Linear Algebra - Symmetric Matrices

$\begin{align*}&\text{Decide which of the given matrices must have real eigenvalues, without calculating eigenvalues or vectors.}\end{align*}$

$\begin{bmatrix}-2&-6&-14&8&-7&-5\-6&20&-14&9&0&13\-14&-14&-15&-14&7&-1\8&9&-14&13&3&-11\-7&0&7&3&-18&5\-5&13&-1&-11&5&-6\end{bmatrix}$

$\begin{bmatrix}-5&1&15&19&12&14\-17&-12&20&3&2&-12\12&5&-9&18&10&18\12&5&-9&18&10&18\-17&-12&20&3&2&-12\-5&1&15&19&12&14\end{bmatrix}$

$\begin{bmatrix}20&-8&-5&-5&-8&20\19&15&-15&-15&15&19\19&12&-17&-17&12&19\-1&-7&-4&-4&-7&-1\6&9&-17&-17&9&6\3&-16&17&17&-16&3\end{bmatrix}$

$\begin{bmatrix}-15&-8&24&14&0&14\-15&6&8&-2&25&-1\16&1&-21&16&-7&18\23&-11&7&-26&0&-15\-7&16&1&9&7&-4\22&-9&9&-7&-13&12\end{bmatrix}$

Explanation

$\begin{align*}&\text{Rather than actually calculating eigenvalues, it is worth noting that symmetric matrices}\&\text{always have real eigenvalues. A symmetric matrix is one that is equal to its transpose, i.e:}\&A=A^{T}\&\text{Therefore, the matrix which must have real eigenvalues is:}\&\begin{bmatrix}-2&-6&-14&8&-7&-5\-6&20&-14&9&0&13\-14&-14&-15&-14&7&-1\8&9&-14&13&3&-11\-7&0&7&3&-18&5\-5&13&-1&-11&5&-6\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Select the symmetric matrix from the following choices:}\end{align*}$

$\begin{bmatrix}-18&8&20\8&-11&14\20&14&3\end{bmatrix}$

$\begin{bmatrix}13&7&-18\5&2&8\-19&9&15\end{bmatrix}$

$\begin{bmatrix}-3&13&23\18&11&3\19&9&-3\end{bmatrix}$

$\begin{bmatrix}11&15&4\23&12&-24\12&-17&-12\end{bmatrix}$

Explanation

$\begin{align*}&\text{A symmetric matrix is equal to its transpose, that is to say:}\&A=A^{T}\&\text{A transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, and so on.}\&\text{The symmetric matrix is:}\&\begin{bmatrix}-18&8&20\8&-11&14\20&14&3\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Choose the symmetric matrix from the four choices given:}\end{align*}$

$\begin{bmatrix}-12&-16&-3&16&17\-16&14&-7&-7&-4\-3&-7&-11&7&-11\16&-7&7&19&7\17&-4&-11&7&-17\end{bmatrix}$

$\begin{bmatrix}13&6&-9&-6&18\-15&-5&15&-19&4\14&-6&9&-19&9\-15&-5&15&-19&4\13&6&-9&-6&18\end{bmatrix}$

$\begin{bmatrix}5&-5&13&-5&5\8&-6&20&-6&8\15&-6&5&-6&15\7&19&-6&19&7\-4&-16&-10&-16&-4\end{bmatrix}$

$\begin{bmatrix}-3&-18&-19&32&-8\-9&5&3&12&4\-10&-5&12&-20&-10\25&20&-11&-6&-7\1&11&-1&2&-27\end{bmatrix}$

Explanation

$\begin{align*}&\text{A symmetric matrix is one in which all values are mirrored across the diagonal. In other words}\text{it is equal to its transpose:}\&A=A^{T}\&\text{Recall that a transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, etc.}\&\begin{bmatrix}-12&-16&-3&16&17\-16&14&-7&-7&-4\-3&-7&-11&7&-11\16&-7&7&19&7\17&-4&-11&7&-17\end{bmatrix}\&\text{Is a symmetric matrix.}\end{align*}$

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

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

Both skew-symmetric and Hermitian

Skew-symmetric

Hermitian

Symmetric

Both 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.

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

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

Skew-symmetric

Hermitian

Symmetric

Both skew-symmetric and Hermitian

Both symmetric and Hermitian

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.

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

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

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

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

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

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

$A^{*}$ does not exist.

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

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

True or False: All skew-symmetric matrices are also symmetric matrices.

False

True

Explanation

If is skew-symmetric, then . But if were symmetric, then . Both conditions would only hold if was the zero matrix, which is not always the case.

$\begin{align*}&\text{Decide which of the given matrices must have real eigenvalues, without calculating eigenvalues or vectors.}\end{align*}$

$\begin{bmatrix}-8&-17&3&15&-12&4\-17&-9&6&-4&5&2\3&6&13&6&-19&0\15&-4&6&14&-8&-17\-12&5&-19&-8&-6&-6\4&2&0&-17&-6&9\end{bmatrix}$

$\begin{bmatrix}14&-1&2&15&10&-13\-12&-5&-3&7&-13&11\-2&-4&-7&17&8&-1\-2&-4&-7&17&8&-1\-12&-5&-3&7&-13&11\14&-1&2&15&10&-13\end{bmatrix}$

$\begin{bmatrix}10&5&19&19&5&10\-19&13&-10&-10&13&-19\-18&8&17&17&8&-18\-15&16&9&9&16&-15\-11&12&4&4&12&-11\15&-12&-14&-14&-12&15\end{bmatrix}$

$\begin{bmatrix}-18&-17&9&34&33&11\-10&-15&-12&17&-11&-29\1&-4&22&-36&-21&9\26&10&-28&2&32&5\25&-2&-12&24&-9&30\3&-20&16&-3&23&10\end{bmatrix}$

Explanation

$\begin{align*}&\text{Rather than actually calculating eigenvalues, it is worth noting that symmetric matrices}\&\text{always have real eigenvalues. A symmetric matrix is one that is equal to its transpose, i.e:}\&A=A^{T}\&\text{Therefore, the matrix which must have real eigenvalues is:}\&\begin{bmatrix}-8&-17&3&15&-12&4\-17&-9&6&-4&5&2\3&6&13&6&-19&0\15&-4&6&14&-8&-17\-12&5&-19&-8&-6&-6\4&2&0&-17&-6&9\end{bmatrix}\end{align*}$

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

Evaluate so that is a skew-Hermitian matrix.

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

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

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

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

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

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.

Which of the following dimensions cannot be that of a symmetric matrix?

2x3

1x1

2x2

3x3

27x27

Explanation

A symmetric matrix is one that equals its transpose. This means that a symmetric matrix can only be a square matrix: transposing a matrix switches its dimensions, so the dimensions must be equal. Therefore, the option with a non square matrix, 2x3, is the only impossible symmetric matrix.

Symmetric Matrices

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