Find all $2\times 2$ symmetric matrices $A$ satisfying $A\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=
\begin{bmatrix}
2 \\
3
\end{bmatrix}$? Express your solution using free variable(s).

Solution.

Let $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ be a $2\times 2$ matrix satisfying the conditions. Then as $A$ is symmetric, we have $A^{\trans}=A$. This yields that $b=c$.
So, we find all matrices $A=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}$ satisfying $A\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=
\begin{bmatrix}
2 \\
3
\end{bmatrix}$.
We have
\begin{align*}
\begin{bmatrix}
2 \\
3
\end{bmatrix}=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=\begin{bmatrix}
a-b \\
b-d
\end{bmatrix}.
\end{align*}
Hence, we need $a-b=2$ and $b-d=3$.
Equivalently, $a=b+2, d=b-3$. So, we have
\begin{align*}
A=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}=\begin{bmatrix}
b+2 & b\\
b& b-3
\end{bmatrix}=b\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}+\begin{bmatrix}
2 & 0\\
0& -3
\end{bmatrix},
\end{align*}
where $b$ is a free variable.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is not using the assumption that $A$ is symmetric or using wrongly.
A matrix $A$ is symmetric if $A^{\trans}=A$. For a 2 by 2 matrix, this yields that the off-diagonal entries must be the same.
However, note that the diagonal entries can be distinct. Some students assumed the same diagonal entries and concluded that there are no matrices satisfying the conditions.

The post Find All Symmetric Matrices satisfying the Equation first appeared on Problems in Mathematics.

Recall that a matrix $A$ is symmetric if $A^\trans = A$, where $A^\trans$ is the transpose of $A$.

Is it true that if $A$ is a symmetric matrix and in reduced row echelon form, then $A$ is diagonal? If so, prove it.

Otherwise, provide a counterexample.

Proof.

This is true.

If $A$ is in reduced row echelon form, then every term below the diagonal must be 0.
That is, the entry $a_{i j} = 0$ for all $i > j$.

If $A$ is, additionally, symmetric, then for $i < j$ we also have $a_{j i} = a_{i j} = 0$.

These two facts together means that $a_{i j} = 0$ whenever $i \neq j$.

In this case, the only non-zero terms in the matrix $A$ lie on the diagonal, and so $A$ is a diagonal matrix.

The post If a Symmetric Matrix is in Reduced Row Echelon Form, then Is it Diagonal? first appeared on Problems in Mathematics.

Let $\mathbf{v}$ be an $n \times 1$ column vector.

Prove that $\mathbf{v} \mathbf{v}^\trans$ is a symmetric matrix.

Definition (Symmetric Matrix).

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

In terms of entries, an $n\times n$ matrix $A=(a_{ij})$ is symmetric if $a_{ij}=a_{ji}$ for all $1 \leq i, j \leq n$.

Proof.

Let $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}$. Then we have
\begin{align*}
\mathbf{v} \mathbf{v}^\trans = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}\begin{bmatrix}
v_1 & v_2 & \cdots & v_n
\end{bmatrix}=
\begin{bmatrix} v_1 v_1 & v_1 v_2 & \cdots & v_1 v_n \\ v_2 v_1 & v_2 v_2 & \cdots & v_2 v_n \\ \vdots & \vdots & \vdots & \vdots \\ v_n v_1 & v_n v_2 & \cdots & v_n v_n \end{bmatrix}.
\end{align*}

In particular, the the $(i, j)$-th component is
\[(\mathbf{v} \mathbf{v}^\trans)_{i j} = v_i v_j = v_j v_i = (\mathbf{v} \mathbf{v}^\trans)_{j i}.\] This shows that the matrix $\mathbf{v} \mathbf{v}^\trans$ is symmetric.

The post Prove that $\mathbf{v} \mathbf{v}^\trans$ is a Symmetric Matrix for any Vector $\mathbf{v}$ first appeared on Problems in Mathematics.

Let $A$ be a $2\times 2$ real symmetric matrix.
Prove that all the eigenvalues of $A$ are real numbers by considering the characteristic polynomial of $A$.

Proof.

