf{v}-A\mathbf{x}(\epsilon) \,\right)\\
&=\mathbf{v}-\lim_{\epsilon \to 0^+}\left(\, A\mathbf{x}(\epsilon) \,\right)\\
&= \sum_{i=1}^nc_i\mathbf{v}_i-\lim_{\epsilon \to 0^+}\left(\, \sum_{i=2}^n c_i\frac{\lambda_i}{\lambda_i+\epsilon}\mathbf{v}_i \,\right)\\
&=\sum_{i=1}c_i \mathbf{v}_i-\sum_{i=2}^n c_i \mathbf{v}_i\\
&=c_1\mathbf{v}_1.
\end{align*}
Using the orthonormality of the basis $\{\mathbf{v}_i\}$, we have
\[\langle\mathbf{v}, \mathbf{v}_1 \rangle=\sum_{i=1}^n \langle c_i\mathbf{v}_i, \mathbf{v}_1 \rangle=c_1.\]

Hence the required expression is
\[\lim_{\epsilon \to 0^+} \epsilon \mathbf{x}(\epsilon)=\langle\mathbf{v}, \mathbf{v}_1 \rangle\mathbf{v}_1,\] where $\mathbf{v}_1$ is the unit eigenvector corresponding to the eigenvalue $0$.

The post A Matrix Equation of a Symmetric Matrix and the Limit of its Solution first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
2 & -1 & -1 \\
-1 &2 &-1 \\
-1 & -1 & 2
\end{bmatrix}.\] Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$.
That is, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

 

We give two solutions.
The first solution is a standard method of diagonalization.
For a review of the process of diagonalization, see the post “How to diagonalize a matrix. Step by step explanation.”

The second solution is a more indirect method to find eigenvalues and eigenvectors.

Solution 1.

We claim that the matrix $A$ is diagonalizable.
One way to see this is to note that $A$ is a real symmetric matrix, and hence it is diagonalizable.

Alternatively, we can compute eigenspaces and check whether $A$ is not defective (namely, the algebraic multiplicity and the geometric multiplicity of each eigenvalue of $A$ are the same.

To diagonalize the matrix $A$, we need to find eigenvalues and three linearly independent eigenvectors.

We compute the characteristic polynomial of $A$ as follows:
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{bmatrix}
2-t & -1 & -1 \\
-1 &2-t &-1 \\
-1 & -1 & 2-t
\end{bmatrix}\\
&=(2-t)\begin{bmatrix}
2-t & -1\\
-1& 2-t
\end{bmatrix}
-(-1)\begin{bmatrix}
-1 & -1\\
-1& 2-t
\end{bmatrix}+(-1)\begin{bmatrix}
-1 & 2-t\\
-1& -1
\end{bmatrix} \\
&\text{(by the first row cofactor expansion)}\\
&=-t(t-3)^2.
\end{align*}
Since eigenvalues are the roots of the characteristic polynomial, eigenvalues of $A$ are $0$ and $3$ with algebraic multiplicity $1$ and $2$, respectively.

(If you did not confirm that $A$ is diagonalizable yet, then at this point we know that the geometric multiplicity of $\lambda=0$ is $1$ since the geometric multiplicity is alway greater than $0$ and less than or equal to the algebraic multiplicity. However the geometric multiplicity of $\lambda=3$ is either $1$ or $2$.)

Next, we determine the eigenspace $E_{\lambda}$ and its basis for each eigenvalue of $A$.

For $\lambda=3$, we find solutions of $(A-3I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A-3I&=\begin{bmatrix}
-1 & -1 & -1 \\
-1 &-1 &-1 \\
-1 & -1 & -1
\end{bmatrix}
\xrightarrow{\substack{R_2-R_1\\R_3-R_1}}
\begin{bmatrix}
-1 & -1 & -1 \\
0 &0 &0\\
0&0&0
\end{bmatrix}
\xrightarrow{-R_1}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &0\\
0&0&0
\end{bmatrix}.
\end{align*}
Hence a solution must satisfy $x_1=-x_2-x_3$, and thus the eigenspace is
\[ E_3=\left\{\, \mathbf{x} \in \R^3 \quad \middle| \quad \mathbf{x}=x_2\begin{bmatrix}
-1\\
1\\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix} \text{ for any } x_2, x_3 \in \C \,\right\}.\] From this expression, it is straightforward to check that the set
\[\left\{\begin{bmatrix}
-1\\
1\\
0
\end{bmatrix}, \begin{bmatrix}
-1\\
0\\
1
\end{bmatrix} \,\right\}\] is a basis of $E_3$.
(Hence the geometric multiplicity of $\lambda=3$ is $2$, and $A$ is not defective and diagonalizable.)

