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Linear Algebra : Orthogonal Matrices

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Example Questions

Example Question #21 : Orthogonal Matrices

Given an orthogonal matrix , does Q^2 have an inverse?

Possible Answers:

Q^T

It has no inverse.

(Q^T)^2

Correct answer:

(Q^T)^2

Explanation:

When is an orthogonal matrix, Q^2 certainly has an inverse as well. has the property: QQ^T=I since it is orthogonal. So we can multiply Q^2 by (Q^T)^2 to get:

Q^2(Q^T)^2=QQQ^TQ^T=QIQ^T=QQ^T=I

so the inverse is (Q^T)^2 .

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Example Question #22 : Orthogonal Matrices

$A= \begin{bmatrix} 0. 8 & 0.6 \\ -0.6 & -0.8 \end{bmatrix}$

True or false:: is an example of an orthogonal matrix.

Possible Answers:

False

True

Correct answer:

False

Explanation:

A matrix is orthogonal if and only if $A A^{T} = I$ , the identity matrix, so test this condition.

$A= \begin{bmatrix} 0. 8 & 0.6 \\ -0.6 & -0.8 \end{bmatrix}$

$A^{T}$ , its transpose, is found by interchanging rows and columns:

$A^{T}= \begin{bmatrix} 0. 8 & -0.6 \\ 0.6 & -0.8 \end{bmatrix}$

Find $A A^{T}$ multiplying rows by columns - adding products of entries in corresponding positions:

$A A^{T}= \begin{bmatrix} 0. 8 & 0.6 \\ -0.6 & -0.8 \end{bmatrix} \begin{bmatrix} 0. 8 & -0.6 \\ 0.6 & -0.8 \end{bmatrix}$

$= \begin{bmatrix} 0. 8(0.8)+ 0.6(0.6) & 0.8(-0.6)+ 0.6(-0.8)\\ -0.6(0.8)+(-0.8)(0.6) & (-0.6)(-0.6)+ (-0.8)(-0.8)\end{bmatrix}$

$= \begin{bmatrix} 0.64+0.36 & -0.48 - 0.48 \\ -0.48 - 0.48 & 0.36+0.64\end{bmatrix}$

$= \begin{bmatrix} 1& -0.96 \\ -0.96& 1\end{bmatrix}$

$\ne I$ .

is not an orthogonal matrix.

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Example Question #23 : Orthogonal Matrices

is an unitary matrix.

True or false: $A^{2}$ must be an unitary matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is an orthogonal matrix, if, by definition, $A A^{*}= A^{*}A = I$ , where $A^{*}$ is the conjugate transpose of . Also, for any square matrix , it holds that $(A ^{2})^{*} = (A^{*})^{2}$ .

Let be orthogonal. Since

$(A ^{2})^{*} = (A^{*})^{2}$ ,

it follows that

$A^{2}( A^{2})^{*} = A^{2}( A^{*})^{2}$

$A^{2}( A^{2})^{*} =(AA )( A^{*} A^{*})$

Matrix multiplication is associative, so

$A^{2}( A^{2})^{*} = A(A A^{*}) A^{*}$

$A^{2}( A^{2})^{*} = AI A^{*}$

$A^{2}( A^{2})^{*} = (AI )A^{*}$

$A^{2}( A^{2})^{*} = AA^{*}$

$A^{2}( A^{2})^{*} =I$

By similar reasoning, it can be demonstrated that $( A^{2})^{*}A^{2} =I$ . $A^{2}$ is therefore unitary.

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Example Question #22 : Orthogonal Matrices

$A = \begin{bmatrix} \frac{3}{5} & \frac{12}{25} & \frac{16 }{25} \\ \\ -\frac{4}{5} & \frac{9}{25} & \frac{12}{25} \\ \\ 0 & - \frac{4}{5} & \frac{3}{5} \end{bmatrix}$

True or false: is an example of an orthogonal matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

A matrix is orthogonal if and only if $AA^{T} = I$ .