Let $A=\begin{bmatrix}
a& b \\
c& d
\end{bmatrix}$.
Then as $A$ is a symmetric matrix, we have $A^{\trans}=A$.
This implies that
\[\begin{bmatrix}
a& c \\
b& d
\end{bmatrix}=\begin{bmatrix}
a& b \\
c& d
\end{bmatrix}.\] Hence we have $b=c$ by comparing entries.

Now, we find the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
p(t)&=\det(A-t I)=\begin{vmatrix}
a-t & b\\
b& d-t
\end{vmatrix}\\[6pt] &=(a-t)(d-t)-b^2\\
&=t^2-(a+d)t+ad-b^2.
\end{align*}

Note that the eigenvalues of $A$ are roots of the characteristic polynomial $p(t)$. Hence, it suffices to show that the roots of $p(t)$ are real numbers.
The quadratic polynomial has only real roots if and only if its discriminant is non-negative.
The discriminant of $p(t)$ is given by
\begin{align*}
(a+d)^2-4(ad-b^2)&=a^2+2ad+d^2-4ad+4b^2\\
&=a^2-2ad+d^2+4b^2\\
&=(a-d)^2+4b^2. \end{align*}
Observe that the last expression is the sum of two squares of real numbers. Hence the discriminant of $p(t)$ is nonnegative.

We conclude that every $2\times 2$ symmetric matrix has only real eigenvalues.

Remark

We also could find the eigenvalues directly. By the quadratic formula, the eigenvalues of $A$ are
\[\frac{a+d\pm\sqrt{(a+d)^2-4(ad-b^2)}}{2}=\frac{a+d\pm \sqrt{(a-d)^2+4b^2}}{2}\] and as the number inside the square root (discriminant) is positive, we conclude that the eigenvalues are real.

The post Eigenvalues of \times 2$ Symmetric Matrices are Real by Considering Characteristic Polynomials first appeared on Problems in Mathematics.

Let $A$ be a real symmetric matrix whose diagonal entries are all positive real numbers.

Is it true that the all of the diagonal entries of the inverse matrix $A^{-1}$ are also positive?
If so, prove it. Otherwise, give a counterexample.

Solution.

The statement is in general false. We give a counterexample.

Let us consider the following $2\times 2$ matrix:
\[A=\begin{bmatrix}
1 & 2\\
2& 1
\end{bmatrix}.\] The matrix $A$ satisfies the required conditions, that is, $A$ is symmetric and its diagonal entries are positive.

The determinant $\det(A)=(1)(1)-(2)(2)=-3$ and the inverse of $A$ is given by
\[A^{-1}=\frac{1}{-3}\begin{bmatrix}
1 & -2\\
-2& 1
\end{bmatrix}=\begin{bmatrix}
-1/3 & 2/3\\
2/3& -1/3
\end{bmatrix}\] by the formula for the inverse matrix for $2\times 2$ matrices.

This shows that the diagonal entries of the inverse matrix $A^{-1}$ are negative.

The post The Inverse Matrix of a Symmetric Matrix whose Diagonal Entries are All Positive first appeared on Problems in Mathematics.

Let $V$ be the vector space over $\R$ of all real $2\times 2$ matrices.
Let $W$ be the subset of $V$ consisting of all symmetric matrices.

(a) Prove that $W$ is a subspace of $V$.

(b) Find a basis of $W$.

(c) Determine the dimension of $W$.

Proof.

Recall that $A$ is symmetric if $A^{\trans}=A$.

(a) Prove that $W$ is a subspace of $V$.

We verify the following subspace criteria:

The zero vector in $V$ is the $2\times 2$ zero matrix $O$.
It is clear that $O^{\trans}=O$, and hence $O$ is symmetric.
Thus $O\in W$ and condition 1 is met.

Let $A, B$ be arbitrary elements in $W$.
That is, $A$ and $B$ are symmetric matrices.
We show that the sum $A+B$ is also symmetric.
We have
\begin{align*}
(A+B)^{\trans}=A^{\trans}+B^{\trans}=A+B.
\end{align*}
The second equality follows as $A, B$ are symmetric.
Hence $A+B$ is symmetric and $A+B\in W$.
Condition 2 is met.

To check condition 3, let $A\in W$ and $r\in \R$.
We have
\begin{align*}
(rA)^{\trans}=rA^{\trans}=rA,
\end{align*}
where the second equality follows since $A$ is symmetric.
This implies that $rA$ is symmetric, and hence $rA\in W$.
So condition 3 is met, and we conclude that $W$ is a subspace of $V$ by subspace criteria.

