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Linear Algebra : Symmetric Matrices

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Example Question #81 : Operations And Properties

Matrix A is a symmetric matrix and is given below. What is x?

$\textbf{A}=\begin{bmatrix} 3& -1& 2\\ -1& 4& 7\\ 2& x& 8 \end{bmatrix}$

Possible Answers:

There is not enough information to determine x.

Correct answer:

Explanation:

A symmetric matrix M must follow the following condition:

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

We can find the transpose of A and compare to find x:

$\textbf{A}=\begin{bmatrix} 3& -1& 2\\ -1& 4& 7\\ 2& x& 8 \end{bmatrix} \rightarrow \textbf{A}^T=\begin{bmatrix} 3& -1& 2\\ -1& 4& x\\ 2& 7& 8 \end{bmatrix}$

We can see that x must be equal to 7.

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Example Question #82 : Operations And Properties

Matrix P is given below. Is P a symmetric matrix?

$\textbf{P} =\begin{bmatrix} 3& -2& 3\\ -9& 4& -9\\ 0& 1& 0 \end{bmatrix}$

Possible Answers:

No, P is not a symmetric matrix.

There is not enough information to determine whether P is a symmetric matrix.

Yes, P is a symmetric matrix.

Correct answer:

No, P is not a symmetric matrix.

Explanation:

A matrix M is symmetric if it satisfies the condition:

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

We can find the transpose of P and see if it satisfies this condition:

$\textbf{P} =\begin{bmatrix} 3& -2& 3\\ -9& 4& -9\\ 0& 1& 0 \end{bmatrix} \rightarrow \textbf{P}^T =\begin{bmatrix} 3& -9& 0\\ -2& 4& 1\\ 3& -9& 0 \end{bmatrix}$

Comparing the equations, we can see:

$\begin{bmatrix} 3& -2& 3\\ -9& 4& -9\\ 0& 1& 0 \end{bmatrix} \neq \begin{bmatrix} 3& -9& 0\\ -2& 4& 1\\ 3& -9& 0 \end{bmatrix} \rightarrow \textbf{P} \neq \textbf{P}^T$

And so we can determine that matrix P is not symmetric.

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Example Question #83 : Operations And Properties

Which of the following matricies are symmetric?

Possible Answers:

$\begin{bmatrix} 4& -2& 4\\ -2&3 &-2 \\ 7& 1&7 \end{bmatrix}$

$\begin{bmatrix} 4& -2& 7\\ 4& -2& 7\\\4& -2& 7\ \end{bmatrix}$

$\begin{bmatrix} 4& -2& 7\\ -2&3 &1 \\ 7& 1&0 \end{bmatrix}$

None of these matricies are symmetric.

$\begin{bmatrix} 4& -2& 7\\ -2&3 &1 \\ 4& -2&7\end{bmatrix}$

Correct answer:

$\begin{bmatrix} 4& -2& 7\\ -2&3 &1 \\ 7& 1&0 \end{bmatrix}$

Explanation:

A matrix M is symmetric if it satisfies the following condition:

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

The only matrix that satisfies this condition is:

$\begin{bmatrix} 4& -2& 7\\ -2&3 &1 \\ 7& 1&0 \end{bmatrix}$

Reversing the rows and columns of this matrix (finding the transpose) results in the same matrix. Therefore, it is symmetric.

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Example Question #81 : Operations And Properties

$A = \begin{bmatrix} 0 & \ln 2 \\ \ln x & 0 \end{bmatrix}$

Which value of makes skew-symmetric?

Possible Answers:

x = 2

$x = - \frac{1}{2}$

There is no such value of .

x = -2

$x = \frac{1}{2}$

Correct answer:

$x = \frac{1}{2}$

Explanation:

A skew-symmetric matrix is one whose transpose $A^{T}$ is equal to its additive inverse .

