Let $\mathbf{v}$ and $\mathbf{w}$ be two $n \times 1$ column vectors.

(a) Prove that $\mathbf{v}^\trans \mathbf{w} = \mathbf{w}^\trans \mathbf{v}$.

(b) Provide an example to show that $\mathbf{v} \mathbf{w}^\trans$ is not always equal to $\mathbf{w} \mathbf{v}^\trans$.

Solution.

(a) Prove that $\mathbf{v}^\trans \mathbf{w} = \mathbf{w}^\trans \mathbf{v}$.

Suppose the vectors have component
\[\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} \, \mbox{ and } \mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \\ \vdots \\ w_n \end{bmatrix}.\] Then,
\[\mathbf{v}^\trans \mathbf{w} = \begin{bmatrix} v_1 & v_2 & \cdots & v_n \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ \vdots \\ w_n \end{bmatrix} = \sum_{i=1}^n v_i w_i,\] while
\[\mathbf{w}^\trans \mathbf{v} = \begin{bmatrix} w_1 & w_2 & \cdots & w_n \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} = \sum_{i=1}^n w_i v_i.\] We can see that they are equal because $v_i w_i = w_i v_i$.

(b) Provide an example to show that $\mathbf{v} \mathbf{w}^\trans$ is not always equal to $\mathbf{w} \mathbf{v}^\trans$.

For the counterexample, let $\mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\mathbf{w} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Then
\[\mathbf{v} \mathbf{w}^\trans = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\] while
\[\quad \mathbf{w} \mathbf{v}^\trans = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}.\]

Comment.

Recall that for two vectors $\mathbf{v}, \mathbf{w} \in \R^n$, the dot product (or inner product) of $\mathbf{v}, \mathbf{w}$ is defined to be
\[\mathbf{v}\cdot \mathbf{w}:=\mathbf{v}^{\trans} \mathbf{w}.\]

Part (a) of the problem deduces that the dot product is commutative. This means that we have
\[\mathbf{v}\cdot \mathbf{w}= \mathbf{w} \cdot \mathbf{v}.\]

In fact, we have
\begin{align*}
\mathbf{v}\cdot \mathbf{w}= \mathbf{v}^\trans \mathbf{w} \stackrel{\text{(a)}}{=} \mathbf{w}^\trans \mathbf{v} \mathbf{w} \cdot \mathbf{v}.
\end{align*}

Also, notice that while $\mathbf{v} \mathbf{w}^\trans$ is not always equal to $\mathbf{w} \mathbf{v}^\trans$, we know that $(\mathbf{v} \mathbf{w}^\trans)^\trans = \mathbf{w} \mathbf{v}^\trans$.

The post Prove that the Dot Product is Commutative: $\mathbf{v}\cdot \mathbf{w}= \mathbf{w} \cdot \mathbf{v}$ first appeared on Problems in Mathematics.

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

Is it always true that $\tr (A B) = \tr (A) \tr (B) $?

If it is true, prove it. If not, give a counterexample.

Solution.

There are many counterexamples.

For one, take
\[A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \text{ and } B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.\]

Then $\tr(A)=1, \tr(B)=1$, and hence $\tr(A) \tr(B) = 1$, while $\tr(AB) = 0$ as $AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

The post Does the Trace Commute with Matrix Multiplication? Is $\tr (A B) = \tr (A) \tr (B) $? first appeared on Problems in Mathematics.

A square matrix $A$ is called nilpotent if some power of $A$ is the zero matrix.
Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.

Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample.

Solution.

We claim that the matrix $B-A$ is not necessarily invertible.
Consider the matrix
\[A=\begin{bmatrix}
0 & -1 \\0& 0
\end{bmatrix}.\] This matrix is nilpotent as we have
\[A^2=\begin{bmatrix}
0 & -1 \\0& 0
\end{bmatrix}
\begin{bmatrix}
0 & -1 \\0& 0
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 \\0& 0
\end{bmatrix}.\]

Also consider the matrix
\[B=\begin{bmatrix}
1 & 0 \\1& 1
\end{bmatrix}.\] Since the determinant of the matrix $B$ is $1$, it is invertible.

