Linear Algebra - Orthogonal Matrices

$\begin{align*}&\text{Conclude whether or not the matrix }A=\begin{bmatrix}-0.60587&0.79556\0.79556&0.60587\end{bmatrix}\&\text{is orthogonal.}\end{align*}$

$\begin{align*}&\text{The matrix is orthogonal.}\end{align*}$

$\begin{align*}&\text{The matrix is not orthogonal.}\end{align*}$

Explanation

$\begin{align*}&\text{There are multiple ways to determine whether or not a matrix is orthogonal.}\&\text{For instance, a matrix is orthogonal if its transpose equals its inverse:}\&A^{-1}=A^{T}\&\text{Another method is to see if the product of it and its tranpose is the identity matrix:}\&A^{T}A=AA^{T}=I\&\text{If so, it is orthgonal.}\&\text{Finally, if the matrix's determinant is }\pm1\&\text{then it is orthgonal. Using this last method:}\&det\begin{bmatrix}-0.60587&0.79556\0.79556&0.60587\end{bmatrix}= \&\text{The matrix is orthogonal.}\end{align*}$

Assume M is an orthogonal matrix. Which of the following is not always true?

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

$\textbf{M}^T=\textbf{M}^{-1}$

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

$(\textup{det}(\textbf{M}))^2=1$

All of these options are always true.

Explanation

Let us examine each of the options:

$\textbf{M}^T\textbf{M}=\textbf{I}$ This is the definition of an orthogonal matrix; it is always true.

$\textbf{M}^T=\textbf{M}^{-1}$ This can be directly proved from the previous statment. If you subtitute the inverse for the transpose in the definition equation, it is still true.

$(\textup{det}(\textbf{M}))^2=1$ The determinant of any orthogonal matrix is either 1 or -1. This statment can be proved in the following way:

$1=\textup{det}(\textbf{I})=\textup{det}(\textbf{M}^T \textbf{M}) =\textup{det}(\textbf{M}^T)\textup{det}(\textbf{M})=(\textup{det}(\textbf{M}))^2$

The incorrect statment is $\textbf{M}=\textbf{M}^T$ . Consider an example matrix:

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

which has a transpose

$\textbf{M}^T=\begin{bmatrix} 0& 1\ -1&0 \end{bmatrix}$

M and its transpose are clearly not equal. However, if we multiply them, we can see that their product is the identity matrix and they are therefore orthogonal.

$\textbf{M}\textbf{M}^T=\begin{bmatrix} 0& -1\ 1&0 \end{bmatrix} \begin{bmatrix} 0& 1\ -1&0 \end{bmatrix}=\begin{bmatrix} 1&0 \ 0& 1 \end{bmatrix} =\textbf{I}$

is an unitary matrix.

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

True

False

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.

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

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

True

False

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.

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

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

True

False

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.

$\begin{align*}&\text{Conclude whether or not the matrix }A=\begin{bmatrix}0.68567&0.12975&-0.71626\0.14807&0.93855&0.31176\0.71269&-0.31982&0.62433\end{bmatrix}\&\text{is orthogonal.}\end{align*}$

$\begin{align*}\text{The matrix is orthogonal.}\end{align*}$

$\begin{align*}\text{The matrix is not orthogonal.}\end{align*}$

Explanation

$\begin{align*}&\text{There are multiple ways to determine whether or not a matrix is orthogonal.}\&\text{For instance, a matrix is orthogonal if its transpose equals its inverse:}\&A^{-1}=A^{T}\&\text{Another method is to see if the product of it and its tranpose is the identity matrix:}\&A^{T}A=AA^{T}=I\&\text{If so, it is orthgonal.}\&\text{Finally, if the matrix's determinant is }\pm1\&\text{then it is orthgonal. Using this last method:}\&det\begin{bmatrix}0.68567&0.12975&-0.71626\0.14807&0.93855&0.31176\0.71269&-0.31982&0.62433\end{bmatrix}=1\&\text{The matrix is orthogonal.}\end{align*}$

The matrix M given below is orthogonal. What is x?

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

There is not enough information to determine x.

Explanation

We know that for any orthogonal matrix:

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

So, we can set up an equation with our matrix. First, let's find the transpose of M:

$\textbf{M}=\begin{bmatrix} x& -1\ 1&0 \end{bmatrix} \rightarrow\textbf{M}^T =\begin{bmatrix} x& 1\ -1& 0 \end{bmatrix}$

Now, let's set up the equation based on the definition:

$\textbf{MM}^T=\begin{bmatrix} x& -1\ 1&0 \end{bmatrix} \begin{bmatrix} x& 1\ -1& 0 \end{bmatrix} =\begin{bmatrix} x^2+1& x\ x&1 \end{bmatrix}=\begin{bmatrix} 1&0 \ 0& 1 \end{bmatrix}=\textbf{I}$

Comparing the last two matricies, one can see that x=0.

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

Is an orthogonal matrix or a unitary matrix?

Unitary

Orthogonal

Both

Neither

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:

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

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

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

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

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:

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

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

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

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

is unitary.

$\begin{align*}&\text{Is }A=\begin{bmatrix}-0.85616&0.46933\0.46933&0.96236\end{bmatrix}\&\text{orthogonal?}\end{align*}$

$\begin{align*}\text{The matrix is not orthogonal.}\end{align*}$

$\begin{align*}\text{The matrix is orthogonal.}\end{align*}$

Explanation

$\begin{align*}&\text{There are multiple ways to determine whether or not a matrix is orthogonal.}\&\text{For instance, a matrix is orthogonal if its transpose equals its inverse:}\&A^{-1}=A^{T}\&\text{Another method is to see if the product of it and its tranpose is the identity matrix:}\&A^{T}A=AA^{T}=I\&\text{If so, it is orthgonal.}\&\text{Finally, if the matrix's determinant is }\pm1\&\text{then it is orthgonal. Using this last method:}\&det\begin{bmatrix}-0.85616&0.46933\0.46933&0.96236\end{bmatrix}=-1.0442\&\text{The matrix is not orthogonal.}\end{align*}$

The matrix A is given below. Is it orthogonal?

$\textbf{A}=\begin{bmatrix} 2& 1&0\ -1& 3&1 \ 1&0 &-1 \end{bmatrix}$

No, A is not orthogonal.

Yes, A is orthogonal.

There is not enough intformation to determine whether or not A is orthogonal.

Explanation

For a matrix M to be orthogonal, it has to satisfy the following condition:

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

We can find the transpose of A and multiply it by A to determine whether or not it is orthogonal:

$\textbf{A}=\begin{bmatrix} 2& 1&0\ -1& 3&1 \ 1&0 &-1 \end{bmatrix} \rightarrow \textbf{A}^T= \begin{bmatrix} 2& -1&1\ 1& 3&0 \ 0&1&-1 \end{bmatrix}$

$\textbf{AA}^T= \begin{bmatrix} 2& 1&0\ -1& 3&1 \ 1&0 &-1 \end{bmatrix} \begin{bmatrix} 2& -1&1\ 1& 3&0 \ 0&1&-1 \end{bmatrix}= \begin{bmatrix} 5& 5&-2 \ 5&11 &-2 \ -2& -2& 2 \end{bmatrix} eq \begin{bmatrix} 1& 0& 0\ 0& 1& 0\ 0& 0&1 \end{bmatrix}$

Therefore, A is not an orthogonal matrix.

Orthogonal Matrices

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