Symmetric Matrices - Linear Algebra

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Question

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

True or false: is a skew-Hermitian matrix.

Answer

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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Question

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

True or false: is a skew-Hermitian matrix.

Answer

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:

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

$A e -A^{}$ , so is not skew-Hermitian.

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Question

$\begin{align}&\text{Which of the following matrices is symmetric?}\end{align}$

Answer

$\begin{align}&\text{A symmetric matrix is one that is equal to its transpose, i.e:}\&A=A^{T}\&\text{To review, a transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, etc.}\&\text{Therefore, the symmetric matrix is:}\&\begin{bmatrix}-3&1&-4&19&8&-5&20\1&-4&-5&5&-11&0&19\-4&-5&15&-11&20&-18&14\19&5&-11&6&3&-9&-4\8&-11&20&3&1&6&-4\-5&0&-18&-9&6&2&20\20&19&14&-4&-4&20&-15\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Which of the following matrices is symmetric?}\end{align}$

Answer

$\begin{align}&\text{A symmetric matrix is one that is equal to its transpose, i.e:}\&A=A^{T}\&\text{To review, a transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, etc.}\&\text{Therefore, the symmetric matrix is:}\&\begin{bmatrix}17&10&13&-18&10&5&3\10&-14&-7&5&-7&8&-8\13&-7&20&1&15&-20&-7\-18&5&1&-15&9&16&-6\10&-7&15&9&18&-11&3\5&8&-20&16&-11&-3&11\3&-8&-7&-6&3&11&20\end{bmatrix}\end{align}$

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Question

Which matrix is symmetric?

Answer

A symmetric matrix is symmetrical across the main diagonal. The numbers in the main diagonal can be anything, but the numbers in corresponding places on either side must be the same. In the correct answer, the matching numbers are the 3's, the -2's, and the 5's.

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Question

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

Answer

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

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Question

$\begin{align}&\text{Which of the following matrices is symmetric?}\end{align}$

Answer

$\begin{align}&\text{A symmetric matrix is one that is equal to its transpose, i.e:}\&A=A^{T}\&\text{To review, a transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, etc.}\&\text{Therefore, the symmetric matrix is:}\&\begin{bmatrix}-19&-7&14\-7&-7&2\14&2&1\end{bmatrix}\end{align}$

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Question

$A = \begin{bmatrix} 1& -3 &4 \ 3& 1& 5\ -4&-5 & 1\end{bmatrix}$

True or false: is an example of a skew-symmetric matrix.

Answer

A square matrix is defined to be skew-symmetric if its transpose $A^{T}$ - the matrix resulting from interchanging its rows and its columns - is equal to its additive inverse; that is, if

$A^{T} = -A$ .

Interchanging rows and columns, we see that if

$A = \begin{bmatrix} 1& -3 &4 \ 3& 1& 5\ -4&-5 & 1\end{bmatrix}$ ,

then

$A^{T} = \begin{bmatrix} 1& 3 &-4 \ -3& 1&-5\ 4&5 & 1 \end{bmatrix}$ .

can be determined by changing each element in to its additive inverse:

$A = \begin{bmatrix} -1& 3 &-4 \- 3&- 1& -5\ 4& 5 &- 1\end{bmatrix}$

$A^{T} e -A$ , since not every element in corresponding positions is equal; in particular, the three elements in the main diagonal differ. is not a skew-symmetric matrix.

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Question

$\begin{align}&\text{Which of the following matrices is symmetric?}\end{align}$

Answer

$\begin{align}&\text{A symmetric matrix is one that is equal to its transpose, i.e:}\&A=A^{T}\&\text{To review, a transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, etc.}\&\text{Therefore, the symmetric matrix is:}\&\begin{bmatrix}-12&-16&-18&11&11&13&-20&1\-16&-18&10&-1&-18&18&-16&6\-18&10&4&-17&2&-14&-11&-16\11&-1&-17&-1&-19&5&8&-20\11&-18&2&-19&-13&8&5&-4\13&18&-14&5&8&10&-19&19\-20&-16&-11&8&5&-19&1&-3\1&6&-16&-20&-4&19&-3&15\end{bmatrix}\end{align}$

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Question

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

Answer

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

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Question

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

Answer

$\begin{align}&\text{A symmetric matrix is one in which all values are mirrored across the diagonal. In other words}\text{it is equal to its transpose:}\&A=A^{T}\&\text{Recall that a transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, etc.}\&\begin{bmatrix}-2&18&4&-12\18&11&15&12\4&15&-9&-4\-12&12&-4&9\end{bmatrix}\&\text{Is a symmetric matrix.}\end{align}$

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Question

$\begin{align}&\text{Which of the following matrices is symmetric?}\end{align}$

Answer

$\begin{align}&\text{A symmetric matrix is one that is equal to its transpose, i.e:}\&A=A^{T}\&\text{To review, a transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, etc.}\&\text{Therefore, the symmetric matrix is:}\&\begin{bmatrix}-7&16&-4&18&4&1\16&-6&3&9&11&18\-4&3&-17&3&-18&-3\18&9&3&0&4&1\4&11&-18&4&-4&-6\1&18&-3&1&-6&16\end{bmatrix}\end{align}$

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Question

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

True or false: is a skew-Hermitian matrix.