So the matrix $A$ and $B$ satisfy the assumption of the problem.
However the matrix
\[B-A=\begin{bmatrix}
1 & 1 \\1& 1
\end{bmatrix}\] is not invertible as its determinant is $0$.
Hence we found a counterexample.

Related Question.

Here is another problem about a nilpotent matrix.

The solution is given in the post ↴
Nilpotent Matrices and Non-Singularity of Such Matrices

The post Is the Sum of a Nilpotent Matrix and an Invertible Matrix Invertible? first appeared on Problems in Mathematics.

Is every diagonalizable matrix invertible?

Solution.

The answer is No.

Counterexample

We give a counterexample. Consider the $2\times 2$ zero matrix.
The zero matrix is a diagonal matrix, and thus it is diagonalizable.
However, the zero matrix is not invertible as its determinant is zero.

More Theoretical Explanation

Let us give a more theoretical explanation.
If an $n\times n$ matrix $A$ is diagonalizable, then there exists an invertible matrix $P$ such that
\[P^{-1}AP=\begin{bmatrix}
\lambda_1 & 0 & \cdots & 0 \\
0 & \lambda_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n
\end{bmatrix},\] where $\lambda_1, \dots, \lambda_n$ are eigenvalues of $A$.
Then we consider the determinants of the matrices of both sides.
The determinant of the left hand side is
\begin{align*}
\det(P^{-1}AP)=\det(P)^{-1}\det(A)\det(P)=\det(A).
\end{align*}
On the other hand, the determinant of the right hand side is the product
\[\lambda_1\lambda_2\cdots \lambda_n\] since the right matrix is diagonal.
Hence we obtain
\[\det(A)=\lambda_1\lambda_2\cdots \lambda_n.\] (Note that it is always true that the determinant of a matrix is the product of its eigenvalues regardless diagonalizability.
See the post “Determinant/trace and eigenvalues of a matrix“.)

Hence if one of the eigenvalues of $A$ is zero, then the determinant of $A$ is zero, and hence $A$ is not invertible.

The true statement is:

Is Every Invertible Matrix Diagonalizable?

Note that it is not true that every invertible matrix is diagonalizable.

For example, consider the matrix
\[A=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}.\] The determinant of $A$ is $1$, hence $A$ is invertible.
The characteristic polynomial of $A$ is
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
1-t & 1\\
0& 1-t
\end{vmatrix}=(1-t)^2.
\end{align*}
Thus, the eigenvalue of $A$ is $1$ with algebraic multiplicity $2$.
We have
\[A-I=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}\] and thus eigenvectors corresponding to the eigenvalue $1$ are
\[a\begin{bmatrix}
1 \\
0
\end{bmatrix}\] for any nonzero scalar $a$.
Thus, the geometric multiplicity of the eigenvalue $1$ is $1$.
Since the geometric multiplicity is strictly less than the algebraic multiplicity, the matrix $A$ is defective and not diagonalizable.

Is There a Matrix that is Not Diagonalizable and Not Invertible?

Finally, note that there is a matrix which is not diagonalizable and not invertible.
For example, the matrix $\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}$ is such a matrix.

Summary

There are all possibilities.

1. Diagonalizable, but not invertible.
   Example: \[\begin{bmatrix}
   0 & 0\\
   0& 0
   \end{bmatrix}.\]
2. Invertible, but not diagonalizable.
   Example: \[\begin{bmatrix}
   1 & 1\\
   0& 1
   \end{bmatrix}\]
3. Not diagonalizable and Not invertible.
   Example: \[\begin{bmatrix}
   0 & 1\\
   0& 0
   \end{bmatrix}.\]
4. Diagonalizable and invertible
   Example: \[\begin{bmatrix}
   1 & 0\\
   0& 1
   \end{bmatrix}.\]

The post True or False. Every Diagonalizable Matrix is Invertible first appeared on Problems in Mathematics.

Let $G$ be a group. Let $a$ and $b$ be elements of $G$.
If the order of $a, b$ are $m, n$ respectively, then is it true that the order of the product $ab$ divides $mn$? If so give a proof. If not, give a counterexample.

