Find a nonzero $3\times 3$ matrix $A$ such that $A^2\neq O$ and $A^3=O$, where $O$ is the $3\times 3$ zero matrix.

(Such a matrix is an example of a nilpotent matrix. See the comment after the solution.)

Solution.

For example, let $A$ be the following $3\times 3$ matrix.
\[A=\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}.\] Then $A$ is a nonzero matrix and we have
\[A^2=\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}
=\begin{bmatrix}
0 & 0 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}\neq O.\]

The third power of $A$ is
\[A^3=A^2A=\begin{bmatrix}
0 & 0 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}=
\begin{bmatrix}
0 & 0 & 0 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}=O.\] Thus, the nonzero matrix $A$ satisfies the required conditions $A^2\neq O, A^3=O$.

Comment.

A square matrix $A$ is called nilpotent if there is a non-negative integer $k$ such that $A^k$ is the zero matrix.
The smallest such an integer $k$ is called degree or index of $A$.

The matrix $A$ in the solution above gives an example of a $3\times 3$ nilpotent matrix of degree $3$.

The post Example of a Nilpotent Matrix $A$ such that $A^2\neq O$ but $A^3=O$. first appeared on Problems in Mathematics.

Let $A$ be a $3\times 3$ singular matrix.

Then show that there exists a nonzero $3\times 3$ matrix $B$ such that
\[AB=O,\] where $O$ is the $3\times 3$ zero matrix.

Proof.

Since $A$ is singular, the equation $A\mathbf{x}=\mathbf{0}$ has a nonzero solution.
Let $\mathbf{x}_1$ be a nonzero solution of $A\mathbf{x}=\mathbf{0}$.
We define the $3\times 3$ matrix $B$ to be
\[B=[\mathbf{x}_1, \mathbf{0}, \mathbf{0}],\] that is, the first column of $B$ is the vector $x_1$ and the second and the third column vectors are $3$-dimensional zero vectors.

Then since $x_1\neq \mathbf{0}$, the matrix $B$ is not the zero matrix.
We have
\begin{align*}
AB&=A[\mathbf{x}_1, \mathbf{0}, \mathbf{0}]\\
&=[A\mathbf{x}_1, A\mathbf{0}, A\mathbf{0}]\\
&=[\mathbf{0}, \mathbf{0}, \mathbf{0}]=O,
\end{align*}
since $A\mathbf{x}_1=\mathbf{0}$.

Thus we have found the nonzero matrix $B$ such that the product $AB=O$.

Comment.

This is one of the midterm exam 1 problems of linear algebra (Math 2568) at the Ohio State University.

In the exam the following hint was given:

Hint: Let $B=[\mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3]$, where $\mathbf{x}_i$ is the $i$-th column vector of $B$ for $i=1,2,3$. Then $AB=[A\mathbf{x}_1, A\mathbf{x}_2, A\mathbf{x}_3]$.

Common Mistake

Several students wrote “Since $A$ is singular, $A$ has a solution”.
This does not make sense. A solution of what? You need to remember the definition correctly.

Also, note that $\mathbf{x}$ is a vector, not a number.

Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
2. Problem 2 and its solution: The vector form of the general solution of a system
3. Problem 3 and its solution: Matrix operations (transpose and inverse matrices)
4. Problem 4 and its solution: Linear combination
5. Problem 5 and its solution: Inverse matrix
6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
7. Problem 7 and its solution: Solve a system by the inverse matrix
8. Problem 8 and its solution (The current page): A proof problem about nonsingular matrix

The post If a Matrix $A$ is Singular, then Exists Nonzero $B$ such that $AB$ is the Zero Matrix first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ singular matrix.
Then prove that there exists a nonzero $n\times n$ matrix $B$ such that
\[AB=O,\] where $O$ is the $n\times n$ zero matrix.

Definition.

Recall that an $n \times n$ matrix $A$ is called singular if the equation
\[A\mathbf{x}=\mathbf{0}\] has a nonzero solution $\mathbf{x}\in \R^n$.

Proof.

Since $A$ is singular, there exists a nonzero vector $\mathbf{b} \in \R^n$ such that
\[A\mathbf{b}=\mathbf{0}.\]

Then we define the $n\times n$ matrix $B$ whose first column is the vector $\mathbf{b}$ and the other entries are zero. That is
\[B=\begin{bmatrix}
\mathbf{b} & \mathbf{0} & \cdots & \mathbf{0}
\end{bmatrix}.\] Since the vector $\mathbf{b}$ is nonzero, the matrix $B$ is nonzero.

With this choice of $B$, we have
\begin{align*}
AB&=A\begin{bmatrix}
\mathbf{b} & \mathbf{0} & \cdots & \mathbf{0}
\end{bmatrix} \\
&=\begin{bmatrix}
A\mathbf{b} & A\mathbf{0} & \cdots & A\mathbf{0}
\end{bmatrix}\\
&=\begin{bmatrix}
\mathbf{0} & \mathbf{0} & \cdots & \mathbf{0}
\end{bmatrix}
=O.
\end{align*}
Hence we have proved that $AB=O$ with the nonzero matrix $B$.

