Let $\mathbf{v} = \begin{bmatrix} 2 & -5 & -1 \end{bmatrix}$.

Find all $3 \times 1$ column vectors $\mathbf{w}$ such that $\mathbf{v} \mathbf{w} = 0$.

Solution.

Let $\mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}$.

Then we want
\[\mathbf{v} \mathbf{w} =\begin{bmatrix} 2 & -5 & -1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}=2 w_1 – 5 w_2 – w_3 = 0.\]

Letting $w_1$ and $w_2$ be free variables, we solve $w_3 = 2w_1 – 5w_2$.
Then every solution to the equation $\mathbf{v} \mathbf{w} = 0$ is of the form
\[\mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \\ 2 w_1 – 5 w_2 \end{bmatrix} = w_1 \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + w_2 \begin{bmatrix} 0 \\ 1 \\ -5 \end{bmatrix}.\] Click here if solved 29

The post Find all Column Vector $\mathbf{w}$ such that $\mathbf{v}\mathbf{w}=0$ for a Fixed Vector $\mathbf{v}$ 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.

Let
\[A=\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\] be a $3\times 3$ matrix. Then find the formula for $A^n$ for any positive integer $n$.

Proof.

We first compute several powers of $A$ and guess the general formula.
We have
\begin{align*}
A^2=\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 3 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix},
\end{align*}
\begin{align*}
A^3=A^2A=\begin{bmatrix}
1 & 1 & 3 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 5 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix},
\end{align*}
\begin{align*}
A^4=A^3A=\begin{bmatrix}
1 & 1 & 5 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 1 & 7 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

From these computations, we guess the general formula of $A^n$ is
\[A^n=\begin{bmatrix}
1 & 1 & 2n-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.\]

We prove this formula by mathematical induction on $n$.
The base case $n=1$ follows from the definition of $A$.

Suppose that the formula is true for $n=k$.
We prove the formula for $n=k+1$.
We have
\begin{align*}
A^{k+1}&=A^{k}A\\
&=\begin{bmatrix}
1 & 1 & 2k-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
&& \text{by the induction hypothesis}\\
&=\begin{bmatrix}
1 & 1 & 2k+1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
1 & 1 & 2(k+1)-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

Thus the formula holds for $n=k+1$.
Hence the formula is true for any positive integer $n$ by induction.

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Let $V$ be the vector space of all $3\times 3$ real matrices.
Let $A$ be the matrix given below and we define
\[W=\{M\in V \mid AM=MA\}.\] That is, $W$ consists of matrices that commute with $A$.
Then $W$ is a subspace of $V$.

Determine which matrices are in the subspace $W$ and find the dimension of $W$.

(a) \[A=\begin{bmatrix}
a & 0 & 0 \\
0 &b &0 \\
0 & 0 & c
\end{bmatrix},\] where $a, b, c$ are distinct real numbers.

(b) \[A=\begin{bmatrix}
a & 0 & 0 \\
0 &a &0 \\
0 & 0 & b
\end{bmatrix},\] where $a, b$ are distinct real numbers.

Solution.

(a) Diagonal matrix with distinct diagonal entries

Let us first determine when a matrix $M$ commutes with $A$.
Let
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}\] and suppose that $AM=MA$:
\[\begin{bmatrix}
a & 0 & 0 \\
0 & b &0 \\
0 & 0 & c
\end{bmatrix}
\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}
\begin{bmatrix}
a & 0 & 0 \\
0 &b &0 \\
0 & 0 & c
\end{bmatrix}.\] Computing matrix products, we obtain
\[\begin{bmatrix}
aa_{1 1} & aa_{1 2} & aa_{1 3} \\
ba_{2 1} & ba_{2 2} & ba_{2 3} \\
ca_{3 1} & ca_{3 2} &c a_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1}a & a_{1 2}b & a_{1 3}c \\
a_{2 1}a & a_{2 2}b & a_{2 3}c\\
a_{3 1}a & a_{3 2}b & a_{3 3}c
\end{bmatrix}. \tag{*}\] Compare the $(1,2)$ entries and we have $aa_{1 2}=ba_{1 2}$.
Since $a\neq b$, we must have $a_{1 2}=0$.