Proof.

We claim that it is not true. As a counterexample, consider $G=S_3$, the symmetric group of three letters.
Let $a=(1\, 2), b=(1 \,3)$ be transposition elements in $S_3$.
The orders of $a$ and $b$ are both $2$.

Consider the product
\[ab=(1\, 2)(1 \,3)=(1 \, 3 \, 2).\] Then it is straightforward to check that the order of $ab$ is $3$, which does not divide $4$ (the product of orders of $a$ and $b$).

Therefore, the group $G=S_3$ and elements $a=(1\, 2), b=(1 \,3)\in G$ serve as a counterexample.

Remark. (Abelian group case)

If we further assume that $G$ is an abelian group, then the statement is true.
Here is the proof if $G$ is abelian.

Let $e$ be the identity element of $G$.
\begin{align*}
(ab)^{mn} &=a^{mn}b^{mn} && \text{ since $G$ is abelian}\\
&=(a^m)^n(b^n)^m\\
&=e^n e^m && \text{since the order of $a, b$ are $m, n$ respectively}\\
&=e.
\end{align*}
Thus the order of $ab$ divides $mn$.

Related Question.

If the group is abelian, then the statement is true.

See the post “Order of the Product of Two Elements in an Abelian Group” for a proof of this problem.

More generally, we can prove the following.

A proof of this problem is given in the post “The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements“.

The post Order of Product of Two Elements in a Group first appeared on Problems in Mathematics.

Each of the following sets are not a subspace of the specified vector space. For each set, give a reason why it is not a subspace.
(1) \[S_1=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1\geq 0 \,\right \}\] in the vector space $\R^3$.

(2) \[S_2=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1-4x_2+5x_3=2 \,\right \}\] in the vector space $\R^3$.

(3) \[S_3=\left \{\, \begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2 \quad \middle | \quad y=x^2 \quad \,\right \}\] in the vector space $\R^2$.

(4) Let $P_4$ be the vector space of all polynomials of degree $4$ or less with real coefficients.
\[S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}\] in the vector space $P_4$.

(5) \[S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}\] in the vector space $P_4$.

(6) Let $M_{2 \times 2}$ be the vector space of all $2\times 2$ real matrices.
\[S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\} \] in the vector space $M_{2\times 2}$.

(7) \[S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\} \] in the vector space $M_{2\times 2}$.

(Linear Algebra Exam Problem, the Ohio State University)

(8) Let $C[-1, 1]$ be the vector space of all real continuous functions defined on the interval $[a, b]$.
\[S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\} \] in the vector space $C[-2, 2]$.

(9) \[S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$.

(10) Let $C^2[a, b]$ be the vector space of all real-valued functions $f(x)$ defined on $[a, b]$, where $f(x), f'(x)$, and $f^{\prime\prime}(x)$ are continuous on $[a, b]$. Here $f'(x), f^{\prime\prime}(x)$ are the first and second derivative of $f(x)$.
\[S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$.

(11) Let $S_{11}$ be the set of real polynomials of degree exactly $k$, where $k \geq 1$ is an integer, in the vector space $P_k$.

(12) Let $V$ be a vector space and $W \subset V$ a vector subspace. Define the subset $S_{12}$ to be the complement of $W$,
\[ V \setminus W = \{ \mathbf{v} \in V \mid \mathbf{v} \not\in W \}.\]

Solution.

Recall the following subspace criteria.
A subset $W$ of a vector space $V$ over the scalar field $K$ is a subspace of $V$ if and only if the following three criteria are met.

Thus, to prove a subset $W$ is not a subspace, we just need to find a counterexample of any of the three criteria.
  

Solution (1). $S_1=\{ \mathbf{x} \in \R^3 \mid x_1\geq 0 \}$

The subset $S_1$ does not satisfy condition 3. For example, consider the vector
\[\mathbf{x}=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}.\] Then since $x_1=1\geq 0$, the vector $\mathbf{x}\in S_1$. Then consider the scalar product of $\mathbf{x}$ and the scalar $-1$. Then we have
\[(-1)\cdot\mathbf{x}=\begin{bmatrix}
-1 \\
0 \\
0
\end{bmatrix},\] and the first entry is $-1$, hence $-\mathbf{x}$ is not in $S_1$. Thus $S_1$ does not satisfy condition 3 and it is not a subspace of $\R^3$.
(You can check that conditions 1, 2 are met.)
  