The post If a Matrix $A$ is Singular, There Exists Nonzero $B$ such that the Product $AB$ is the Zero Matrix first appeared on Problems in Mathematics.

Let $A_1, A_2, \dots, A_m$ be $n\times n$ Hermitian matrices. Show that if
\[A_1^2+A_2^2+\cdots+A_m^2=\calO,\] where $\calO$ is the $n \times n$ zero matrix, then we have $A_i=\calO$ for each $i=1,2, \dots, m$.

Hint.

Recall that a complex matrix $A$ is Hermitian if the conjugate transpose of $A$ is $A$ itself.
Namely, $A$ is Hermitian if
\[\bar{A}^{\trans}=A.\]

We also use the length of a vector in the proof below.
Let $\mathbf{v}$ be $n$-dimensional complex vector. Then the length of $\mathbf{v}$ is defined to be
\[\|\mathbf{v}\|=\sqrt{\bar{\mathbf{v}}^{\trans}\mathbf{v}}.\] The length of a complex vector $\mathbf{v}$ is a non-negative real number.

The length is also called norm or magnitude.

Proof.

Let $\mathbf{x}$ be an $n$-dimensional vector, that is, $\mathbf{x}\in \R^n$.

Then for each $i$, we have
\[\bar{\mathbf{x}}^{\trans}A_i^2\mathbf{x}=\bar{\mathbf{x}}^{\trans}\bar{A}_i^{\trans}A_i\mathbf{x}=(\overline{A_i\mathbf{x}})^{\trans}(A_i\mathbf{x})=\|A_i\mathbf{x}\|^2\geq 0.\] Here, the first equality follows from the definition of a Hermitian matrix.

Now we compute
\begin{align*}
0&=\bar{\mathbf{x}}^{\trans}\calO \mathbf{x}=\bar{\mathbf{x}}^{\trans}(A_1^2+A_2^2+\cdots+A_m^2) \mathbf{x}\\
&=\bar{\mathbf{x}}^{\trans}A_1^2\mathbf{x}+\bar{\mathbf{x}}^{\trans}A_2^2\mathbf{x}+\cdots+\bar{\mathbf{x}}^{\trans}A_m^2 \mathbf{x}\\
&=\|A_1\mathbf{x}\|^2+\|A_2\mathbf{x}\|^2+\cdots +\|A_m\mathbf{x}\|^2.
\end{align*}

Since each length $\|A_i\mathbf{x}\|$ is a non-negative real number, this implies that we have $A_i\mathbf{x}=\mathbf{0}$ for all $\mathbf{x \in \R^n}$. Hence we must have $A_i=\calO$ for each $i=1,2,\dots, m$.

The post Sum of Squares of Hermitian Matrices is Zero, then Hermitian Matrices Are All Zero first appeared on Problems in Mathematics.

Test your understanding of basic properties of matrix operations.

There are 10 True or False Quiz Problems.

These 10 problems are very common and essential.
So make sure to understand these and don’t lose a point if any of these is your exam problems.
(These are actual exam problems at the Ohio State University.)

You can take the quiz as many times as you like.

The solutions will be given after completing all the 10 problems.
Click the View question button to see the solutions.

Notations: $I$ denotes an identity matrix and $O$ denotes a zero matrix. The sizes of these matrices should be determined from the context.

(The Ohio State University, linear algebra exam)

Related Question.

Check out “10 True of False Problems about Nonsingular / Invertible Matrices” for True or False problems about nonsingular matrices, invertible matrices, and linearly independent vectors

The post 10 True or False Problems about Basic Matrix Operations first appeared on Problems in Mathematics.

Problem 98

Let $A$ and $B$ be $n\times n$ matrices. Suppose that the matrix product $AB=O$, where $O$ is the $n\times n$ zero matrix.

Is it true that the matrix product with opposite order $BA$ is also the zero matrix?
If so, give a proof. If not, give a counterexample.

Solution.

The statement is in general not true. We give a counter example.
Consider the following $2\times 2$ matrices.
\[A=\begin{bmatrix}
0 & 1\\
0& 1
\end{bmatrix} \text{ and } \begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}.\]

Then we compute
\[AB=\begin{bmatrix}
0 & 1\\
0& 1
\end{bmatrix}
\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}
=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}=O.\] Thus the matrix product $AB$ is the $2\times 2$ zero matrix $O$.

On the other hand, we compute
\[BA=\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
0& 1
\end{bmatrix}=\begin{bmatrix}
0 & 2\\
0& 0
\end{bmatrix}.\]

Thus the matrix product $BA$ is not the zero matrix.
Therefore the statement is not true in general.

The post If the Matrix Product $AB=0$, then is $BA=0$ as Well? first appeared on Problems in Mathematics.