Similarly, comparing the off-diagonal entries and noting $a, b, c$ are distinct, we find that off diagonal entries $a_{i j} , i\neq j$ must be $0$.

Thus, $M$ commutes with $A$ if and only if
\[M=\begin{bmatrix}
a_{1 1} & 0 & 0 \\
0 & a_{2 2} & 0 \\
0 & 0 & a_{3 3}
\end{bmatrix}.\]

Therefore, the subspace $W$ consists of all $3\times 3$ diagonal matrices:
\[W=\{W\in V\mid W \text{ is diagonal}\}.\] Then it is easy to see that the set $\{E_{1 1}, E_{2 2}, E_{3 3}\}$ is a basis for $W$, where $E_{i j}$ is the $3\times 3$ matrix whose $(i,j)$-entry is $1$ and the other entries are zero. Thus the dimension of $W$ is $3$.

(b) Diagonal matrix two diagonal entries are the same

Now consider the case
\[A=\begin{bmatrix}
a & 0 & 0 \\
0 &a &0 \\
0 & 0 & b
\end{bmatrix}.\] Let
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}\] and compute $AM=MA$ as in part (a) (or you just need to replace $b, c$ in (*) by $a, b$, respectively) and obtain
\[\begin{bmatrix}
aa_{1 1} & aa_{1 2} & aa_{1 3} \\
aa_{2 1} & aa_{2 2} & aa_{2 3} \\
ba_{3 1} & ba_{3 2} & ba_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1}a & a_{1 2}a & a_{1 3}b \\
a_{2 1}a & a_{2 2}a & a_{2 3}b\\
a_{3 1}a & a_{3 2}a & a_{3 3}b
\end{bmatrix}. \] Comparing entries and noting $a\neq b$, we have
\[a_{1 3}=0, a_{2 3}=0, a_{3 1}=0, a_{3 2}=0.\]

Thus, $M$ commutes with $A$ is and only if
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & 0 \\
a_{2 1} & a_{2 2} & 0 \\
0 & 0 & a_{3 3}
\end{bmatrix},\] and hence the subspace $W$ consists of such matrices.
From this, we see that the set
\[\{E_{1 1}, E_{1 2}, E_{2 1}, E_{2 2}, E_{3 3}\}\] is a basis for $W$, and we conclude that the dimension of $W$ is $5$.

The post Basis For Subspace Consisting of Matrices Commute With a Given Diagonal Matrix first appeared on Problems in Mathematics.

Let $A$ and $B$ are matrices such that the matrix product $AB$ is defined and $AB$ is a square matrix.
Is it true that the matrix product $BA$ is also defined and $BA$ is a square matrix? If it is true, then prove it. If not, find a counterexample.

Definition/Hint.

Let $A$ be an $m\times n$ matrix.
This means that the matrix $A$ has $m$ rows and $n$ columns.

Let $B$ be an $r \times s$ matrix.
Then the matrix product $AB$ is defined if $n=r$, that is, if the number of columns of $A$ is equal to the number of rows of $B$.

Definition. A matrix $C$ is called a square matrix if the size of $C$ is $n\times n$ for some positive integer $n$.
(The number of rows and the number of columns are the same.)

Proof.

We prove that the matrix product $BA$ is defined and it is a square matrix.

Let $A$ be an $m\times n$ matrix and $B$ be an $r\times s$ matrix.

Since the matrix product $AB$ is defined, we must have $n=r$ and the size of $AB$ is $m\times s$.
Since $AB$ is a square matrix, we have $m=s$.

Thus the size of the matrix $B$ is $n \times m$.
From this, we see that the product $BA$ is defined and its size is $n \times n$, hence it is a square matrix.

The post If matrix product $AB$ is a square, then is $BA$ a square matrix? first appeared on Problems in Mathematics.

