An $n\times n$ matrix $A$ is called orthogonal if $A^{\trans}A=I$.
Let $V$ be the vector space of all real $2\times 2$ matrices.

Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$.

Solution.

We claim that $W$ is not a subspace of $V$.

One way to see that $W$ is not a subspace of $V$ is to note that the zero vector $O$ in $V$, which is the $n\times n$ zero matrix, is not in $W$ as we have $O^{\trans}O=O\neq I$.
Thus, $W$ is not a subspace of $V$.

Another approach

You may also show that scalar multiplication (or addition) is not closed in $W$.

For example, the identity matrix $I$ is orthogonal as $I^{\trans}I=I$, and thus $I$ is an element in $W$.
However, the scalar product $2I$ is not orthogonal since
\[(2I)^{\trans}(2I)=4I\neq I.\] Click here if solved 62

The post Is the Set of All Orthogonal Matrices a Vector Space? first appeared on Problems in Mathematics.

Let $A$ be a $3\times 3$ real orthogonal matrix with $\det(A)=1$.

(a) If $\frac{-1+\sqrt{3}i}{2}$ is one of the eigenvalues of $A$, then find the all the eigenvalues of $A$.

(b) Let
\[A^{100}=aA^2+bA+cI,\] where $I$ is the $3\times 3$ identity matrix.
Using the Cayley-Hamilton theorem, determine $a, b, c$.

(Kyushu University, Linear Algebra Exam Problem)
 

Solution.

(a) Find the all the eigenvalues of $A$.

Since $A$ is a real matrix and $\frac{-1+\sqrt{3}i}{2}$ is a complex eigenvalue, its conjugate $\frac{-1-\sqrt{3}i}{2}$ is also an eigenvalue of $A$.
As $A$ is a $3\times 3$ matrix, it has one more eigenvalue $\lambda$.

Note that the product of all eigenvalues of $A$ is the determinant of $A$.
Thus, we have
\[\frac{-1+\sqrt{3}i}{2} \cdot \frac{-1-\sqrt{3}i}{2}\cdot \lambda =\det(A)=1.\] Solving this, we obtain $\lambda=1$.
Therefore, the eigenvalues of $A$ are
\[\frac{-1+\sqrt{3}i}{2}, \frac{-1-\sqrt{3}i}{2}, 1.\]

(a) Using the Cayley-Hamilton theorem, determine $a, b, c$.

To use the Cayley-Hamilton theorem, we first need to determine the characteristic polynomial $p(t)=\det(A-tI)$ of $A$.
Since we found all the eigenvalues of $A$ in part (a) and the roots of characteristic polynomials are the eigenvalues, we know that
\begin{align*}
p(t)&=-\left(\, t-\frac{-1+\sqrt{3}i}{2} \,\right)\left(\, t-\frac{-1-\sqrt{3}i}{2} \,\right)(t-1) \tag{*}\\
&=-(t^2+t+1)(t-1)\\
&=-t^3+1.
\end{align*}
(Remark that if your definition of the characteristic polynomial is $\det(tI-A)$, then the first negative sign in (*) should be omitted.)

Then the Cayley-Hamilton theorem yields that
\[P(A)=-A^3+I=O,\] where $O$ is the $3\times 3$ zero matrix.

Hence we have $A^3=I$.
We compute
\begin{align*}
A^{100}=(A^3)^{33}A=I^{33}A=IA=A.
\end{align*}

Thus, we conclude that $a=0, b=1, c=0$.

Comment.

Observe that we did not use the assumption that $A$ is orthogonal.

The post Use the Cayley-Hamilton Theorem to Compute the Power $A^{100}$ first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ real skew-symmetric matrix.

(a) Prove that the matrices $I-A$ and $I+A$ are nonsingular.

(b) Prove that
\[B=(I-A)(I+A)^{-1}\] is an orthogonal matrix.

Proof.

(a) Prove that the matrices $I-A$ and $I+A$ are nonsingular.

The eigenvalues of a skew-symmetric matrix are either $0$ or purely imaginary numbers.
(See the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even” for a proof of this fact.)

Namely, the eigenvalues of $A$ are of the form $ib$, where $i=\sqrt{-1}$ and $b$ is a real number.