For $\lambda=0$, we solve $(A-0I)\mathbf{x}=\mathbf{0}$, thus $A\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A&=\begin{bmatrix}
2 & -1 & -1 \\
-1 &2 &-1 \\
-1 & -1 & 2
\end{bmatrix}
\xrightarrow{R_1 \leftrightarrow R_2}
\begin{bmatrix}
-1 &2 &-1 \\
2 & -1 & -1 \\
-1 & -1 & 2
\end{bmatrix}
\xrightarrow{-R_1}
\begin{bmatrix}
1 & -2 & 1 \\
2 & -1 & -1 \\
-1 & -1 & 2
\end{bmatrix}\\
&\xrightarrow{\substack{R_2-2R_1\\ R_3+R_1}}
\begin{bmatrix}
1 & -2 & 1 \\
0 & 3 & -3 \\
0 & -3 & 3
\end{bmatrix}
\xrightarrow{R_3+R_2}
\begin{bmatrix}
1 & -2 & 1 \\
0 & 3 & -3 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{\frac{1}{3}R_2}
\begin{bmatrix}
1 & -2 & 1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}\\
&\xrightarrow{R_1+2R_2}
\begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
Hence any solution satisfies
\[x_1=x_3 \text{ and } x_2=x_3.\] Therefore, the eigenspace is
\[E_0=\left\{\, \mathbf{x} \in \R^3 \quad \middle| \quad x_3\begin{bmatrix}
1\\
1\\
1
\end{bmatrix} \text{ for any } x_3 \in \C \,\right\},\] and a basis of $E_0$ is
\[\left\{\, \begin{bmatrix}
1\\
1\\
1
\end{bmatrix} \,\right\}.\]

Let
\[\mathbf{u}_1=\begin{bmatrix}
-1\\
1\\
0
\end{bmatrix}, \mathbf{u}_2=\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix}, \mathbf{u}_3=\begin{bmatrix}
1\\
1\\
1
\end{bmatrix}.\] Note that $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a basis of $E_3$ and $\{\mathbf{u}_3\}$ is a basis of $E_0$.
Thus, it follows that the vectors $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ are linearly independent eigenvectors.
Put
\[S=[\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3]= \begin{bmatrix}
-1 & -1 & 1\\
1& 0& 1\\
0& 1 & 1
\end{bmatrix}.\] Since the columns of $S$ are linearly independent, the matrix $S$ is nonsingular. Then the procedure of the diagonalization yields that
\[S^{-1}AS=\begin{bmatrix}
\mathbf{3} & 0 & 0\\
0& \mathbf{3}& 0\\
0 & 0& \mathbf{0}
\end{bmatrix},\] where diagonal entries are eigenvalues corresponding to the eigenvector $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ with this order.

Solution 2.

The second solution uses a different method to find eigenvalues and eigenvectors.
Let $B=A-3I$. Then every entry of $B$ is $-1$.
We find eigenvalues $\lambda$ and eigenvectors of $B$. Then the eigenvalues of $A$ are $\lambda+3$ and eigenvectors are the same.