Let $A$ be an $m \times n$ matrix.
Let $\calN(A)$ be the null space of $A$. Suppose that $\mathbf{u} \in \calN(A)$ and $\mathbf{v} \in \calN(A)$.
Let $\mathbf{w}=3\mathbf{u}-5\mathbf{v}$.

Then find $A\mathbf{w}$.
 

Hint.

Recall that the null space of an $m\times n$ matrix $A$ is a subspace of $\R^n$ defined by
\[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\] Here $\mathbf{0}_m$ is the $m$-dimensional zero vector in $\R^m$.

Solution.

Since $\mathbf{u}, \mathbf{v} \in \calN(A)$, we have
\[A\mathbf{u}=\mathbf{0}_m \text{ and } A\mathbf{v}=\mathbf{0}_m,\] where $\mathbf{0}_m$ is the $m$-dimensional zero vector in $\R^m$.

Now using the properties of the matrix multiplication, we have
\begin{align*}
A\mathbf{w}&=A(3\mathbf{u}-5\mathbf{v})\\
&=A(3\mathbf{u})+A(-5\mathbf{v})\\
&=3A\mathbf{u}-5A\mathbf{v}\\
&=3\mathbf{0}_m-5\mathbf{0}_m=\mathbf{0}_m.
\end{align*}
Therefore we obtained
\[A\mathbf{w}=\mathbf{0}_m\in \R^m.\]

Remark.

Note that a map $T:\R^n \to \R^m$ defined by $T(\mathbf{x})=A\mathbf{x}$, where $A$ is an $m\times n$ matrix is a linear transformation.
That is the map $T$ satisfies:

1. $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$ for any $\mathbf{u}, \mathbf{v} \in \R^n$, and
2. $T(c\mathbf{v})=cT(\mathbf{v})$ for any $\mathbf{v}\in \R^n$ and $c \in \R$.

From this point of view, the above problem can be classified into a problem of linear transformation.

The post Linear Properties of Matrix Multiplication and the Null Space of a Matrix first appeared on Problems in Mathematics.

Let $A$ and $B$ be $n \times n$ real symmetric matrices. Prove the followings.

(a) The product $AB$ is symmetric if and only if $AB=BA$.

(b) If the product $AB$ is a diagonal matrix, then $AB=BA$.

Hint.

A matrix $A$ is called symmetric if $A=A^{\trans}$.

In this problem, we need the following property of transpose:
Let $A$ be an $m\times n$ and $B$ be an $n \times r$ matrix. Then we have
\[(AB)^{\trans}=B^{\trans}A^{\trans}.\] (When you distribute transpose over the product of two matrices, then you need to reverse the order of the matrix product.)

Solution.

(a) The product $AB$ is symmetric if and only if $AB=BA$.

Suppose $AB$ is symmetric. This means that we have
\[ AB=(AB)^{\trans}=B^{\trans}A^{\trans}=BA.\] The second equality is a general property of transpose. In the last equality we used the assumption that $A, B$ are symmetric, that is, $A=A^{\trans}, B=B^{\trans}$. Thus we have $AB=BA$.

On the other hand, if $AB=BA$, then we have
\[AB=BA=B^{\trans}A^{\trans}=(AB)^{\trans}.\] Thus, we have $AB=(AB)^{\trans}$, hence $AB$ is symmetric.

If the product $AB$ is a diagonal matrix, then $AB=BA$.

Note that a diagonal matrix is symmetric. Hence the result follows from part (a).

10 True or False Problems about Matrices

Test your understanding about matrices with 10 True or False questions given in the post “10 True or False Problems about Basic Matrix Operations“.

Or try True or False problems about nonsingular, invertible matrices, and linearly independent vectors at “10 True of False Problems about Nonsingular / Invertible Matrices“.

The post Symmetric Matrices and the Product of Two Matrices 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.

Let $A$ and $B$ be $2\times 2$ matrices.

Prove or find a counterexample for the statement that $(A-B)(A+B)=A^2-B^2$.

Hint.

In general, matrix multiplication is not commutative: $AB$ and $BA$ might be different.

Solution.