$A^{T}$ is the transpose of . Find this by interchanging rows with columns: $A = \begin{bmatrix} \frac{3}{5} & \frac{12}{25} & \frac{16 }{25} \\ \\ -\frac{4}{5} & \frac{9}{25} & \frac{12}{25} \\ \\ 0 & - \frac{4}{5} & \frac{3}{5} \end{bmatrix}$ ,

$A^{T} = \begin{bmatrix} \frac{3}{5} & -\frac{4}{5} & 0 \\ \\ \frac{12}{25} & \frac{9}{25} & - \frac{4}{5} \\ \\ \frac{16 }{25} & \frac{12}{25} & \frac{3}{5} \end{bmatrix}$

Multiply the matrices by multiplying rows by columns, adding the products of entries in corresponding positions:

$A A^{T}= \begin{bmatrix} \frac{3}{5} & \frac{12}{25} & \frac{16 }{25} \\ \\ -\frac{4}{5} & \frac{9}{25} & \frac{12}{25} \\ \\ 0 & - \frac{4}{5} & \frac{3}{5} \end{bmatrix} \begin{bmatrix} \frac{3}{5} & -\frac{4}{5} & 0 \\ \\ \frac{12}{25} & \frac{9}{25} & - \frac{4}{5} \\ \\ \frac{16 }{25} & \frac{12}{25} & \frac{3}{5} \end{bmatrix}$

$=\begin{bmatrix} \frac{3}{5} \cdot \frac{3}{5} +\frac{12}{25} \cdot \frac{12}{25} + \frac{16}{25} \cdot \frac{16}{25} & \frac{3}{5} \cdot (- \frac{4}{5}) +\frac{12}{25} \cdot \frac{9}{25} + \frac{16}{25} \cdot \frac{12}{25} & \frac{3}{5} \cdot 0 +\frac{12}{25} \cdot( - \frac{4}{5} )+ \frac{16}{25} \cdot \frac{3}{5} \\ \\- \frac{4}{5} \cdot \frac{3}{5} +\frac{9}{25} \cdot \frac{12}{25} + \frac{12}{25} \cdot \frac{16}{25} & -\frac{4}{5} \cdot (- \frac{4}{5}) +\frac{9}{25} \cdot \frac{9}{25} + \frac{12}{25} \cdot \frac{12}{25} & -\frac{4}{5} \cdot 0 +\frac{9}{25} \cdot( - \frac{4}{5} )+ \frac{12}{25} \cdot \frac{3}{5} \\ \\0 \cdot \frac{3}{5}+(-\frac{4}{5}) \cdot \frac{12}{25} + \frac{3}{5} \cdot \frac{16}{25} &0 \cdot (- \frac{4}{5}) +(-\frac{4}{5}) \cdot \frac{9}{25} + \frac{3}{5} \cdot \frac{12}{25} & 0\cdot 0 +( - \frac{4}{5} ) \cdot( - \frac{4}{5} )+ \frac{3}{5} \cdot \frac{3}{5} \end{bmatrix}$

$=\begin{bmatrix} \frac{9}{25} +\frac{144}{625} + \frac{256}{625} &- \frac{12}{25} +\frac{108}{625} + \frac{192}{625} & 0 -\frac{48}{125} + \frac{48}{125} \\ \\- \frac{12}{25} +\frac{108}{625} + \frac{192}{625} & \frac{16}{25} +\frac{81}{625} + \frac{144}{625} & 0 -\frac{36}{125} + \frac{36}{125} \\ \\0 -\frac{48}{125} + \frac{48}{125} &0 -\frac{36}{125} + \frac{36}{125} & 0 + \frac{16}{25} + \frac{9}{25} \end{bmatrix}$

$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$AA^{T}= I$ . is indeed an orthogonal matrix.

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Example Question #21 : Orthogonal Matrices

$A = \begin{bmatrix}\frac{\sqrt{2} }{2}i& 0 &\frac{\sqrt{2} }{2} i \\ 0 & 1 & 0 \\ -\frac{\sqrt{2} }{2}i & 0 & \frac{\sqrt{2} }{2}i \end{bmatrix}$

Is an orthogonal matrix or a unitary matrix?

Possible Answers:

Unitary

Both

Orthogonal

Neither

Correct answer:

Unitary

Explanation:

is an orthogonal matrix if and only if $AA^{T} = I$ ; it is a unitary matrix if and only if $AA^{*} = I$ .

$A^{T}$ , the transpose of , can be found by interchanging rows with columns:

$A^{T}= \begin{bmatrix}\frac{\sqrt{2} }{2}i& 0 &-\frac{\sqrt{2} }{2} i \\ 0 & 1 & 0 \\ \frac{\sqrt{2} }{2}i & 0 & \frac{\sqrt{2} }{2}i \end{bmatrix}$

Multiply:

$AA^{T} =\begin{bmatrix}\frac{\sqrt{2} }{2}i& 0 &\frac{\sqrt{2} }{2} i \\ 0 & 1 & 0 \\- \frac{\sqrt{2} }{2}i & 0 & \frac{\sqrt{2} }{2}i \end{bmatrix}\begin{bmatrix}\frac{\sqrt{2} }{2}i& 0 &-\frac{\sqrt{2} }{2} i \\ 0 & 1 & 0 \\ \frac{\sqrt{2} }{2}i & 0 & \frac{\sqrt{2} }{2}i \end{bmatrix}$