First we reduce the matrix $B$ as follows:
\begin{align*}
B&=\begin{bmatrix}
-1&-1&-1\\
-1&-1&-1\\
-1&-1&-1
\end{bmatrix}
\xrightarrow{\substack{R_2-R_1\\ R_3-R_1}}
\begin{bmatrix}
-1&-1&-1\\
0&0&0\\
0&0&0
\end{bmatrix}
\xrightarrow{-R_1}
\begin{bmatrix}
1&1&1\\
0&0&0\\
0&0&0
\end{bmatrix}.
\end{align*}
Hence the rank of $B$ is $1$, and the nullity of $B$ is $2$ by the rank-nullity theorem. It follows that $\lambda=0$ is an eigenvalue of $B$.
Note that the eigenspace $E_0$ corresponding to $\lambda=0$ is the null space of $A$. From the reduction above, we see that the null space consists of vectors
\[\mathbf{x}=x_2\begin{bmatrix}
-1\\1\\0
\end{bmatrix}
+x_3\begin{bmatrix}
-1\\0\\1
\end{bmatrix}\] for any complex numbers $x_2, x_3$.
It follows that
\[\begin{bmatrix}
-1\\1\\0
\end{bmatrix},
\begin{bmatrix}
-1\\0\\1
\end{bmatrix}\] are basis vectors of eigenspace $E_0$. Hence the geometric multiplicity of $\lambda=0$ is $2$.
(The algebraic multiplicity is either $2$ or $3$. We will see it must be $2$.)

Since all the entries of $B$ are $-1$, by inspection, we find that the vector
\[\begin{bmatrix}
1\\1\\1
\end{bmatrix}\] is an eigenvector corresponding to the eigenvalue $-3$.
In fact, we have
\begin{align*}
B\begin{bmatrix}
1\\1\\1
\end{bmatrix}
&=\begin{bmatrix}
-1&-1&-1\\
-1&-1&-1\\
-1&-1&-1
\end{bmatrix}
\begin{bmatrix}
1\\1\\1
\end{bmatrix}
=\begin{bmatrix}
-3\\-3\\-3
\end{bmatrix}
=-3\begin{bmatrix}
1\\1\\1
\end{bmatrix}.
\end{align*}

Since the algebraic multiplicity of $\lambda=0$ is either $2$ or $3$, and the sum of all the algebraic multiplicities is equal to $3$, the algebraic multiplicity of $\lambda=-3$ must be $1$ and that of $\lambda=0$ is $2$.
Hence the geometric multiplicity of $\lambda=-3$ is $1$. Thus
\[\begin{bmatrix}
1\\1\\1
\end{bmatrix}\] is a basis vector of $E_{-3}$.

In a nutshell, we have obtained that eigenvalues of $B$ are $0$ and $-3$ and basis vectors of $E_0$ and $E_{-3}$ are
\[\begin{bmatrix}
-1\\1\\0
\end{bmatrix},
\begin{bmatrix}
-1\\0\\1
\end{bmatrix} \text{ and } \begin{bmatrix}
1\\1\\1
\end{bmatrix},\] respectively.

Since $A=B+3I$, the eigenvalues of $A$ are $0+3=3$ and $-3+3=0$ and the corresponding eigenvectors are the same. Thus
\[\begin{bmatrix}
-1\\1\\0
\end{bmatrix},
\begin{bmatrix}
-1\\0\\1
\end{bmatrix} \text{ and } \begin{bmatrix}
1\\1\\1
\end{bmatrix}\] are the basis vectors of the eigenspace $E_3$ and $E_0$ of $A$, respectively.

As in solution 1, we put
\[S=\begin{bmatrix}
-1 & -1 & 1\\
1& 0& 1\\
0& 1 & 1
\end{bmatrix}.\] Then we have
\[S^{-1}AS=\begin{bmatrix}
\mathbf{3} & 0 & 0\\
0& \mathbf{3}& 0\\
0 & 0& \mathbf{0}
\end{bmatrix},\] where diagonal entries are eigenvalues corresponding to the eigenvector $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ with this order.

Comment.

This is the first problem of Quiz 13 (Take Home Quiz) for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 13 (Part 1) Diagonalize a Matrix first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ matrix with the characteristic polynomial
\[p(t)=t^3(t-1)^2(t-2)^5(t+2)^4.\] Assume that the matrix $A$ is diagonalizable.

(a) Find the size of the matrix $A$.

(b) Find the dimension of the eigenspace $E_2$ corresponding to the eigenvalue $\lambda=2$.

(c) Find the nullity of $A$.

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

Hint/Definition.

• Recall that when a matrix is diagonalizable, the algebraic multiplicity of each eigenvalue is the same as the geometric multiplicity.
• The geometric multiplicity of an eigenvalue $\lambda$ is the dimension of the eigenspace $E_{\lambda}=\calN(A-\lambda I)$ corresponding to $\lambda$.
• The nullity of $A$ is the dimension of the null space $\calN(A)$ of $A$.