The eigenvalues of the matrix $I\pm A$ are of the form $1\pm \lambda$, where $\lambda$ is an eigenvalue of $A$.
Since $\lambda=ib$, we have $1\pm \lambda \neq 0$.
Thus, $I\pm A$ do not have $0$ as an eigenvalue.

Since the determinant is the product of all eigenvalues, it follows that the determinants of the matrices $I\pm A$ are not zero, hence they are nonsingular.

(b) Prove that $B=(I-A)(I+A)^{-1}$ is an orthogonal matrix.

Note that by part (a), the matrix $I+A$ is nonsingular, hence it is invertible.
Thus the expression $B=(I-A)(I+A)^{-1}$ is well-defined.

Our goal is to show that $B^{\trans}B=I$.
Recall the following basic properties of transpose and inverse matrices.

1. $(AB)^{\trans}=B^{\trans} A^{\trans}$
2. $(A^{-1})^{\trans}=(A^{\trans})^{-1}$ if $A$ is invertible.
3. $(A+B)^{\trans}=A^{\trans}+B^{\trans}$.

We have
\begin{align*}
B^{\trans}&=\left(\, (I-A)(I+A)^{-1} \,\right)^{\trans}\\
&=\left(\, (I+A)^{-1} \,\right)^{\trans}(I-A)^{\trans} && \text{by property 1}\\
&=\left(\, (I+A)^{\trans} \,\right)^{-1}(I-A)^{\trans} && \text{by property 2}\\
&=(I^{\trans}+A^{\trans})^{-1}(I^{\trans}-A^{\trans}) && \text{by property 3}\\
&=(I+A^{\trans})^{-1}(I-A^{\trans}) && \text{since } I^{\trans}=I\\
&=(I-A)^{-1}(I+A),
\end{align*}
where the last step follows since $A$ is skew-symmetric: $A^{\trans}=-A$.

Hence we have
\begin{align*}
B^{\trans} B&=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}.
\end{align*}

We note that the middle two matrices $I+A$ and $I-A$ commutes.
In fact we have
\begin{align*}
(I+A)(I-A)&=(I+A)I-(I+A)A=I+A-A-A^2=I-A^2 \text{ and }\\
(I-A)(I+A)&=(I-A)I+(I-A)A=I-A+A-A^2=I-A^2.
\end{align*}
It yields that
\[(I+A)(I-A)=(I-A)(I+A) \tag{*}.\]

Hence we have
\begin{align*}
&B^{\trans} B\\
&=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}\\
&=(I-A)^{-1}(I-A)(I+A)(I+A)^{-1} && \text{ by (*)}\\
&=I\cdot I=I,
\end{align*}
and we have obtained $B^{\trans}B=I$.
Therefore, the matrix $B$ is an orthogonal matrix.

The post If $A$ is a Skew-Symmetric Matrix, then $I+A$ is Nonsingular and $(I-A)(I+A)^{-1}$ is Orthogonal 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 $A$ be an $n\times n$ real symmetric matrix.
Prove that there exists an eigenvalue $\lambda$ of $A$ such that for any vector $\mathbf{v}\in \R^n$, we have the inequality
\[\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2.\]

Proof.

Recall that all the eigenvalues of a symmetric matrices are real numbers.
Let $\lambda_1, \dots, \lambda_n$ be eigenvalues of $A$.

Since these eigenvalues are real numbers, there is the largest one.
Let $\lambda$ be the largest eigenvalue of $A$.
With this choice of $\lambda$ we show that the inequality
\[\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2\] holds for any $\mathbf{v}\in \R^n$.

Also recall that for a real symmetric matrix, there are eigenvalues $\mathbf{v}_1, \dots, \mathbf{v}_n$ corresponding to $\lambda_1, \dots, \lambda_n$ such that
\[B=\{\mathbf{v}_1, \dots, \mathbf{v}_n\}\] form an orthonormal basis of $\R^n$.
(This statement is equivalent to that every real symmetric matrix is diagonalizable by an orthogonal matrix.)