$=\begin{bmatrix} \frac{\sqrt{2} }{2}i \cdot \frac{\sqrt{2} }{2}i + 0\cdot 0+ \frac{\sqrt{2} }{2}i \cdot \frac{\sqrt{2} }{2}i & \frac{\sqrt{2} }{2}i \cdot 0 + 0 \cdot 1+ \frac{\sqrt{2} }{2}i \cdot 0& \frac{\sqrt{2} }{2}i \cdot \left (-\frac{\sqrt{2} }{2}i \right )+ 0 \cdot 0+ \frac{\sqrt{2} }{2}i \cdot \frac{\sqrt{2}}{2}i \\ 0 \cdot \frac{\sqrt{2} }{2}i + 1 \cdot 0+0\cdot \frac{\sqrt{2} }{2}i &0 \cdot 0 +1 \cdot 1+ 0 \cdot 0&0\cdot \left (-\frac{\sqrt{2} }{2}i \right )+ 1 \cdot 0+0\cdot \frac{\sqrt{2}}{2}i \\-\frac{\sqrt{2} }{2}i \cdot \frac{\sqrt{2} }{2}i + 0\cdot 0+ \frac{\sqrt{2} }{2}i \cdot \frac{\sqrt{2} }{2}i &- \frac{\sqrt{2} }{2}i \cdot 0 + 0 \cdot 1+ \frac{\sqrt{2} }{2}i \cdot 0& -\frac{\sqrt{2} }{2}i \cdot \left (-\frac{\sqrt{2} }{2}i \right )+ 0 \cdot 0+ \frac{\sqrt{2} }{2}i \cdot \frac{\sqrt{2}}{2}i \end{bmatrix}$

$=\begin{bmatrix}-\frac{1}{2}+ 0 + \left ( -\frac{1}{2}\right )& 0+0+0& \frac{1}{2}+ 0+ \left ( - \frac{1}{2}\right )\\ 0+0+0 &0 +1 + 0 &0+0+0 \\ \frac{1}{2}+ 0 + \left ( -\frac{1}{2}\right ) &0+0+0& -\frac{1}{2}+ 0+ \left ( -\frac{1}{2}\right ) \end{bmatrix}$

$=\begin{bmatrix}-1& 0 & 0\\ 0 &1&0 \\0 & 0&-1 \end{bmatrix}$

$\ne I$

is not orthogonal.

$A^{*}$ , the conjugate transpose, can be found by changing each entry in $A^{T}$ to its complex conjugate:

$A^{*}= \begin{bmatrix}-\frac{\sqrt{2} }{2}i& 0 &\frac{\sqrt{2} }{2} i \\ 0 & 1 & 0 \\- \frac{\sqrt{2} }{2}i & 0 & -\frac{\sqrt{2} }{2}i \end{bmatrix}$

Multiply:

$AA^{*} =\begin{bmatrix}\frac{\sqrt{2} }{2}i& 0 &\frac{\sqrt{2} }{2} i \\ 0 & 1 & 0 \\- \frac{\sqrt{2} }{2}i & 0 & \frac{\sqrt{2} }{2}i \end{bmatrix}\begin{bmatrix}-\frac{\sqrt{2} }{2}i& 0 &\frac{\sqrt{2} }{2} i \\ 0 & 1 & 0 \\ -\frac{\sqrt{2} }{2}i & 0 &- \frac{\sqrt{2} }{2}i \end{bmatrix}$

$=\begin{bmatrix} \frac{\sqrt{2} }{2}i \cdot (-\frac{\sqrt{2} }{2}i) + 0\cdot 0+ \frac{\sqrt{2} }{2}i \cdot( -\frac{\sqrt{2} }{2}i )& \frac{\sqrt{2} }{2}i \cdot 0 + 0 \cdot 1+ \frac{\sqrt{2} }{2}i \cdot 0& \frac{\sqrt{2} }{2}i \cdot \left (\frac{\sqrt{2} }{2}i \right )+ 0 \cdot 0+ \frac{\sqrt{2} }{2}i \cdot (\frac{-\sqrt{2}}{2}i )\\ 0 \cdot (-\frac{\sqrt{2} }{2}i )+ 1 \cdot 0+0\cdot( -\frac{\sqrt{2} }{2}i )&0 \cdot 0 +1 \cdot 1+ 0 \cdot 0&0\cdot \frac{\sqrt{2} }{2}i + 1 \cdot 0+0\cdot(- \frac{\sqrt{2}}{2}i )\\\frac{\sqrt{2} }{2}i \cdot (- \frac{\sqrt{2} }{2}i) + 0\cdot 0+ - \frac{\sqrt{2} }{2}i \cdot (- \frac{\sqrt{2} }{2}i )&- \frac{\sqrt{2} }{2}i \cdot 0 + 0 \cdot 1+ \frac{\sqrt{2} }{2}i \cdot 0& -\frac{\sqrt{2} }{2}i \cdot \frac{\sqrt{2} }{2}i + 0 \cdot 0+ \frac{\sqrt{2} }{2}i \cdot( -\frac{\sqrt{2}}{2}i) \end{bmatrix}$