(b) Find a basis of $W$.

Let
\[A=\begin{bmatrix}
a_{11} & a_{12}\\
a_{21}& a_{22}
\end{bmatrix}\] be an arbitrary element in the subspace $W$.
Then since $A^{\trans}=A$, we have
\[\begin{bmatrix}
a_{11} & a_{21}\\
a_{12}& a_{22}
\end{bmatrix}=\begin{bmatrix}
a_{11} & a_{12}\\
a_{21}& a_{22}
\end{bmatrix}.\] This implies that $a_{12}=a_{21}$, and hence
\begin{align*}
A&=\begin{bmatrix}
a_{11} & a_{12}\\
a_{12}& a_{22}
\end{bmatrix}\\[6pt] &=a_{11}\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}+a_{12}\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}+a_{22}\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.
\end{align*}

Let $B=\{v_1, v_2, v_3\}$, where $v_1, v_2, v_3$ are $2\times 2$ matrices appearing in the above linear combination of $A$.
Note that these matrices are symmetric.
Hence we showed that any element in $W$ is a linear combination of matrices in $B$.
Thus $B$ is a spanning set for the subspace $W$.

We show that $B$ is linearly independent.
Suppose that we have
\[c_1v_1+c_2v_2+c_3v_3=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}.\] Then it follows that
\[\begin{bmatrix}
c_1 & c_2\\
c_2& c_3
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}.\] Thus $c_1=c_2=c_3=$ and the set $B$ is linearly independent.

As $B$ is a linearly independent spanning set, we conclude that $B$ is a basis for the subspace $W$.

(c) Determine the dimension of $W$.

Recall that the dimension of a subspace is the number of vectors in a basis of the subspace.

In part (b), we found that $B=\{v_1, v_2, v_3\}$ is a basis for the subspace $W$.
As $B$ consists of three vectors, the dimension of $W$ is $3$.

Related Question (Skew-Symmetric Matrices)

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

The solution is given in the post ↴
Subspace of Skew-Symmetric Matrices and Its Dimension

The post The set of \times 2$ Symmetric Matrices is a Subspace first appeared on Problems in Mathematics.

The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).
The time limit was 55 minutes.

This post is Part 3 and contains Problem 7, 8, and 9.
Check out Part 1 and Part 2 for the rest of the exam problems.

Problem 7. Let $A=\begin{bmatrix}
-3 & -4\\
8& 9
\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}
-1 \\
2
\end{bmatrix}$.

(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.

(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.

Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.

Problem 9.
Determine whether each of the following sentences is true or false.

(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.

(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.

(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.

(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.

(e) The vectors
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\] are linearly independent.

Solution of Problem 7.

(a) Calculate $A\mathbf{v}$.

We calculate
\[A\mathbf{v}=\begin{bmatrix}
-3 & -4\\
8& 9
\end{bmatrix}\begin{bmatrix}
-1 \\
2
\end{bmatrix}=\begin{bmatrix}
-5 \\
10
\end{bmatrix}=5\begin{bmatrix}
-1 \\
2
\end{bmatrix}.\] Thus we see that $\lambda=5$.

(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.

From part (a), we know that $A\mathbf{v}=5\mathbf{v}$.
Using the properties of matrix multiplication, we compute
\begin{align*}
A^3\mathbf{v}&=A^2(A\mathbf{v})=A^2(5\mathbf{v})=5A^2\mathbf{v}=5A(A\mathbf{v})=5A(5\mathbf{v})\\
&=5^2A\mathbf{v}=5^2(5\mathbf{v})=5^3\mathbf{v}\\
&=5^3\begin{bmatrix}
-1 \\
2
\end{bmatrix}=\begin{bmatrix}
-125 \\
250
\end{bmatrix}.
\end{align*}
Hence we obtain $A^3\mathbf{v}=\begin{bmatrix}
-125 \\
250
\end{bmatrix}$.

Proof of Problem 8.

Let $A$ and $B$ be nonsingular matrices.
Suppose that $(AB)\mathbf{v}=\mathbf{0}$ for some $n$-dimensional vector $\mathbf{v}$.
Then we have
\[A(B\mathbf{v})=\mathbf{0}.\] It follows that the vector $B\mathbf{v}$ is a solution of $A\mathbf{x}=\mathbf{0}$.
As the matrix $A$ is nonsingular, any solution must be the zero vector.
Hence we have $B\mathbf{v}=\mathbf{0}$.