$A^{T}$ can be found by interchanging its rows with its columns:

$A^{T} = \begin{bmatrix} 0 & \ln x \\ \ln 2 & 0 \end{bmatrix}$

Also,

$-A = \begin{bmatrix} 0 &- \ln 2 \\ -\ln x & 0 \end{bmatrix}$

For to be skew-symmetric, it must hold that

$\begin{bmatrix} 0 & \ln x \\ \ln 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 &- \ln 2 \\ -\ln x & 0 \end{bmatrix}$

That is,

$\ln x = -\ln 2$

$\ln x = -1 \ln 2$

Using a property of logarithms:

$\ln x = \ln 2^{-1}$

$\ln x = \ln \frac{1}{2}$

$x = \frac{1}{2}$

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Example Question #81 : Operations And Properties

$A^{T} = \begin{bmatrix} 2+ i & 5 \\ 3 - 6i & 2i \\ 1 + 7i & 0 \end{bmatrix}$

Which of the following is equal to $A^{*}$

Possible Answers:

$A^{*} = \begin{bmatrix} 2- i & 3 + 6i & 1 - 7i \\ 5 & -2i & 0 \end{bmatrix}$

$A^{*} = \begin{bmatrix} -2+ i & -5 \\ -3 - 6i & 2i \\ -1 + 7i & 0 \end{bmatrix}$

$A^{*}$ does not exist.

$A^{*} = \begin{bmatrix} 2- i & 5 \\ 3 + 6i & -2i \\ 1 - 7i & 0 \end{bmatrix}$

$A^{*} = \begin{bmatrix} - 2+i & -3 - 6i & -1 + 7i \\ -5 & 2i & 0 \end{bmatrix}$

Correct answer:

$A^{*} = \begin{bmatrix} 2- i & 5 \\ 3 + 6i & -2i \\ 1 - 7i & 0 \end{bmatrix}$

Explanation:

$A^{T}$ is the transpose of - the result of interchanging the rows of with its columns. $A^{*}$ is the conjugate transpose of - the result of changing each entry of $A^{T}$ to its complex conjugate. Therefore, if

$A^{T} = \begin{bmatrix} 2+ i & 5 \\ 3 - 6i & 2i \\ 1 + 7i & 0 \end{bmatrix}$ ,

then

$A^{*} = \begin{bmatrix} \overline{2+ i} & \overline{5} \\ \overline{3 - 6i} & \overline{2i} \\ \overline{1 + 7i} & \overline{0} \end{bmatrix} = \begin{bmatrix} 2- i & 5 \\ 3 + 6i & -2i \\ 1 - 7i & 0 \end{bmatrix}$ .

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Example Question #21 : Symmetric Matrices

$A = \begin{bmatrix} 0 & 6i & 3i \\ -6i & 0 & 2i \\ -3i & -2i & 0 \end{bmatrix}$

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

Possible Answers:

Both skew-symmetric and Hermitian

Symmetric

Skew-symmetric

Both symmetric and Hermitian

Hermitian

Correct answer:

Both skew-symmetric and Hermitian

Explanation:

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.

$A = \begin{bmatrix} 0 & 6i & 3i \\ -6i & 0 & 2i \\ -3i & -2i & 0 \end{bmatrix}$

The transpose of is

$A^{T} = \begin{bmatrix} 0 & -6i & -3i \\ 6i & 0 & -2i \\ 3i & 2i & 0 \end{bmatrix}$

A matrix is symmetric if and only if $A = A^{T}$ . This can be seen to not be the case.

A matrix is skew-symmetric if $-A = A^{T}$ . Taking the additive inverse of each entry in , it can be seen that

$-A = \begin{bmatrix} 0 & -6i & -3i \\ 6i & 0 & -2i \\ 3i & 2i & 0 \end{bmatrix} = A^{T}$ .

is therefore skew-symmetric.

A matrix is Hermitian if it is equal to its conjugate transpose $A^{*}$ . Find this by changing each entry in $A^{T}$ to its complex conjugate:

$A^{*} = \begin{bmatrix} 0 & 6i & 3i \\- 6i & 0 & 2i \\ -3i & -2i & 0 \end{bmatrix} =A$

is also Hermitian.

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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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Linear Algebra : Symmetric Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #81 : Operations And Properties

Example Question #82 : Operations And Properties

Example Question #83 : Operations And Properties

Example Question #81 : Operations And Properties

Example Question #81 : Operations And Properties

Example Question #21 : Symmetric Matrices

Example Question #21 : Symmetric Matrices

Example Question #22 : Symmetric Matrices

Example Question #21 : Symmetric Matrices

Example Question #21 : Symmetric Matrices

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