Let us calculate $(A-B)(A+B)$ as follows using the fact that the matrix product is distributive.
\begin{align*}
(A-B)(A+B)&=A(A+B)-B(A+B)\\
&=A^2+AB-BA-B^2\\
&=A^2-B^2+(AB-BA).
\end{align*}
Thus if $(A-B)(A+B)=A^2-B^2$ then $AB-BA=O$, the zero matrix. Equivalently, $AB=BA$.

Note that matrix multiplication is not commutative, namely, $AB\neq BA$ in general.
Thus we can disprove the statement if we find matrices $A$ and $B$ such that $AB \neq BA$.

For example, let
\[A=\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}
\text{ and }
B=\begin{bmatrix}
0 & 1\\
0& 1
\end{bmatrix}.\] Then we have
\[AB=\begin{bmatrix}
0 & 2\\
0& 0
\end{bmatrix} \text{ and }
BA=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}.\]

Since $AB \neq BA$, we have $(A-B)(A+B) \neq A^2-B^2$ for \[A=\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}
\text{ and }
B=\begin{bmatrix}
0 & 1\\
0& 1
\end{bmatrix}.\]

Hence we found a counterexample for the statement.

10 True or False Quiz Problems about Matrix Operations

Check out the post “10 True or False Problems about Basic Matrix Operations” and take a quiz about basic properties of matrix operations.

There are 10 True or False problems about basic properties of matrix operations (matrix product, transpose, etc.).

The complete solutions are given as well.

The post True or False: $(A-B)(A+B)=A^2-B^2$ for Matrices $A$ and $B$ first appeared on Problems in Mathematics.

Let $A=(a_{i j})$ and $B=(b_{i j})$ be $n\times n$ real matrices for some $n \in \N$. Then answer the following questions about the trace of a matrix.

(a) Express $\tr(AB^{\trans})$ in terms of the entries of the matrices $A$ and $B$. Here $B^{\trans}$ is the transpose matrix of $B$.

(b) Show that $\tr(AA^{\trans})$ is the sum of the square of the entries of $A$.

(c) Show that if $A$ is nonzero symmetric matrix, then $\tr(A^2)>0$.

Hint.

Review

1. the definition of the transpose of a matrix
2. the definition of matrix multiplication
3. the definition of a symmetric matrix

Then the proofs of these statement is straightforward computations.

Proof.

(a) Express $\tr(AB^{\trans})$ in terms of the entries of $A$ and $B$.

Here we use the following notation for an entry of a matrix: the $(i, j)$-entry of a matrix $C$ is denoted by $(C)_{i,j}$.

Then the $(i,j)$-entry of $AB^{\trans}$ is $(AB^{\trans})_{ij}=\sum_{k=1}^n a_{ik}b_{jk}$.
Thus we have
\[\tr(AB^{\trans})=\sum_{l=1}^n (AB^{\trans})_{ll}=\sum_{l=1}^n \sum_{k=1}^n a_{lk}b_{lk}.\]

(b) Show that $\tr(AA^{\trans})$ is the sum of the square of the entries of $A$

By the formula obtained in part (a), we have
\[ \tr(AA^{\trans})=\sum_{l=1}^n \sum_{k=1}^n a_{lk}^2.\] This is the sum of the squares of entries of $A$.

(c) Show that if $A$ is nonzero symmetric matrix, then $\tr(A^2)>0$.

Since $A$ is a symmetric matrix, we have $A^{\trans}=A$.
Thus by the result of part (b), we have

\[ \tr(A^2)=\tr(AA^{\trans})=\sum_{l=1}^n \sum_{k=1}^n a_{lk}^2>0.\] The last sum is strictly positive since $A$ is not the zero matrix, there is a nonzero entry of $A$ (and of course the square of a real number is nonnegative).

Comment.

The results we proved in this article can be extended to complex matrices, matrices with complex number entries.
In this case, the condition in (c) that $A$ is symmetric is replaced by the condition that $A$ is a hermitian matrix.

The post Questions About the Trace of a Matrix first appeared on Problems in Mathematics.