Solution (2). $S_2= \{ \mathbf{x}\in \R^3\mid x_1-4x_2+5x_3=2 \}$

The zero vector of the vector space $\R^3$ is
\[\mathbf{0}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.\] Since the zero vector $\mathbf{0}$ does not satisfy the defining relation $x_1-4x_2+5x_3=2$, it is not in $S_2$. Hence condition 1 is not met, hence $S_2$ is not a subspace of $\R^3$.
(You can check that conditions 2, 3 are not met as well.)
  

Solution (3). $S_3=\{\mathbf{x}\in \R^2 \mid y=x^2 \quad \}$

Consider vectors
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
-1 \\
1
\end{bmatrix}.\] These are vectors in $S_3$ since both vectors satisfy the defining relation $y=x^2$.

However, their sum
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} + \begin{bmatrix}
-1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
0 \\
1
\end{bmatrix}\] is not in $S_3$ since $1\neq 0^2$.
Hence condition 2 is not met, and thus $S_3$ is not a subspace of $\R^2$.
(You can check that condition 1 is fulfilled yet condition 3 is not met.)
  

Solution (4). $S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}$

Consider the polynomial $f(x)=x$. Since the degree of $f(x)$ is $1$ and $f(1)=1$ is an integer, it is in $S_4$. Consider the scalar product of $f(x)$ and the scalar $1/2\in \R$.
Then we evaluate the scalar product at $x=1$ and we have
\begin{align*}
\frac{1}{2}f(1)=\frac{1}{2},
\end{align*}
which is not an integer.
Thus $(1/2)f(x)$ is not in $S_4$, hence condition 3 is not met. Thus $S_4$ is not a subspace of $P_4$.
(You can check that conditions 1, 2 are met.)
  

Solution (5). $S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}$

Let $f(x)=x$. Then $f(x)$ is a degree $1$ polynomial and $f(1)=1$ is a rational number.
However, the scalar product $\sqrt{2} f(x)$ of $f(x)$ and the scalar $\sqrt{2} \in \R$ is not in $S_5$ since
\[\sqrt{2}f(1)=\sqrt{2},\] which is not a rational number. Hence condition 3 is not met and $S_5$ is not a subspace of $P_4$.
(You can check that conditions 1, 2 are met.)
  

Solution (6). $S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\}$

The zero vector of the vector space $M_{2 \times 2}$ is the $2\times 2$ zero matrix $O$.
Since the determinant of the zero matrix $O$ is $0$, it is not in $S_6$. Thus, condition 1 is not met and $S_6$ is not a subspace of $M_{2 \times 2}$.
(You can check that conditions 2, 3 are not met as well.)
  

Solution (7). $S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\}$

Consider the matrices
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.\] The determinants of $A$ and $B$ are both $0$, hence they belong to $S_7$.
However, their sum
\[A+B=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\] has the determinant $1$, hence the sum $A+B$ is not in $S_7$.
So condition 2 is not met and $S_7$ is not a subspace of $M_{2 \times 2}$.
(You can check that conditions 1, 3 are met.)
  

Solution (8). $S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\}$

Consider the continuous functions
\[f(x)=x-1 \text{ and } g(x)=x+1.\] (These are polynomials, hence they are continuous.)
We have
\begin{align*}
&f(-1)f(1)=(-2)\cdot(0)=0 \text{ and }\\
&g(-1)g(1)=(0)\cdot 2=0.
\end{align*}
So these functions are in $S_8$.

However, their sum $h(x):=f(x)+g(x)$ does not belong to $S_8$ since we have
\begin{align*}
h(-1)h(1)&=\big(f(-1)+g(-1)\big) \big(f(1)+g(1) \big)\\
&=(-2+0)(0+2)=-4\neq 0.
\end{align*}
Therefore, condition 2 is not met and $S_8$ is not a subspace of $C[-1, 1]$.
(You can check that conditions 1, 3 are met.)
  

