Question

How do I know if a matrix A,is symmetric, unitary , hermitian,skew hermitian and nonsingular

Answer 1

Answer:

very simple bro

Step-by-step explanation:

you just need to know two concept that are transpose and conjugate.

please first this two definition and then ask help.

I'll help you sure

Answer 2

$\large\underline{\sf{Solution-}}$

1. Symmetric Matrix

A square matrix 'A' is said to be symmetric if and only iff A' = A.

where A' means transpose of matrix A.

For example,

If

$\rm :\longmapsto\:A = \begin{gathered}\sf\left[\begin{array}{ccc}1&-1&2\\ - 1&4&-5\\2&-5&3\end{array}\right]\end{gathered}$

then

$\rm :\longmapsto\: {A}^{T} \: or \: {A}^{'} = \begin{gathered}\sf\left[\begin{array}{ccc}1&-1&2\\ - 1&4&-5\\2&-5&3\end{array}\right]\end{gathered} = A$

2. Hermitian Matrix

A square matrix 'A' is said to be Hermitian iff

$\boxed{ \sf{ \: {A}^{\theta} = A}}$

$\rm :\longmapsto\:where \: {A}^{ \theta} \: means \: transpose \: conjugate.$

For example :-

If

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}1&-i&1 - i\\i&4&1 + i\\1 + i&1 - i&3\end{array}\right]\end{gathered}$

then

$\rm :\longmapsto\: {A}^{\theta} = \begin{gathered}\sf \left[\begin{array}{ccc}1&-i&1 - i\\i&4&1 + i\\1 + i&1 - i&3\end{array}\right]\end{gathered} = A$

3. Skew - Hermitian Matrix

A square matrix A is said to be skew - hermitian iff

$\boxed{ \sf{ \: {A}^{\theta} = - \: A}}$

Remark :- The main diagonal elements of skew - hermitian matrix are always 0 or purely imaginary.

For example

If

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}0&i&2 - i\\ - i&0&1 - i\\ - 2 + i& - 1 + i&0\end{array}\right]\end{gathered}$

then

$\bf\implies \: {A}^{\theta} = - A$

For example

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}3i&1 + i \\ - 1 + i&2i \end{matrix} \bigg]$

4. Unitary matrix

A square matrix A said to be Unitary matrix iff

$\rm :\longmapsto\:A {A}^{\theta} = {A}^{\theta}A = I$

For example

If

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}\dfrac{1}{ \sqrt{2} } & \dfrac{1}{ \sqrt{2} }i \\ \dfrac{1}{ \sqrt{2} }& - \: \dfrac{1}{ \sqrt{2} }i \end{matrix} \bigg]$

Now,

$\rm :\longmapsto\: {A}^{\theta} A = \bigg[ \begin{matrix}\dfrac{1}{ \sqrt{2} } & - \dfrac{1}{ \sqrt{2} }i \\ \dfrac{1}{ \sqrt{2} }&\dfrac{1}{ \sqrt{2} }i \end{matrix} \bigg] \bigg[ \begin{matrix}\dfrac{1}{ \sqrt{2} } & \dfrac{1}{ \sqrt{2} }i \\ \dfrac{1}{ \sqrt{2} }& - \: \dfrac{1}{ \sqrt{2} }i \end{matrix} \bigg]$

$\rm \: = \: \:\bigg[ \begin{matrix}\dfrac{1}{2} - \dfrac{1}{2} {i}^{2} &\dfrac{1}{2} + \dfrac{1}{2} {i}^{2} \\ \dfrac{1}{2} + \dfrac{1}{2} {i}^{2} &\dfrac{1}{2} - \dfrac{1}{2} {i}^{2} \end{matrix} \bigg]$

$\rm \: = \: \:\bigg[ \begin{matrix}\dfrac{1}{2} + \dfrac{1}{2}&\dfrac{1}{2} - \dfrac{1}{2}\\ \dfrac{1}{2} - \dfrac{1}{2}&\dfrac{1}{2} + \dfrac{1}{2} \end{matrix} \bigg]$

$\rm \: = \: \:\bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg]$

$\rm \: = \: \:I$