Let $\mathbf{v}$ be an arbitrary vector in $\R^n$.
Then since $B$ is a basis of $\R^n$, we can write
\[\mathbf{v}=c_1\mathbf{v}_1+\dots+c_n\mathbf{v}_n\] for some $c_1, \dots, c_n\in \R$.

Then we calculate
\begin{align*}
A\mathbf{v}&=A(c_1\mathbf{v}_1+\dots+c_n\mathbf{v}_n)\\
&=c_1A\mathbf{v}_1+\dots+c_nA\mathbf{v}_n\\
&=c_1\lambda_1\mathbf{v}_1+\dots+c_n\lambda_n\mathbf{v}_n
\end{align*}
since $A\mathbf{v}_i=\lambda_i\mathbf{v}_i$ for $i=1, \dots, n$.

Using this, we have
\begin{align*}
\mathbf{v}\cdot A\mathbf{v}&=(c_1\mathbf{v}_1+\dots+c_n\mathbf{v}_n)\cdot (c_1\lambda_1\mathbf{v}_1+\dots+c_n\lambda_n\mathbf{v}_n)\\
&=c_1^2\lambda_1+\cdots+c_n^2\lambda_n.
\end{align*}

Here, we used that $B=\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ is an orthonormal basis of $\R^3$.
That is, we used the properties
\begin{align*}
\mathbf{v}_i\cdot \mathbf{v}_j=\begin{cases}
1 & \text{if } i=j\\
0 & \text{if } i\neq j.
\end{cases}
\end{align*}

Since $\lambda$ is the largest eigenvalue of $A$, we have
\begin{align*}
\mathbf{v}\cdot A\mathbf{v}&=c_1^2\lambda_1+\cdots+c_n^2\lambda_n\\
& \leq c_1^2\lambda+\cdots+c_n^2\lambda\\
&=\lambda(c_2^2+\cdots+c_n^2)\\
&=\lambda \|\mathbf{v}\|^2.
\end{align*}
Hence the required inequality holds.

The post Inequality about Eigenvalue of a Real Symmetric Matrix first appeared on Problems in Mathematics.

(a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.

(b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue.

Proof.

(a) Prove that the length (magnitude) of each eigenvalue of $A$ is $1$

Let $A$ be a real orthogonal $n\times n$ matrix.
Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{v}$ be a corresponding eigenvector.
Then we have
\[A\mathbf{v}=\lambda \mathbf{v}.\] It follows from this we have
\[\|A\mathbf{v}\|^2=\|\lambda \mathbf{v}\|^2=|\lambda|^2\|\mathbf{v}\|^2.\] The left hand side becomes
\begin{align*}
&\|A\mathbf{v}\|^2\\
&=\overline{(A\mathbf{v})}^{\trans}(A\mathbf{v}) && \text{by definition of the length}\\
&=\bar{\mathbf{v}}^{\trans}A^{\trans}A\mathbf{v} && \text{because $A$ is real}\\
&=\bar{\mathbf{v}}^{\trans}\mathbf{v} && \text{because $A^{\trans}A=I$ as $A$ is orthogonal}\\
&=\|\mathbf{v}\|^2 && \text{by definition of the length.}
\end{align*}

It follows that we obtain
\[\|\mathbf{v}\|^2=|\lambda|^2\|\mathbf{v}\|^2.\] Since $\mathbf{v}$ is an eigenvector, it is non-zero, and hence $\|\mathbf{v}\|\neq 0$.
Canceling $\|\mathbf{v}\|$, we have
\[|\lambda|^2=1.\] Since the length is non-negative, we obtain
\[|\lambda|=1,\] as required.

(b) Prove that $A$ has $1$ as an eigenvalue.

Let $A$ be a real orthogonal $3\times 3$ matrix with $\det(A)=1$.
Let us consider the characteristic polynomial $p(t)=\det(A-tI)$ of $A$.
The roots of $p(t)$ are eigenvalues of $A$.

Since $A$ is a real $3\times 3$ matrix, the degree of the polynomial $p(t)$ is $3$ and the coefficients are real.
Thus, there are two cases to consider:

1. there are three real eigenvalues $\alpha, \beta, \gamma$, and
2. there is one real eigenvalue $\alpha$ and a complex conjugate pair $\beta, \bar{\beta}$ of eigenvalues.