Solution (9). $S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}$

Let $f(x)=x^2$, an open-up parabola.
Then $f(x)$ is continuous and non-negative for $-1 \leq x \leq 1$. Hence $f(x)=x^2$ is in $S_9$.
However, the scalar product $(-1)f(x)$ of $f(x)$ and the scalar $-1$ is not in $S_9$ since, say,
\[(-1)f(1)=-1\] is negative.
So condition 3 is not met and $S_9$ is not a subspace of $C[-1, 1]$.
(You can check that conditions 1, 2 are met.)
  

Solution (10). $S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}$

The zero vector of the vector space $C^2[-1, 1]$ is the zero function $\theta(x)=0$.
The second derivative of the zero function is still the zero function.
Thus,
\[\theta^{\prime\prime}(x)+\theta(x)=0\] and since $\sin(x)$ is not the zero function, $\theta(x)$ is not in $S_{10}$.
Hence $S_{10}$ is not a subspace of $C^2[-1, 1]$.

(You can check that conditions 2, 3 are not met as well.
For example, consider the function $f(x)=-\frac{1}{2}x\cos(x)\in S_{10}$.)

Solution (11). Let $S_{11}$ be the set of real polynomials of degree exactly $k$.

The set $S_{11}$ is not a vector subspace of $\mathbf{P}_k$. One reason is that the zero function $\mathbf{0}$ has degree $0$, and so does not lie in $S_{11}$. The set $S_{11}$ is also not closed under addition. Consider the two polynomials $f(x) = x^k + 1$ and $g(x) = – x^k + 1$. Both of these polynomials lie in $S_{11}$, however $f(x) + g(x) = 2$ has degree $0$ and so does not lie in $S_{11}$.

Solution (12). The complement

The complement $S_{12}= V \setminus W$ is not a vector subspace. Specifically, if $\mathbf{0} \in V$ is the zero vector, then we know $\mathbf{0} \in W$ because $W$ is a subspace. But then $\mathbf{0} \not\in V \setminus W$, and so $V \setminus W$ cannot be a vector subspace.

Linear Algebra Midterm Exam 2 Problems and Solutions

• True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
• Problem 1 and its solution (current problem): See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
• Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
• Problem 3 and its solution: Orthonormal basis of null space and row space
• Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
• Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
• Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
• Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
• Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

The post 12 Examples of Subsets that Are Not Subspaces of Vector Spaces first appeared on Problems in Mathematics.

Determine whether the following is true or false. If it is true, then give a proof. If it is false, then give a counterexample.

Let $W_1$ and $W_2$ be subspaces of the vector space $\R^n$.
If $B_1$ and $B_2$ are bases for $W_1$ and $W_2$, respectively, then $B_1\cap B_2$ is a basis of the subspace $W_1\cap W_2$.

Solution.

The statement is false. We give a counterexample.
Let us consider the vector space $\R^2$, the plane.

Let $W_1$ and $W_2$ be $\R^2$ itself. Then they are subspaces of $\R^2$.
Let
\[B_1=\left\{\begin{bmatrix}
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1
\end{bmatrix} \right\}\text{ and }
B_2=\left\{ \begin{bmatrix}
-1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
-1
\end{bmatrix} \right\}\] be bases of $W_1$ and $W_2$, respectively.

(Note. $B_1$ consists of two linearly independent vectors in the 2-dimensional vector space $W_1=\R^2$, hence $B_1$ is a basis of $W_1$. Similarly, $B_2$ is a basis of $W_2$.)

Since $W_1$ and $W_2$ are both $\R^2$, we have $W_1\cap W_2=\R^2$.
However, the intersection $B_1\cap B_2$ is the empty set, and the empty set is not a basis of $W_1\cap W_2=\R^2$. Thus, we have found a counterexample to the statement.

The post True or False. The Intersection of Bases is a Basis of the Intersection of Subspaces first appeared on Problems in Mathematics.