This equation says that the vector $\mathbf{v}$ is a solution of $B\mathbf{x}=\mathbf{0}$.
As the matrix $B$ is nonsingular, any solution must be the zero vector.
This implies that $\mathbf{v}=\mathbf{0}$.

This proves that if $(AB)\mathbf{v}=\mathbf{0}$, then $\mathbf{v}=\mathbf{0}$.
This is equivalent to say that the matrix $AB$ is nonsingular by definition.

Solution of Problem 9.

True or False: (a) There is a $3\times 3$ homogeneous system that has exactly three solutions.

False. Every system of linear equations has no solution at all or it has either one or infinitely many solutions.

True or False: (b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.

True. Since $A$ and $B$ are symmetric, we have $A^{\trans}=A$ and $B^{\trans}=B$.
It follows that
\[(A+B)^{\trans}=A^{\trans}+B^{\trans}=A+B.\] Thus the sum $A+B$ is symmetric.

True or False: (c) If $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is linearly dependent.

True. Since the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, there exists scalars $c_1, c_2, c_3$, not all of them are zero, such that
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}.\] Let $\mathbf{v}_4$ be any $n$-dimensional vector.
Then we have the linear combination
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3+0\mathbf{v}_4=\mathbf{0}\] whose coefficient is not trivial as at least one of $c_1, c_2, c_3$ is nonzero.
This implies that the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent.

True or False: (d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.

False. The system could be consistent even though the coefficient matrix is singular.
For example, consider the system
\begin{align*}
x_1+2x_2&=3\\
2x_1+4x_2&=6.
\end{align*}
The coefficient matrix of the system is $A=\begin{bmatrix}
1 & 2\\
2& 4
\end{bmatrix}$.
This is singular since, for example, it is row equivalent to $\begin{bmatrix}
1 & 2\\
0& 0
\end{bmatrix}$.
However, the system has a solution, for example, $x_1=1, x_2=1$.
Hence the system is consistent even though its coefficient matrix is singular.

True or False: (e) The vectors
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\] are linearly independent.

True. Consider the linear combination
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}\] for some scalars $c_1, c_2, c_3$.
Then this can be written as
\[A\begin{bmatrix}
c_1 \\
c_2 \\
c_3
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix},\] where
\[A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
1 & 0 & 1
\end{bmatrix}.\] So the scalars $c_1, c_2, c_3$ is a solution of the system $A\mathbf{x}=\mathbf{0}$.

The augmented matrix of the system is
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
1 & 0 & 1 & 0
\end{array} \right] \xrightarrow{R_3-R_1}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array} \right].
\end{align*}
Thus the solution is $c_1=0, c_2=0, c_3=0$.
It follows that the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent.

Go to Part 1 and Part 2

Go to Part 1 for Problem 1, 2, and 3.

Go to Part 2 for Problem 4, 5, and 6.

The post Linear Algebra Midterm 1 at the Ohio State University (3/3) first appeared on Problems in Mathematics.

Let $A$ and $B$ be $n\times n$ skew-symmetric matrices. Namely $A^{\trans}=-A$ and $B^{\trans}=-B$.

(a) Prove that $A+B$ is skew-symmetric.

(b) Prove that $cA$ is skew-symmetric for any scalar $c$.

(c) Let $P$ be an $m\times n$ matrix. Prove that $P^{\trans}AP$ is skew-symmetric.

(d) Suppose that $A$ is real skew-symmetric. Prove that $iA$ is an Hermitian matrix.

(e) Prove that if $AB=-BA$, then $AB$ is a skew-symmetric matrix.

(f) Let $\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\mathbf{v}^{\trans}A\mathbf{v}=0$.

(g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\mathbf{v}=\mathbf{0}$ for some vector $\mathbf{v}\in \R^n$. Then prove that $A\mathbf{v}=\mathbf{0}$.

Proof.

(a) Prove that $A+B$ is skew-symmetric.

We have
\begin{align*}
(A+B)^{\trans}=A^{\trans}+B^{\trans}=(-A)+(-B)=-(A+B).
\end{align*}
Hence $A+B$ is skew-symmetric.