Let us first deal with case 1.
By part (a), the lengths of eigenvalues $\alpha, \beta, \gamma$ are $1$. Since they are real numbers, we have
\[\alpha=\pm 1, \beta=\pm 1, \gamma=\pm 1.\] Recall that the product of all eigenvalues of $A$ is the determinant of $A$.
(For a proof, see the post “Determinant/trace and eigenvalues of a matrix“.)
Thus we have
\[\alpha \beta \gamma=\det(A)=1.\] Thus, at least one of $\alpha, \beta, \gamma$ is $1$.

Next, we consider case 2. Again the lengths of eigenvalues $\alpha, \beta, \bar{\beta}$ are $1$.
Then we have
\begin{align*}
1&=\det(A)=\alpha \beta \bar{\beta}\\
&=\alpha |\beta|^2=\alpha.
\end{align*}

Therefore, in either case, we see that $A$ has $1$ as an eigenvalue.

The post Eigenvalues of Orthogonal Matrices Have Length 1. Every \times 3$ Orthogonal Matrix Has 1 as an Eigenvalue first appeared on Problems in Mathematics.

Suppose that $A$ is a real $n\times n$ matrix.

(a) Is it true that $A$ must commute with its transpose?

(b) Suppose that the columns of $A$ (considered as vectors) form an orthonormal set.
Is it true that the rows of $A$ must also form an orthonormal set?

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

Solution.

(a) Is it true that $A$ must commute with its transpose?

The answer is no.

We give a counterexample. Let
\[A=\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}.\] Then the transpose of $A$ is
\[A^{\trans}=\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}.\] We compute
\[AA^{\trans}=\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}
=
\begin{bmatrix}
2 & -2\\
-2& 4
\end{bmatrix},\] and
\[A^{\trans}A=
\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}
\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}
=
\begin{bmatrix}
1 & -1\\
-1& 5
\end{bmatrix}.
\] Therefore, we see that
\[AA^{\trans}\neq A^{\trans} A,\] that is, $A$ does not commute with its transpose $A^{\trans}$.

(b) Is it true that the rows of $A$ must also form an orthonormal set?

The answer is yes.

Note that in general the column vectors of a matrix $M$ form an orthonormal set if and only if $M^{\trans}M=I$, where $I$ is the identity matrix. (Such a matrix is called orthogonal matrix.)

Thus, by assumption we have $A^{\trans} A=I$. Let $B=A^{\trans}$.
Then the column vectors of $B$ is the row vectors of $A$. Hence it suffices to show that $B^{\trans}B=I$.

Since $A^{\trans} A=I$, we know that $A$ is invertible and the inverse $A^{-1}=A^{\trans}$.
In particular, we have $A^{\trans} A=A A^{\trans}=I$.

We have
\begin{align*}
B^{\trans}B=(A^{\trans})^{\trans}A^{\trans}=(AA^{\trans})^{\trans}=I^{\trans}=I.
\end{align*}
Thus, we obtain $B^{\trans}B=I$ and by the general fact stated above, the column vectors of $B$ form an orthonormal set.
Hence the row column vectors of $A$ form an orthonormal set.

The post If Column Vectors Form Orthonormal set, is Row Vectors Form Orthonormal Set? first appeared on Problems in Mathematics.

For a real number $0\leq \theta \leq \pi$, we define the real $3\times 3$ matrix $A$ by
\[A=\begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}.\]

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

(b) Show that $A$ is an orthogonal matrix.

(c) Find the eigenvalues of $A$.

Solution.

(a) The determinant of the matrix $A$

By the cofactor expansion corresponding to the third row, we compute
\begin{align*}
\det(A)&=\begin{vmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{vmatrix}\\
&=0\cdot \begin{vmatrix}
-\sin \theta & 0\\
\cos \theta& 0
\end{vmatrix}-0\cdot \begin{vmatrix}
\cos \theta & 0\\
\sin \theta& 0
\end{vmatrix}+1\cdot \begin{vmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{vmatrix}\\
&=\cos^2 \theta +\sin^2 \theta\\
&=1.
\end{align*}
The last step follows from the famous trigonometry identity
\[\cos^2 \theta +\sin^2 \theta=1.\] Thus we have
\[\det(A)=1.\]

(b) The matrix $A$ is an orthogonal matrix

We give two solutions for part (b).