Answer

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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Question

$\begin{align}&\text{Which of the following matrices is symmetric?}\end{align}$

Answer

$\begin{align}&\text{A symmetric matrix is one that is equal to its transpose, i.e:}\&A=A^{T}\&\text{To review, a transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, etc.}\&\text{Therefore, the symmetric matrix is:}\&\begin{bmatrix}-13&6&13&2&-17\6&19&17&14&-7\13&17&7&-10&-13\2&14&-10&-6&-15\-17&-7&-13&-15&4\end{bmatrix}\end{align}$

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Question

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

Answer

$\begin{align}&\text{A symmetric matrix is equal to its transpose, that is to say:}\&A=A^{T}\&\text{A transpose of a matrix is one in which the first row becomes the first column,}\&\text{the second row the second column, and so on.}\&\text{The symmetric matrix is:}\&\begin{bmatrix}14&6&-12&-3&-4&4\6&17&17&-13&1&-4\-12&17&-13&-20&4&-17\-3&-13&-20&-2&16&9\-4&1&4&16&-10&-13\4&-4&-17&9&-13&-13\end{bmatrix}\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

$A = \begin{bmatrix} 0& -3 &4 \ 3& 0& 5\ -4&-5 & 0 \end{bmatrix}$

True or false: is an example of a skew-symmetric matrix.

Answer

A square matrix is defined to be skew-symmetric if its transpose $A^{T}$ - the matrix resulting from interchanging its rows and its columns - is equal to its additive inverse; that is, if

$A^{T} = -A$ .

Interchanging rows and columns, we see that if

$A = \begin{bmatrix} 0& -3 &4 \ 3& 0& 5\ -4&-5 & 0 \end{bmatrix}$ ,

then

$A^{T} = \begin{bmatrix} 0& 3 &-4 \ -3& 0&-5\ 4&5 & 0 \end{bmatrix}$ .

We see that each element of $A^{T}$ is the additive inverse of the corresponding element in , so $A^{T} = -A$ , and is skew-symmetric.

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Question

$A = \begin{bmatrix} a & 2& -3\ -2&a & 5 \ 3& -5& a \end{bmatrix}$

Which of the following must be true of for to be a skew-symmetric matrix?

Answer

A square matrix is defined to be skew-symmetric if its transpose $A^{T}$ - the matrix resulting from interchanging its rows and its columns - is equal to its additive inverse; that is, if

$A^{T} = -A$ .

Interchanging rows with columns in , we see that if

$A = \begin{bmatrix} a & 2& -3\ -2&a & 5 \ 3& -5& a \end{bmatrix}$

then

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

Also, by changing each entry in to its additive inverse, we see that

$-A = \begin{bmatrix} -a & -2& 3\ 2&-a & -5 \- 3& 5& -a \end{bmatrix}$

Setting the two equal to each other, we see that:

$\begin{bmatrix} a &- 2& 3\ 2&a &- 5 \ - 3& 5& a \end{bmatrix}= \begin{bmatrix} -a & -2& 3\ 2&-a & -5 \- 3& 5& -a \end{bmatrix}$

The non-diagonal elements - all constants - are all equal. Looking at the diagonal elements, we see that it is necessary and sufficient for ; that is, must be its own additive inverse. The only such number is 0, so .

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Question

is a three-by-three nonsingular skew-symmetric matrix

Then which of the following must be equal to $A^{T}A^{-1}$ ?

Answer

A square matrix is defined to be skew-symmetric if its transpose $A^{T}$ - the matrix resulting from interchanging its rows and its columns - is equal to its additive inverse; that is, if

$A^{T} = -A$ .

Therefore, by substitution,

$A^{T}A^{-1} = (-A)A^{-1}$

$A^{T}A^{-1} = (-1 A)A^{-1}$

$A^{T}A^{-1} = -1(AA^{-1})$

$A^{T}A^{-1} = -1I$

$A^{T}A^{-1}$ must be equal to the opposite of the three-by-three identity matrix, which is $\begin{bmatrix} -1& 0& 0\ 0&-1 &0 \ 0&0 &-1 \end{bmatrix}$ .

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