Solution.

(a) Find the size of the matrix $A$.

In general, if $A$ is an $n\times n$ matrix, then its characteristic polynomials has degree $n$.
Since the degree of $p(t)$ is $14$, the size of $A$ is $14 \times 14$.

(b) Find the dimension of the eigenspace $E_2$ corresponding to the eigenvalue $\lambda=2$.

Note that the dimension of the eigenspace $E_2$ is the geometric multiplicity of the eigenvalue $\lambda=2$ by definition.

From the characteristic polynomial $p(t)$, we see that $\lambda=2$ is an eigenvalue of $A$ with algebraic multiplicity $5$.
Since $A$ is diagonalizable, the algebraic multiplicity of each eigenvalue is the same as the geometric multiplicity.

It follows that the geometric multiplicity of $\lambda=2$ is $5$, hence the dimension of the eigenspace $E_2$ is $5$.

(c) Find the nullity of $A$.

We first observe that $\lambda=0$ is an eigenvalue of $A$ with algebraic multiplicity $3$ from the characteristic polynomial.

By definition, the nullity of $A$ is the dimension of the null space $\calN(A)$, and furthermore the null space $\calN(A)$ is the eigenspace $E_0$.
Thus, the nullity of $A$ is the same as the geometric multiplicity of the eigenvalue $\lambda=0$.

Since $A$ is diagonalizable, the algebraic and geometric multiplicities are the same. Hence the nullity of $A$ is $3$.

The post Determine Dimensions of Eigenspaces From Characteristic Polynomial of Diagonalizable Matrix first appeared on Problems in Mathematics.

A complex square ($n\times n$) matrix $A$ is called normal if
\[A^* A=A A^*,\] where $A^*$ denotes the conjugate transpose of $A$, that is $A^*=\bar{A}^{\trans}$.
A matrix $A$ is said to be nilpotent if there exists a positive integer $k$ such that $A^k$ is the zero matrix.

(a) Prove that if $A$ is both normal and nilpotent, then $A$ is the zero matrix.
You may use the fact that every normal matrix is diagonalizable.

(b) Give a proof of (a) without referring to eigenvalues and diagonalization.

(c) Let $A, B$ be $n\times n$ complex matrices. Prove that if $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$, where $I$ is the $n\times n$ identity matrix.

Proof.

(a) If $A$ is normal and nilpotent, then $A=O$

Since $A$ is normal, it is diagonalizable. Thus there exists an invertible matrix $P$ such that $P^{-1}AP=D$, where $D$ is a diagonal matrix whose diagonal entries are eigenvalues of $A$.

Since $A$ is nilpotent, all the eigenvalues of $A$ are $0$. (See the post “Nilpotent matrix and eigenvalues of the matrix” for the proof.)
Hence the diagonal entries of $D$ are zero, and we have $D=O$, the zero matrix.

It follows that we have
\begin{align*}
A=PDP^{-1}=POP^{-1}=O.
\end{align*}
Therefore, every normal nilpotent matrix must be a zero matrix.

(b) Give a proof of (a) without referring to eigenvalues and diagonalization.

Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$.
We prove by induction on $k$ that $A=O$.
The base case $k=1$ is trivial.

Suppose $k>1$ and the case $k-1$ holds. Let $B=A^{k-1}$. Note that since $A$ is normal, the matrix $B$ is also normal.
For any vector $x \in \C^n$, we compute the length of the vector $B^*Bx$ as follows.
\begin{align*}
&\|B^*Bx\|=(B^*Bx)^*(B^*Bx) && \text{by definition of the length}\\
&=x^*B^*(BB^*)Bx\\
&=x^*B^*(B^*B)Bx && \text{since $B$ is normal}\\
&=x^* (B^*)^2B^2x,
\end{align*}
and the last expression is $O$ since $B^2=A^{2k-2}=O$ as $k \geq 2$ implies $2k-2 \geq k$.
Hence we have $B^*Bx=\mathbf{0}$ for every $x\in \C^n$.

This yields that
\begin{align*}
\|Bx\|&=(Bx)^*(Bx)\\
&=x^*B^*Bx=0,
\end{align*}
for every $x\in \C^n$, and hence $B=O$.