The first solution of (b)

The first solution computes $A^{\trans}A$ and show that it is the identity matrix $I$.
We have
\begin{align*}
A^{\trans}A&=\begin{bmatrix}
\cos\theta & \sin\theta & 0 \\
-\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
\cos^2 \theta +\sin^2\theta & 0 & 0 \\
0 &\cos^2 \theta+\sin^2 \theta &0 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}=I.
\end{align*}
Similarly, you can check that $AA^{\trans}=I$. Thus $A$ is an orthogonal matrix.

The second solution of (b)

The second proof uses the following fact: a matrix is orthogonal if and only its column vectors form an orthonormal set.
Let
\[A_1=\begin{bmatrix}
\cos \theta \\
\sin \theta \\
0
\end{bmatrix}, A_2=\begin{bmatrix}
-\sin\theta \\
\cos \theta \\
0
\end{bmatrix}, A_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\] be the column vectors of the matrix $A$. The length of these vectors are all $1$. For example, we have
\begin{align*}
||A_1||=\sqrt{(\cos\theta)^2+(\sin \theta)^2+0^2}=\sqrt{1}=1.
\end{align*}
Similarly, we have $||A_2||=||A_3||=1$.
The dot (inner) product of $A_1$ and $A_2$ is
\begin{align*}
A_1\cdot A_2=\cos \theta \cdot (-\sin \theta)+\sin \theta \cdot \cos \theta +0\cdot 0=0.
\end{align*}
Similarly, we have $A_1\cdot A_3=A_2\cdot A_3=0$.
Therefore, the column vectors $A_1, A_2, A_3$ are orthonormal vectors. Hence by the above fact, the matrix $A$ is orthogonal.

(c) The eigenvalues of $A$

We compute the characteristic polynomial $p(t)=\det(A-tI)$ as follows.
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
\cos\theta-t & -\sin\theta & 0 \\
\sin\theta &\cos\theta -t&0 \\
0 & 0 & 1-t
\end{vmatrix}\\
&=(1-t)\begin{vmatrix}
\cos \theta -t & -\sin \theta\\
\sin \theta& \cos \theta-t
\end{vmatrix} \text{ by the third row cofactor expansion}\\
&=(1-t)(\cos^2 \theta -2t \cos \theta +t^2 +\sin^2 \theta)\\
&=(1-t)(t^2-(2\cos \theta)t+1).
\end{align*}

The eigenvalues are roots of the characteristic polynomial $p(t)$, hence we solve
\[p(t)=(1-t)(t^2-(2\cos \theta)t+1)=0.\] One solution is $t=1$. The other solutions come from the quadratic polynomial in $p(t)$.
By the quadratic formula, those solutions are
\begin{align*}
t&=\cos\theta \pm\sqrt{\cos^2 \theta -1}\\
&=\cos\theta \pm \sqrt{-\sin^2 \theta}\\
&=\cos \theta \pm i \sin \theta
\end{align*}
since $\sin \theta\geq 0$ since $0 \leq \theta \leq \pi$.
Therefore the eigenvalues of the matrix $A$ are
\[1, \cos \theta \pm i \sin \theta.\]

Related Question.

The following problem treats the rotation matrix in the plane.

The solution is given in the post ↴
Rotation Matrix in the Plane and its Eigenvalues and Eigenvectors

The post Rotation Matrix in Space and its Determinant and Eigenvalues first appeared on Problems in Mathematics.

Find the inverse matrix of the matrix
\[A=\begin{bmatrix}
\frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\[6 pt] \frac{6}{7} &\frac{2}{7} &-\frac{3}{7} \\[6pt] -\frac{3}{7} & \frac{6}{7} & -\frac{2}{7}
\end{bmatrix}.\]

Hint.

You may use the augmented matrix method to find the inverse matrix.
Here we give an alternative way to find the inverse matrix by noting that $A$ is an orthogonal matrix.

