Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Orthogonal Matrices

Study concepts, example questions & explanations for Linear Algebra

Create An Account

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 Next →

Example Question #21 : Orthogonal Matrices

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

Possible Answers:

It has no inverse.

(Q^T)^2

Q^T

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 .

Report an Error

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.

Report an Error

Example Question #23 : Orthogonal Matrices

is an unitary matrix.

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

Possible Answers:

True

False

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.

Report an Error

Example Question #21 : 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:

True

False

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.

Report an Error

Example Question #25 : 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:

Neither

Unitary

Both

Orthogonal

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.

Report an Error

Example Question #26 : 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.

Report an Error

Example Question #27 : Orthogonal Matrices

A real matrix is both orthogonal and involutory.

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

Possible Answers:

True

False

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.

Report an Error

Example Question #28 : 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:

True

False

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.

Report an Error

Example Question #29 : 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.

Report an Error

← Previous 1 2 3 Next →

View Linear Algebra Tutors

Anna
Certified Tutor

Seattle University, Bachelor of Science, Mathematics.

View Linear Algebra Tutors

Aidan
Certified Tutor

The University of Texas at Austin, Bachelor of Science, Mathematics.

View Linear Algebra Tutors

Aaron
Certified Tutor

Portland State University, Bachelors, Economics. Portland State University, Masters, Environmental and Resource Management.

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Houston, GRE Tutors in Phoenix, Spanish Tutors in Philadelphia, Statistics Tutors in Seattle, Algebra Tutors in Seattle, Algebra Tutors in Miami, Spanish Tutors in San Francisco-Bay Area, Statistics Tutors in New York City, ACT Tutors in New York City, Reading Tutors in Boston

Popular Courses & Classes

LSAT Courses & Classes in Chicago, ISEE Courses & Classes in Washington DC, Spanish Courses & Classes in Washington DC, Spanish Courses & Classes in Philadelphia, MCAT Courses & Classes in Boston, SAT Courses & Classes in Boston, LSAT Courses & Classes in Phoenix, GMAT Courses & Classes in New York City, Spanish Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Seattle

Popular Test Prep

SAT Test Prep in Denver, SAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Seattle, GRE Test Prep in Houston, SSAT Test Prep in Los Angeles, SSAT Test Prep in Houston, LSAT Test Prep in Atlanta, MCAT Test Prep in Houston, SAT Test Prep in Los Angeles, ISEE Test Prep in Washington DC

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Linear Algebra : Orthogonal Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #21 : Orthogonal Matrices

Example Question #22 : Orthogonal Matrices

Example Question #23 : Orthogonal Matrices

Example Question #21 : Orthogonal Matrices

Example Question #25 : Orthogonal Matrices

Example Question #26 : Orthogonal Matrices

Example Question #27 : Orthogonal Matrices

Example Question #28 : Orthogonal Matrices

Example Question #29 : Orthogonal Matrices

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: