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Linear Algebra : Operations and Properties

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Example Questions

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Example Question #21 : Symmetric Matrices

$A= \begin{bmatrix} 2 & \cos \frac{3\pi }{7} + i \sin \frac{3\pi }{7} \\ \cos \frac{11\pi }{7} + i \sin \frac{11\pi }{7} & 2 \end{bmatrix}$

True or false:

is an example of a Hermitian matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

A Hermitian matrix is equal to its conjugate transpose $A^{*}$ , which is the result of interchanging rows and columns, then changing entry to its complex conjugate.

$A= \begin{bmatrix} 2 & \cos \frac{3\pi }{7} + i \sin \frac{3\pi }{7} \\ \cos \frac{11\pi }{7} + i \sin \frac{11\pi }{7} & 2 \end{bmatrix}$

$A^{T}= \begin{bmatrix} 2 & \cos \frac{11\pi }{7} + i \sin \frac{11\pi }{7} \\ \cos \frac{3\pi }{7} + i \sin \frac{3\pi }{7} & 2 \end{bmatrix}$

$A^{*}= \begin{bmatrix} 2 & \cos \frac{11\pi }{7} - i \sin \frac{11\pi }{7} \\ \cos \frac{3\pi }{7} - i \sin \frac{3\pi }{7} & 2 \end{bmatrix}$

For $A = A^{*}$ to be true, all elements in corresponding positions must be equal. The diagonal elements are already equal, so examine the other elements. It must hold that

$\cos \frac{3\pi }{7} + i \sin \frac{3\pi }{7} =\cos \frac{11\pi }{7} - i \sin \frac{11\pi }{7}$

and

$\cos \frac{11\pi }{7} + i \sin \frac{11\pi }{7} = \cos \frac{3\pi }{7} - i \sin \frac{3\pi }{7}$

From both statements, it is necessary and sufficient to show

$\cos \frac{3\pi }{7} =\cos \frac{11\pi }{7}$

and

$\sin \frac{3\pi }{7} =-\sin \frac{11\pi }{7}$

For any $\theta \;$ ,

$\cos \theta = \cos (- \theta) = \cos (2 \pi - \theta )$

and

$\sin\theta = - \sin (- \theta) = - \sin (2 \pi - \theta )$

Set $\theta = \frac{3 \pi}{7}$ in both identities; the resulting statements are

$\cos \frac{3 \pi}{7} = \cos \frac{11 \pi}{7}$

and

$\sin \frac{3 \pi}{7} = - \sin \frac{11 \pi}{7}$ ,

precisely what is needed to be proved. It follows that is indeed Hermitian.

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Example Question #22 : Symmetric Matrices

$A= \begin{bmatrix} 0 & \cos \frac{5\pi }{7} + i \sin \frac{5\pi }{7} \\ \\ \cos \frac{12\pi }{7} + i \sin \frac{12\pi }{7} & 0 \end{bmatrix}$

Is is an example of a symmetric matrix, a skew-symmetric matrix, or a Hermitian matrix?

Possible Answers:

Symmetric

Hermitian

Skew-symmetric

Both symmetric and Hermitian

Both skew-symmetric and Hermitian

Correct answer:

Skew-symmetric

Explanation:

The answer to this question can be found by first comparing $\cos \frac{5\pi }{7} + i \sin \frac{5\pi }{7}$

and

$\cos \frac{12\pi }{7} + i \sin \frac{12\pi }{7}$ .

Note that $\frac{5 \pi}{7} + \pi = \frac{12\pi}{7}$ .

Also note that for all $\theta \;$ ,

$\cos (\theta + \pi) = -\cos \theta$

and

$\sin (\theta + \pi) = -\sin \theta$

Setting $\theta = \frac{5\pi}{7}$ , we get that

$\cos \frac{12\pi }{7} = - \cos \frac{5\pi }{7}$

and

$\sin \frac{12\pi }{7} = - \sin \frac{5\pi }{7}$ .

It follows that

$\cos \frac{12\pi }{7} + i \sin \frac{12\pi }{7} = -\left ( \cos \frac{5\pi }{7} + i \sin \frac{5\pi }{7} \right )$ ,

and that can be rewritten as

$A= \begin{bmatrix} 0 & \cos \frac{5\pi }{7} + i \sin \frac{5\pi }{7} \\ \\- \left (\cos \frac{5\pi }{7} + i \sin \frac{5\pi }{7} \right ) & 0 \end{bmatrix}$

A matrix can be identified as symmetric, skew-symmetric, or Hermitian, if any of these, by comparing the matrix to its transpose, the result of interchanging rows with columns. The transpose of is

$A^{T}= \begin{bmatrix} 0 &- \left ( \cos \frac{5\pi }{7} + i \sin \frac{5\pi }{7} \right )\\ \\ \cos \frac{5\pi }{7} + i \sin \frac{5\pi }{7} & 0 \end{bmatrix}$

Each entry in $A^{T}$ is equal to the additive inverse of the corresponding entry in ; that is, $A^{T}= -A$ .This identifies as skew-symmetric by definition.

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Example Question #21 : Symmetric Matrices

$A = \begin{bmatrix} i & 3 & -4 \\ 3 &- 4 & i \\ -4 & i & 3 \end{bmatrix}$

Which of the following describes : symmetric, skew-symmetric, or Hermitian?

Possible Answers:

Hermitian

Skew-symmetric and Hermitian

Symmetric and Hermitian

Skew-symmetric

Symmetric

Correct answer:

Symmetric

Explanation:

All three types of matrices are defined in terms of how compares to its transpose.

is symmetric if and only if $A = A^{T}$ , so find $A^{T}$ , the transpose, by interchanging its rows and its columns:

$A = \begin{bmatrix} i & 3 & -4 \\ 3 &- 4 & i \\ -4 & i & 3 \end{bmatrix}$

$A^{T} = \begin{bmatrix} i & 3 & -4 \\ 3 &- 4 & i \\ -4 & i & 3 \end{bmatrix}= A$

$A = A^{T}$ , so is symmetric.

is skew-symmetric if and only if $-A = A^{T}$ . Find , the additive inverse of :

$-A = \begin{bmatrix} -i &- 3 & 4 \\ -3 & 4 &- i \\ 4 &- i & -3 \end{bmatrix} \ne A^{T}$

$-A \ne A^{T}$ , so is not skew-symmetric.

is Hermitian if and only if $A = A^{*}$ , its conjugate transpose, so find $A^{*}$ by replacing each entry in $A^{T}$ with its complex conjugate:

$A^{*} = \begin{bmatrix}- i & 3 & -4 \\ 3 &- 4 &- i \\ -4 & -i & 3 \end{bmatrix}\ne A$

$A \ne A^{*}$ , so is not Hermitian.

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Example Question #21 : Symmetric Matrices

$A = \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & -3 \\ 3 & 1 \end{bmatrix}$ , $C = \begin{bmatrix} i & -3 \\ 3 & i \end{bmatrix}$

Which of these matrices is skew-symmetric?

Possible Answers:

only

None of A ,B,C

only

All three of A ,B,C

only

Correct answer:

only

Explanation:

A matrix is skew-symmetric if it is equal to the additive inverse of its transpose. Taking the transpose of each matrix by interchanging rows with columns:

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

$A^{T} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$

$- A^{T} = \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} = A$

is skew-symmetric.

$B = \begin{bmatrix} 1 & -3 \\ 3 & 1 \end{bmatrix}$

$B ^{T}= \begin{bmatrix} 1 & 3 \\- 3 & 1 \end{bmatrix}$

$- B ^{T}= \begin{bmatrix} -1 & -3 \\ 3 &- 1 \end{bmatrix} \ne B$

is not skew-symmetric.

$C = \begin{bmatrix} i & -3 \\ 3 & i \end{bmatrix}$

$C ^{T}= \begin{bmatrix} i & 3 \\- 3 & i \end{bmatrix}$

$- C ^{T}= \begin{bmatrix} -i & -3 \\ 3 &- i \end{bmatrix} \ne C$

is not skew-symmetric.

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Example Question #171 : Linear Algebra

$A= \begin{bmatrix} 0 & 8+7 i \\ -8+7i &0 \end{bmatrix}$

True or false: is a skew-Hermitian matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is a skew-Hermitian matrix if it is equal to the additive inverse of its conjugate transpose - that is, if

$A = -A^{*}$

To determine whether this is the case, first, find the transpose of by exchanging rows with columns in :

$A= \begin{bmatrix} 0 & 8+7 i \\ -8+7i &0 \end{bmatrix}$

$A^{T}= \begin{bmatrix} 0 & -8+7 i \\ 8+7i &0 \end{bmatrix}$

Obtain the conjugate transpose by changing each element to its complex conjugate:

$A^{*}= \begin{bmatrix} 0 & -8-7 i \\ 8-7i &0 \end{bmatrix}$

Now find the additive inverse of this by changing each entry to its additive inverse:

$-A^{*}= \begin{bmatrix} 0 & 8+7 i \\- 8+7i &0 \end{bmatrix} = A$

$A = -A^{*}$ , so is skew-Hermitian.

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Example Question #32 : Symmetric Matrices

$A= \begin{bmatrix} 1 & 8+7 i \\ -8+7i &1 \end{bmatrix}$

True or false: is a skew-Hermitian matrix.

Possible Answers:

True

False

Correct answer:

False

Explanation:

is a skew-Hermitian matrix if it is equal to the additive inverse of its conjugate transpose - that is, if

$A = -A^{*}$

To determine whether this is the case, first, find the transpose of by exchanging rows with columns in :

$A= \begin{bmatrix} 1 & 8+7 i \\ -8+7i &1 \end{bmatrix}$

$A^{T}= \begin{bmatrix}1 & -8+7 i \\ 8+7i &1 \end{bmatrix}$

Obtain the conjugate transpose by changing each element to its complex conjugate:

$A^{*}= \begin{bmatrix} 1 & -8-7 i \\ 8-7i &1\end{bmatrix}$

Now find the additive inverse of this by changing each entry to its additive inverse:

$-A^{*}= \begin{bmatrix} -1 & 8+7 i \\- 8+7i &-1 \end{bmatrix} \ne A$

$A \ne -A^{*}$ , so is not skew-Hermitian.

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Example Question #171 : Linear Algebra

$A= \begin{bmatrix} 1 & 3+ i & -2 - i \\ -3+i &1 & 5i \\ 2-i &5i & 1\end{bmatrix}$

True or false: is a skew-Hermitian matrix.

Possible Answers:

True

False

Correct answer:

False

Explanation:

is a skew-Hermitian matrix if it is equal to the additive inverse of its conjugate transpose - that is, if

$A = -A^{*}$

To determine whether this is the case, first, find the transpose of by exchanging rows with columns in :

$A= \begin{bmatrix} 1 & 3+ i & -2 - i \\ -3+i &1 & 5i \\ 2-i &5i & 1\end{bmatrix}$

$A^{T}= \begin{bmatrix} 1 & -3+i & 2-i \\ 3+ i &1 & 5i \\ -2 - i &5i & 1\end{bmatrix}$

Obtain the conjugate transpose by changing each element to its complex conjugate:

$A^{*}= \begin{bmatrix} 1 &- 3-i & 2+i \\ 3- i &1 & -5i \\ -2 + i &-5i & 1\end{bmatrix}$

Now find the additive inverse of this by changing each entry to its additive inverse:

$-A^{*} = \begin{bmatrix} -1 & 3+ i & -2 - i \\ -3+i &-1 & 5i \\ 2-i &5i & -1\end{bmatrix}$

$A \ne -A^{*}$ , so is not skew-Hermitian.

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Example Question #31 : Symmetric Matrices

$A= \begin{bmatrix} 0 & 3+ i & -2 - i \\ -3+i &0 & 5i \\ 2-i &5i & 0\end{bmatrix}$

True or false: is a skew-Hermitian matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is a skew-Hermitian matrix if it is equal to the additive inverse of its conjugate transpose - that is, if

$A = -A^{*}$

To determine whether this is the case, first, find the transpose of by exchanging rows with columns in :

$A= \begin{bmatrix} 0& 3+ i & -2 - i \\ -3+i &0 & 5i \\ 2-i &5i &0\end{bmatrix}$

$A^{T}= \begin{bmatrix} 0 & -3+i & 2-i \\ 3+ i &0 & 5i \\ -2 - i &5i & 0\end{bmatrix}$

Obtain the conjugate transpose by changing each element to its complex conjugate:

$A^{*}= \begin{bmatrix} 0 &- 3-i & 2+i \\ 3- i &0 & -5i \\ -2 + i &-5i &0\end{bmatrix}$

Now find the additive inverse of this by changing each entry to its additive inverse:

$-A^{*} = \begin{bmatrix} 0& 3+ i & -2 - i \\ -3+i &0 & 5i \\ 2-i &5i & 0\end{bmatrix} = A$

$A = -A^{*}$ , so is skew-Hermitian.

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Example Question #35 : Symmetric Matrices

$A = \begin{bmatrix} 0 & 4 - i\sqrt{2} \\ x & 0\end{bmatrix}$

Evaluate so that is a skew-Hermitian matrix.

Possible Answers:

$x = 4+i\sqrt{2}$

cannot be made skew-Hermitian regardless of the value of .

$x = 4-i\sqrt{2}$

$x = -4-i\sqrt{2}$

$x = -4+i\sqrt{2}$

Correct answer:

$x = -4-i\sqrt{2}$

Explanation:

is a skew-Hermitian matrix if it is equal to the additive inverse of its conjugate transpose - that is, if

$A = -A^{*}$

Therefore, first, take the transpose of :

$A ^{T}= \begin{bmatrix} 0 &x \\ 4 - i\sqrt{2} & 0\end{bmatrix}$

Obtain the conjugate transpose by changing each element to its complex conjugate:

$A ^{*}= \begin{bmatrix} 0 &\overline{x} \\ 4+ i\sqrt{2} & 0\end{bmatrix}$

Now find the additive inverse of this by changing each entry to its additive inverse:

$-A ^{*}= \begin{bmatrix} 0 &-\overline{x} \\ -4- i\sqrt{2} & 0\end{bmatrix}$

For $A = -A^{*}$ , or,

$\begin{bmatrix} 0 & 4 - i\sqrt{2} \\ x & 0\end{bmatrix}= \begin{bmatrix} 0 &-\overline{x} \\ -4- i\sqrt{2} & 0\end{bmatrix}$ i

It is necessary and sufficient that the two equations

$- \overline{x} = 4-i\sqrt{2}$

and

$x =-4- i\sqrt{2}$

These conditions are equivalent, so

$x =-4- i\sqrt{2}$

makes skew-Hermitian.

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Example Question #36 : Symmetric Matrices

Which of the following matrices is "Skew-symmetric"?

Possible Answers:

$\begin{bmatrix} 1&5 \\ 5& 1 \end{bmatrix}$

$\begin{bmatrix} -1&5 \\ -5& -1 \end{bmatrix}$

$\begin{bmatrix} -1&-5 \\ 5& 1 \end{bmatrix}$

$\begin{bmatrix} 0&-5 \\ 5& 0 \end{bmatrix}$

$\begin{bmatrix} 0&5 \\ 5& 0 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 0&-5 \\ 5& 0 \end{bmatrix}$

Explanation:

A skew-symmetric matrix is one that becomes negative once the transpose is taken, or A^T=-A .

We have

$\begin{bmatrix} 0&-5 \\ 5& 0 \end{bmatrix}^T =\begin{bmatrix} 0& 5\\ -5& 0 \end{bmatrix} = -\begin{bmatrix} 0&-5 \\ 5& 0 \end{bmatrix}$ .

Hence $\begin{bmatrix} 0&-5 \\ 5& 0 \end{bmatrix}$ is skew-symmetric.

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Linear Algebra : Operations and Properties

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #21 : Symmetric Matrices

Example Question #22 : Symmetric Matrices

Example Question #21 : Symmetric Matrices

Example Question #21 : Symmetric Matrices

Example Question #171 : Linear Algebra

Example Question #32 : Symmetric Matrices

Example Question #171 : Linear Algebra

Example Question #31 : Symmetric Matrices

Example Question #35 : Symmetric Matrices

Example Question #36 : Symmetric Matrices

All Linear Algebra Resources

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