(b) Prove that $cA$ is skew-symmetric for any scalar $c$.

We compute
\begin{align*}
(cA)^{\trans}=cA^{\trans}=c(-A)=-cA.
\end{align*}
Thus, $cA$ is skew-symmetric.

(c) Let $P$ be an $m\times n$ matrix. Prove that $P^{\trans}AP$ is skew-symmetric.

Using the properties of transpose, we have
\begin{align*}
(P^{\trans}AP)^{\trans}&=P^{\trans}A^{\trans}(P^{\trans})^{\trans}=P^{\trans}A^{\trans}P\\
&=P^{\trans}(-A)P=-(P^{\trans}AP).
\end{align*}
This implies that $P^{\trans}AP$ is skew-symmetric.

(d) Suppose that $A$ is real skew-symmetric. Prove that $iA$ is an Hermitian matrix.

Note that since $A$ is real, we have $\bar{A}=A$.
Then we have
\begin{align*}
(\overline{iA})^{\trans}=(\bar{i}\bar{A})^{\trans}=(-iA)^{\trans}=(-i)A^{\trans}=(-i)(-A)=iA.
\end{align*}
It follows that $iA$ is Hermitian.

(e) Prove that if $AB=-BA$, then $AB$ is a skew-symmetric matrix.

We calculate
\begin{align*}
(AB)^{\trans}&=B^{\trans}A^{\trans}=(-B)(-A)\\
&=BA=-AB,
\end{align*}
where the last step follows from the assumption $AB=-BA$.
This proves that $AB$ is skew-symmetric.

(f) Let $\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\mathbf{v}^{\trans}A\mathbf{v}=0$.

Observe that $\mathbf{v}^{\trans}A\mathbf{v}$ is a $1\times 1$ matrix, or just a number.
So we have
\begin{align*}
\mathbf{v}^{\trans}A\mathbf{v}&=(\mathbf{v}^{\trans}A\mathbf{v})^{\trans}=\mathbf{v}^{\trans}A^{\trans}(\mathbf{v}^{\trans})^{\trans}\\
&=\mathbf{v}^{\trans}A^{\trans}\mathbf{v}=\mathbf{v}^{\trans}(-A)\mathbf{v}=-(\mathbf{v}^{\trans}A\mathbf{v}).
\end{align*}
This yields that $2\mathbf{v}^{\trans}A\mathbf{v}=0$, and hence $\mathbf{v}^{\trans}A\mathbf{v}=0$.

(g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\mathbf{v}=\mathbf{0}$ for some vector $\mathbf{v}\in \R^n$. Then prove that $A\mathbf{v}=\mathbf{0}$.

Let us compute the length of the vector $A\mathbf{v}$.
We have
\begin{align*}
\|A\mathbf{v}\|&=(A\mathbf{v})^{\trans}(A\mathbf{v})=\mathbf{v}^{\trans}A^{\trans}A\mathbf{v}\\
&=\mathbf{v}^{\trans}(-A)A\mathbf{v}=-\mathbf{v}^{\trans}A^2\mathbf{v}\\
&=-\mathbf{v}\mathbf{0} &&\text{by assumption}\\
&=0.
\end{align*}
Since the length $\|A\mathbf{v}\|=0$, we conclude that $A\mathbf{v}=\mathbf{0}$.

The post 7 Problems on Skew-Symmetric Matrices first appeared on Problems in Mathematics.

Let $\mathbf{v}$ be a nonzero vector in $\R^n$.
Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$.
Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by
\[A=I-a\mathbf{v}\mathbf{v}^{\trans},\] where $I$ is the $n\times n$ identity matrix.

Prove that $A$ is a symmetric matrix and $AA=I$.
Conclude that the inverse matrix is $A^{-1}=A$.

Proof.

$A$ is symmetric

We first show that the matrix $A$ is symmetric.
We calculate using properties of transpose
\begin{align*}
A^{\trans}&=(I-a\mathbf{v}\mathbf{v}^{\trans})^{\trans} && \text{by definition of $A$}\\
&=I^{\trans}-(a\mathbf{v}\mathbf{v}^{\trans})^{\trans}\\
&=I-a(\mathbf{v}^{\trans})^{\trans}\mathbf{v}^{\trans}\\
&=I-a\mathbf{v}\mathbf{v}^{\trans}\\
&=A && \text{by definition of $A$}.
\end{align*}
Hence we have $A^{\trans}=A$, and thus $A$ is symmetric.

$AA=I$ and $A^{-1}=A$

Next, we prove that $AA=I$.

We compute
\begin{align*}
AA&=(I-a\mathbf{v}\mathbf{v}^{\trans})(I-a\mathbf{v}\mathbf{v}^{\trans})\\
&=I(I-a\mathbf{v}\mathbf{v}^{\trans})-a\mathbf{v}\mathbf{v}^{\trans}(I-a\mathbf{v}\mathbf{v}^{\trans})\\
&=I-a\mathbf{v}\mathbf{v}^{\trans}-a\mathbf{v}\mathbf{v}^{\trans}+a^2\mathbf{v}\mathbf{v}^{\trans}\mathbf{v}\mathbf{v}^{\trans}\\
&=I-2a\mathbf{v}\mathbf{v}^{\trans}+a^2\mathbf{v}\mathbf{v}^{\trans}\mathbf{v}\mathbf{v}^{\trans}. \tag{*}
\end{align*}

Note that we have
\begin{align*}
\mathbf{v}\mathbf{v}^{\trans}\mathbf{v}\mathbf{v}^{\trans}&=\mathbf{v}(\mathbf{v}^{\trans}\mathbf{v})\mathbf{v}^{\trans}\\
&=\mathbf{v}\left(\, \frac{2}{a} \,\right)\mathbf{v}^{\trans} &&\text{by definition of $a\neq 0$}\\
&=\frac{2}{a}\mathbf{v}\mathbf{v}^{\trans}.
\end{align*}

Plugging this relation into (*), we obtain
\begin{align*}
AA&=I-2a\mathbf{v}\mathbf{v}^{\trans}+a^2\frac{2}{a}\mathbf{v}\mathbf{v}^{\trans}=I.
\end{align*}
Thus we get $AA=I$.
This implies that the inverse matrix of $A$ is $A$ itself: $A^{-1}=A$.

The post Construction of a Symmetric Matrix whose Inverse Matrix is Itself first appeared on Problems in Mathematics.

(a) Suppose that $A$ is an $n\times n$ real symmetric positive definite matrix.
Prove that
\[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}\] defines an inner product on the vector space $\R^n$.

(b) Let $A$ be an $n\times n$ real matrix. Suppose that
\[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}\] defines an inner product on the vector space $\R^n$.

Prove that $A$ is symmetric and positive definite.

Definitions.

Inner Product on a Real Vector Space

Let $V$ be a real vector space. An inner product on $V$ is a function that assigns a real number $\langle \mathbf{u}, \mathbf{v}\rangle$ to each pair of vectors $\mathbf{u}$ and $\mathbf{v}$ in $V$ satisfying the following properties.
For any vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ and a real number $r\in \R$,

A Positive-Definite Matrix

A real symmetric $n\times n$ matrix $A$ is called positive definite if $\mathbf{x}^{\trans}A\mathbf{x} > 0$ for each nonzero vector $\mathbf{x}\in \R^n$.

Proof.

(a) If $A$ is positive definite, then $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ defines an inner product

First of all, note that we can write
\[\langle \mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^{\trans}A\mathbf{y}=\mathbf{x}\cdot (A\mathbf{y}),\] where the “dot” is the dot product of $\R^n$.
Thus, $\langle \mathbf{x}, \mathbf{y}\rangle$ is a real number.

We verify the three properties of an inner product.
Since the dot product is commutative, we have
\begin{align*}
\langle \mathbf{x}, \mathbf{y}\rangle&=\mathbf{x}\cdot (A\mathbf{y})= (A\mathbf{y})\cdot \mathbf{x}\\
&=(A\mathbf{y})^{\trans}\mathbf{x}=\mathbf{y}^{\trans}A^{\trans}\mathbf{x}\\
&=\mathbf{y}^{\trans}A\mathbf{x} &&\text{since $A$ is symmetric}\\
&=\langle \mathbf{y}, \mathbf{x} \rangle.
\end{align*}
Thus, the function $\langle\,,\,\rangle$ is symmetric.