Let $A$ and $B$ be $n\times n$ matrices, where $n$ is an integer greater than $1$.

Is it true that
\[\det(A+B)=\det(A)+\det(B)?\] If so, then give a proof. If not, then give a counterexample.
 

Solution.

We claim that the statement is false.
As a counterexample, consider the matrices
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.\] Then
\[A+B=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\] and we have
\[\det(A+B)=\begin{vmatrix}
1 & 0\\
0& 1
\end{vmatrix}=1.\]

On the other hand, the determinants of $A$ and $B$ are
\[\det(A)=0 \text{ and } \det(B)=0,\] and hence
\[\det(A)+\det(B)=0\neq 1=\det(A+B).\]

Therefore, the statement is false and in general we have
\[\det(A+B)\neq \det(A)+\det(B).\]

Remark.

When we computed the $2\times 2$ matrices, we used the formula
\[\begin{vmatrix}
a & b\\
c& d
\end{vmatrix}=ad-bc.\]

This problem showed that the determinant does not preserve the addition.
However, the determinant is multiplicative.
In general, the following is true:
\[\det(AB)=\det(A)\det(B).\] Click here if solved 22

The post Is the Determinant of a Matrix Additive? 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.

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

(a) If $A$ is a nilpotent $n \times n$ matrix and $B$ is an $n\times n$ matrix such that $AB=BA$. Show that the product $AB$ is nilpotent.

(b) Let $P$ be an invertible $n \times n$ matrix and let $N$ be a nilpotent $n\times n$ matrix. Is the product $PN$ nilpotent? If so, prove it. If not, give a counterexample.

Hint.

For (b), the statement is false. Try to find a counter example.
A typical nilpotent matrix is an upper triangular matrix whose diagonal entries are all zero.

Proof.

(a) Show that $AB$ is nilpotent

Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$. Then we have
\[(AB)^k=(AB)(AB)\cdots (AB)=A^kB^k=OB^k=O.\]

Here in the second equality, we used the assumption that $AB=BA$.
Thus we have $(AB)^k=O$, hence the product matrix $AB$ is nilpotent.

(b) Is $PN$ nilpotent?

In general, the product $PN$ of an invertible matrix $P$ and a nilpotent matrix $N$ is not nilpotent.
Here is a counterexample.
Let
\[P=\begin{bmatrix}
1 & 0 & 0 \\
1 &1 &0 \\
0 & 0 & 1
\end{bmatrix} \text{ and }
N=\begin{bmatrix}
0 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}.\]

Then the matrix $P$ is invertible since $\det(P)=1$.
(Note that $P$ is a lower triangular matrix. So the determinant is the product of diagonal entries.)

Also, it is easy to see by direct computation that $N^3=O$, hence $N$ is nilpotent. Indeed,
\[N^2=\begin{bmatrix}
0 & 0 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix} \] and
\[
N^3=N^2N=\begin{bmatrix}
0 & 0 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}=O.\]

Now the product $PN$ is
\[PN=\begin{bmatrix}
0 & 1 & 1 \\
0 &1 &2 \\
0 & 0 & 0
\end{bmatrix}.\] We show that $PN$ is not nilpotent.
We have
\[(PN)^2=\begin{bmatrix}
0 & 1 & 2 \\
0 &1 &2 \\
0 & 0 & 0
\end{bmatrix}\] \[(PN)^3=(PN)^2(PN)=\begin{bmatrix}
0 & 1 & 2 \\
0 &1 &2 \\
0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
0 & 1 & 1 \\
0 &1 &2 \\
0 & 0 & 0
\end{bmatrix}
=\begin{bmatrix}
0 & 1 & 2 \\
0 &1 &2 \\
0 & 0 & 0
\end{bmatrix}.\]

This calculation shows that
\[(PN)^k=\begin{bmatrix}
0 & 1 & 2 \\
0 &1 &2 \\
0 & 0 & 0
\end{bmatrix}\neq O \text{ for all } k \geq 2.\]

Thus $PN$ is not nilpotent. In conclusion, the product $PN$ of the invertible matrix $P$ and the nilpotent matrix $N$ is not nilpotent.

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