By the induction hypothesis, $A^{k-1}=O$ implies $A=O$, and the induction is completed.
So the matrix $A$ must be the zero matrix.

(c) If $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$

We claim that the matrix $B$ is normal as well. If this claim is proved, then part (a) yields that $B=O$ since $B$ is a nilpotent normal matrix, which implies the result $A=I$.

To prove the claim, we compute
\begin{align*}
B^* B&=(I-A)^* (I-A)\\
&=(I-A^*)(I-A)\\
&=I-A-A^*+A^*A,
\end{align*}
and
\begin{align*}
B B^*&=(I-A) (I-A)^*\\
&=(I-A)(I-A^*)\\
&=I-A^*-A+AA^*\\
&=I-A^*-A+A^*A \qquad \text{ since $A$ is normal.}
\end{align*}
It follows that we have $B^* B=BB^*$, and thus $B$ is normal.
Hence, the claim is proved.

The post Normal Nilpotent Matrix is Zero Matrix first appeared on Problems in Mathematics.

Let $A, B, C$ are $2\times 2$ diagonalizable matrices.

The graphs of characteristic polynomials of $A, B, C$ are shown below. The red graph is for $A$, the blue one for $B$, and the green one for $C$.

From this information, determine the rank of the matrices $A, B,$ and $C$.

Graphs of characteristic polynomials

Hint.

Observe that a null space a matrix $M$ is the same as the eigenspace $E_0$ corresponding to the eigenvalue $\lambda=0$ (if 0 is an eigenvalue).
So what you need to see is whether the graphs pass though the origin.

Also, since the matrices are diagonalizable, the algebraic multiplicities are the same as the geometric multiplicities (the dimension of the eigenspace).
You can determine the algebraic multiplicities from the graph by looking at whether the graphs are tangential to $x$-axis or not.

Solution.

We first determine the nullities of the matrices, and then using the rank-nullity theorem we obtain the ranks.

In general the nullity of any matrix $M$ is the dimension of the null space
\[\calN(M)=\{\mathbf{x}\in \R^2 \mid M\mathbf{x}=\mathbf{0}\}\] and the null space is the same as the eigenspace
\[E_{0}=\{\mathbf{x}\in \R^2\mid M\mathbf{x}=0\mathbf{x}=\mathbf{0}\}\] corresponding to the eigenvalue $\lambda=0$ (if any).

From this observation, we see that the nullity of a matrix $M$ is the geometric multiplicity of the eigenspace $E_0$ associated to the eigenvalue $0$ (if any).

The Nullity of the Matrix $A$

Now we determine the nullity of the matrix $A$.
The graph of the characteristic polynomial $p_A(\lambda)$ of $A$ passes through the origin $(0,0)$.

Thus $\lambda=0$ is a root of $p_A(\lambda)$ and hence $\lambda=0$ is an eigenvalue.
Since the $x$-axis is tangential to the graph of $p_A(\lambda)$, the algebraic multiplicity of $\lambda=0$ is $2$. Since the matrix $A$ is diagonalizable, the geometric multiplicity is the same as the algebraic multiplicity. Therefore the nullity of $A$ is $2$.

The Nullity of the Matrix $B$

Next, the graph of characteristic polynomial $p_B(\lambda)$ of $B$ does not pass through the origin $(0,0)$. Thus $\lambda=0$ is not an eigenvalue of $B$. This yields that the matrix $B=B-0I$ is nonsingular matrix, and hence we have the null space $\calN(B)=\{0\}$ and the nullity of $B$ is zero.

The Nullity of the Matrix $C$

Third, the origin $(0,0)$ is on the graphs of characteristic polynomial $p_C(\lambda)$ of $C$, but the $x$-axis is not tangential to the graph of $p_C(\lambda)$.
Therefore the algebraic (hence geometric) multiplicity of $\lambda=0$ is $1$. Thus the nullity of $C$ is $1$.

Ranks by the Rank-Nullity Theorem

Finally, using the rank-nullity theorem
\[\text{rank}+\text{nullity}=2,\] we obtain the rank of $A, B, C$ are $0, 2, 1$, respectively.

The post Given Graphs of Characteristic Polynomial of Diagonalizable Matrices, Determine the Rank of Matrices first appeared on Problems in Mathematics.