Next, for any vectors $\mathbf{x}, \mathbf{y}, \mathbf{z}$ and any real number $r$, we have
\begin{align*}
\langle r\mathbf{x}, \mathbf{y}\rangle &=(r\mathbf{x})^{\trans}A\mathbf{y}=r\mathbf{x}^{\trans}A\mathbf{y}=r\langle \mathbf{x}, \mathbf{y}\rangle
\end{align*}
and
\begin{align*}
\langle \mathbf{x}+\mathbf{y}, \mathbf{z}\rangle &=(\mathbf{x}+\mathbf{y})^{\trans}A\mathbf{z}=(\mathbf{x}^{\trans}+\mathbf{y}^{\trans})A\mathbf{z}\\
&=\mathbf{x}^{\trans}A\mathbf{z}+\mathbf{y}^{\trans}A\mathbf{z}=\langle \mathbf{x}, \mathbf{z}\rangle+\langle \mathbf{y}, \mathbf{z}\rangle.
\end{align*}
Thus, the linearity in the first argument is satisfied.

If $\mathbf{x}$ is a nonzero vector in $\R^n$, then we have
\begin{align*}
\langle \mathbf{x}, \mathbf{x}\rangle=\mathbf{x}^{\trans}A\mathbf{x} > 0
\end{align*}
since $A$ is positive definite.
We also have
\begin{align*}
\langle \mathbf{0}, \mathbf{0}\rangle=\mathbf{0}^{\trans}A\mathbf{0}=0.
\end{align*}
It follows that $\langle \mathbf{x}, \mathbf{x}\rangle \geq 0$ for any vector $\mathbf{x}\in \R^n$.

Suppose that $\langle \mathbf{x}, \mathbf{x}\rangle=0$.
Then we have
\begin{align*}
\langle \mathbf{x}, \mathbf{x}\rangle=\mathbf{x}^{\trans}A\mathbf{x}=0.
\end{align*}
Since $A$ is positive definite, this happens if and only if $\mathbf{x}=\mathbf{0}$.
Hence $\langle \mathbf{x}, \mathbf{x}\rangle=0$ if and only if $\mathbf{x}=0$.
This proves the positive-definiteness of the function $\langle\,,\,\rangle$.

This completes the verification of the three properties, and hence $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ defines an inner product on $\R^n$.

(b) If $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ defines an inner product, then $A$ is symmetric positive definite

Suppose that $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ is an inner product on $\R^n$.
Let us write $A=(a_{ij})$.

We first prove that $A$ is symmetric.
Let $\mathbf{e}_i$ denote the $i$-th standard unit vectors, that is,
\[\mathbf{e}_i=\begin{bmatrix}
0 \\
\vdots \\
1 \\
\vdots \\
0
\end{bmatrix},\] where $1$ is on the $i$-th columns and other entries are all zero.

We compute
\begin{align*}
\langle \mathbf{e}_i, \mathbf{e}_j\rangle &=\mathbf{e}^{\trans}_iA\mathbf{e}_j\\
&=\begin{bmatrix}
0 & \dots &1 & \dots 0
\end{bmatrix}
\begin{bmatrix}
a_{1 j} \\
a_{2 j} \\
\vdots \\
a_{n j}
\end{bmatrix}=a_{ij}.
\end{align*}
Similarly, $\langle \mathbf{e}_j, \mathbf{e}_i\rangle =a_{j i}$. By symmetry of the inner product, it yields that
\[a_{ij}=\langle \mathbf{e}_i, \mathbf{e}_j\rangle =\langle \mathbf{e}_j, \mathbf{e}_i\rangle =a_{j i}.\] Therefore, the matrix $A$ is symmetric.

Next, we show that $A$ is positive definite.
Let $\mathbf{x}$ be a nonzero vector in $\R^n$.
Then we have
\[\mathbf{x}^{\trans}A\mathbf{x}=\langle \mathbf{x}, \mathbf{x}\rangle > 0\] by positive-definiteness property of the inner product.
This proves that $A$ is a positive definite matrix.

Related Question.

A concrete example of a positive-definite matrix is given in the next problem.

See the post ↴
The Inner Product on $\R^2$ induced by a Positive Definite Matrix and Gram-Schmidt Orthogonalization
for proofs.

The post A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space first appeared on Problems in Mathematics.