Recall that a matrix $B$ is orthogonal if $B^{\trans}B=B^{\trans}B=I$.
Thus, once we know $B$ is an orthogonal matrix, then the inverse matrix $B^{-1}$ is just the transpose matrix $B^{\trans}$.

Also, recall that a matrix $B$ is orthogonal if and only if the column vectors of $B$ form an orthonormal set.

Solution.

We first show that $A$ is an orthogonal matrix.
To do this, it suffices to that the column vectors form an orthonormal set.

Let
\[ \mathbf{v}_1= \begin{bmatrix}
\frac{2}{7} \\[6 pt] \frac{6}{7} \\[6pt] -\frac{3}{7}
\end{bmatrix},
\mathbf{v}_2=\begin{bmatrix}
\frac{3}{7} \\[6 pt] \frac{2}{7} \\[6pt] \frac{6}{7} \end{bmatrix},
\mathbf{v}_3=\begin{bmatrix}
\frac{6}{7} \\[6 pt] -\frac{3}{7} \\[6pt] -\frac{2}{7}
\end{bmatrix}\] be the column vectors of $A$.

Then the length of the vector $\mathbf{v}_1$ is
\[||\mathbf{v}_1||=\sqrt{(2/7)^2+(6/7)^2+(-3/7)^2}=1.\] Similarly, we have $||\mathbf{v}_2||=||\mathbf{v}_3||=1$.
Thus, column vectors are unit vectors.

The dot (inner) product of the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is
\[\mathbf{v}_1\cdot \mathbf{v}_2=\frac{2}{7}\cdot \frac{3}{7}+\frac{6}{7}\cdot \frac{2}{7}+\left( -\frac{3}{7}\right) \cdot \frac{6}{7}=0.\] Similarly, we have
\[\mathbf{v}_1\cdot \mathbf{v}_3=0, \quad \mathbf{v}_2\cdot \mathbf{v}_3=0.\]

Therefore, the column vectors are orthogonal.
Hence the column vectors of $A$ are orthonormal, and this implies that $A$ is an orthogonal matrix. Namely, $A^{\trans}=A^{-1}$.
Thus the inverse matrix of $A$ is
\[A^{-1}=\begin{bmatrix}
\frac{2}{7} & \frac{6}{7} & -\frac{3}{7} \\[6 pt] \frac{3}{7} &\frac{2}{7} &\frac{6}{7} \\[6pt] \frac{6}{7} & -\frac{3}{7} & -\frac{2}{7}
\end{bmatrix}.\] Click here if solved 46

The post Find the Inverse Matrix of a Matrix With Fractions first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ matrix with real number entries.

Show that if $A$ is diagonalizable by an orthogonal matrix, then $A$ is a symmetric matrix.

Proof.

Suppose that the matrix $A$ is diagonalizable by an orthogonal matrix $Q$.
The orthogonality of the matrix $Q$ means that we have
\[Q^{\trans}Q=QQ^{\trans}=I, \tag{*}\] where $Q^{\trans}$ is the transpose matrix of $Q$ and $I$ is the $n\times n$ identity matrix.

Since $Q$ diagonalizes the matrix $A$, we have
\[Q^{-1}AQ=D,\] where $D$ is a diagonal matrix.
Equivalently, we have
\[A=QDQ^{-1} \tag{**}.\] Taking transpose of both sides, we obtain
\begin{align*}
A^{\trans}&=(QDQ^{-1})^{\trans}\\
&=(Q^{-1})^{\trans}D^{\trans} Q^{-1}\\
&=(Q^{-1})^{\trans}D Q^{-1} \text{ since } D \text{ is diagonal.}\tag{***}
\end{align*}

By (*), we observe that the inverse matrix of $Q$ is the transpose $Q^{\trans}$, that is, $Q^{-1}=Q^{\trans}$.
It follows from this observation and (***) that we have
\[A^{\trans}=QDQ^{-1}.\] (Note that $(Q^{-1})^{\trans}=Q^{\trans \trans}=Q$.)

Comparing this with (**), we obtain
\[A^{\trans}=A,\] and hence $A$ is a symmetric matrix.

The post Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix first appeared on Problems in Mathematics.