$=\begin{bmatrix}\frac{1}{2}+ 0 + \frac{1}{2} & 0+0+0& \left ( - \frac{1}{2}\right )+ 0 +\frac{1}{2} \\ 0+0+0 &0 +1 + 0 &0+0+0 \\ \frac{1}{2}+ 0 + \left ( -\frac{1}{2}\right ) &0+0+0& \frac{1}{2}+ 0+ \frac{1}{2} \end{bmatrix}$

$=\begin{bmatrix} 1& 0 & 0\\ 0 &1&0 \\0 & 0& 1 \end{bmatrix}$

= I

is unitary.

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Example Question #24 : Orthogonal Matrices

$A =\frac{ \sqrt{5}}{5} \begin{bmatrix} 0 & 1 +2 i \\ 1 - 2i & 0 \end{bmatrix}$

True or false: is an example of a unitary matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

A matrix is unitary if $AA^{*} = I$ .

$A^{*}$ , the conjugate transpose of , can be found by first interchanging rows with columns:

$A =\frac{ \sqrt{5}}{5} \begin{bmatrix} 0 & 1 +2 i \\ 1 - 2i & 0 \end{bmatrix}$

$A^{T} =\frac{ \sqrt{5}}{5} \begin{bmatrix} 0 & 1 -2 i \\ 1+ 2i & 0 \end{bmatrix}$

Then changing each entry to its complex conjugate:

$A^{*} =\frac{ \sqrt{5}}{5} \begin{bmatrix} 0 & 1 +2 i \\ 1- 2i & 0 \end{bmatrix}$

Find $AA^{*}$ by multiplying rows of by columns, as follows:

$AA^{*} = \frac{ \sqrt{5}}{5} \begin{bmatrix} 0 & 1 +2 i \\ 1 - 2i & 0 \end{bmatrix} \times \frac{ \sqrt{5}}{5} \begin{bmatrix} 0 & 1 +2 i \\ 1- 2i & 0 \end{bmatrix}$

$=\left ( \frac{ \sqrt{5}}{5} \times \frac{ \sqrt{5}}{5} \right ) \begin{bmatrix} 0 & 1 +2 i \\ 1 - 2i & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 +2 i \\ 1- 2i & 0 \end{bmatrix}$

$= \frac{ 1}{5} \begin{bmatrix} 0 & 1 +2 i \\ 1 - 2i & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 +2 i \\ 1- 2i & 0 \end{bmatrix}$

$= \frac{ 1}{5} \begin{bmatrix} 0(0)+ (1+2i)(1-2i) & 0(1+2i) + (1+2i) (0) \\(1-2i) (0)+ 0(1-2i) & (1-2i) (1+2i)+ 0(0) \end{bmatrix}$

$= \frac{ 1}{5} \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

= I

$AA^{*} = I$ , so is a unitary matrix.

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Example Question #22 : Orthogonal Matrices

A real matrix is both orthogonal and involutory.

True or false: It follows that is an identity matrix.

Possible Answers:

False

True

Correct answer:

False

Explanation:

Real matrix is orthogonal if and $MM^{T }= I$ ; it is involutory if $M^{2}= I$ . We show the statement is false by giving a nonidentity matrix that has both properties.

Let $M = \begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \\ - \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix}$ .

The transpose of , the result of switching rows with columns, is

$M^{T} = \begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \\ - \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix}$

It can be seen that $M^{T } = M$ , so $MM^{T }= I$ if and only if $M^{2}= I$ . It follows that $M^{2}= I$ is a sufficient condition for to be both orthogonal and involutory. Square by multiplying the rows of by its columns - adding the products of corresponding entries:

$M^{2} = \begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \\ - \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \\ - \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix}$ \ $= \begin{bmatrix} \left ( \frac{\sqrt{2}}{2} \right )\left ( \frac{\sqrt{2}}{2} \right ) + \left (- \frac{\sqrt{2}}{2} \right )\left (- \frac{\sqrt{2}}{2} \right ) & \left ( \frac{\sqrt{2}}{2} \right )\left (- \frac{\sqrt{2}}{2} \right ) + \left (- \frac{\sqrt{2}}{2} \right )\left (- \frac{\sqrt{2}}{2} \right ) \\ \\\left ( -\frac{\sqrt{2}}{2} \right )\left ( \frac{\sqrt{2}}{2} \right ) + \left (- \frac{\sqrt{2}}{2} \right )\left (- \frac{\sqrt{2}}{2} \right ) & \left (- \frac{\sqrt{2}}{2} \right )\left (- \frac{\sqrt{2}}{2} \right ) + \left (- \frac{\sqrt{2}}{2} \right )\left (- \frac{\sqrt{2}}{2} \right ) \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{2}+ \frac{1}{2} & -\frac{1}{2}+ \frac{1}{2} \\ \\ - \frac{1}{2}+ \frac{1}{2} & \frac{1}{2}+ \frac{1}{2} \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

= I

$M^{2}= MM^{T} = I$

Thus, there exists at least one non-identity matrix that is both involutory and orthogonal, making the statement false.

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Example Question #23 : Orthogonal Matrices

$A = \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$

True or false: is an example of a unitary matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is unitary if $AA^{*} = I$ , where $A^{*}$ is its conjugate transpose.

$A = \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$ ,

so transpose rows and columns to get

$A^{T} = \begin{bmatrix} 0 & 0 &- i \\ 0 &1 & 0 \\ i & 0 & 0 \end{bmatrix}$

Now change each entry to its complex conjugate:

$A^{*} = \begin{bmatrix} 0 & 0 &- i \\ 0 &1 & 0 \\ i & 0 & 0 \end{bmatrix}$

Find $AA^{*}$ by multiplying rows of by columns of $A^{*}$ - adding the products of corresponding entries, as follows:

$AA^{*} = \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$

$= \begin{bmatrix} 0(0)+0(0) + i(-i) & 0(0)+0(1)+ i(0) & 0(i)+0(0)+ i(0) \\ 0(0)+1(0) +0(-i) & 0(0)+1(1)+0(0) & 0(i)+1(0)+0i(0) \\-i(0)+0(0) + 0(-i) & -i(0)+0(1)+0(0) & -i(i)+0(0)+ 0(0) \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 & 0 \\ 0& 1 & 0 \\0& 0 & 1\end{bmatrix}$

$AA^{*} = I$ , so is unitary.

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Example Question #21 : Orthogonal Matrices

Is the following matrix M orthogonal?

$\textbf{M}=\begin{bmatrix} \cos(\theta)& -\sin(\theta)\\ \sin(\theta)& \cos(\theta) \end{bmatrix}$

Possible Answers:

It is impossible to determine from the information given.

No, M is not orthogonal

Yes, M is orthogonal

Correct answer:

Yes, M is orthogonal

Explanation:

An orthogonal matrix is a real, square matrix that satisfies the following condition:

$\textbf{M}^T\textbf{M}=\textbf{M}\textbf{M}^T=\textbf{I}$

where I is the identity matrix.

We can calculate the product of M and its transpose:

$\textbf{M}\textbf{M}^T=\begin{bmatrix} \cos(\theta)& -\sin(\theta)\\ \sin(\theta)& \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\theta)& \sin(\theta)\\ -\sin(\theta)& \cos(\theta) \end{bmatrix}$

$=\begin{bmatrix} \cos(\theta)\cos(\theta)+(-\sin(\theta))(-\sin(\theta))& \cos(\theta)\sin(\theta)+(-\sin(\theta))\cos(\theta)\\ \sin(\theta)\cos(\theta)+\cos(\theta)(-\sin(\theta))& \sin(\theta)\sin(\theta)+\cos(\theta)\cos(\theta) \end{bmatrix}$

$=\begin{bmatrix} \cos^2(\theta)+\sin^2(\theta)& \cos(\theta)\sin(\theta)-\sin(\theta)\cos(\theta)\\ \sin(\theta)\cos(\theta)-\sin(\theta)\cos(\theta)& \sin^2(\theta)+\cos^2(\theta) \end{bmatrix}$

$\begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}=\textbf{I}$

Therefore, M is indeed an orthogonal matrix.

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Linear Algebra : Orthogonal Matrices

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