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3 Questions around this concept.
If A is Hermitian such that A2 = 0, Then
$
If matrix A=\left[\begin{array}{ccc}1 & 0 & \omega^2 \\ 0 & \omega & 0 \\ 1 & 0 & \omega^2\end{array}\right] \text {. Then matrix } \mathrm{A}+A^\theta \text { is (where } \omega \text { is the cube root of unity }
$
Hermitian matrix
A square matrix $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{n} \times \mathrm{n}}$ is said to be Hermitian matrix if $a_{i j}=\overline{a_{j i}} \forall \mathrm{i}, \mathrm{j}$,
i.e. $A=A^\theta, \quad\left[\right.$ where $A^\theta$ is conjugate transpose of matrix $\left.A\right]$
We know that when we take the transpose of a matrix, its diagonal elements remain the same, and while taking conjugate we just change the sign from +ve to -ve and -ve to +ve for the imaginary part of all elements, So to satisfy the condition A? = A diagonal elements must not change, ⇒ all diagonal element must be purely real,
E.g.
Let, $A=\left[\begin{array}{ccc}3 & 3-4 i & 5+2 i \\ 3+4 i & 5 & -2+i \\ 5-2 i & -2-i & 7\end{array}\right]$
Then,
$
\begin{gathered}
\mathrm{A}^{\prime}=\left[\begin{array}{ccc}
3 & 3+4 i & 5-2 i \\
3-4 i & 5 & -2-i \\
5+2 i & -2+i & 7
\end{array}\right] \\
\therefore \mathrm{A}^\theta=\overline{\left(\mathrm{A}^{\prime}\right)}=\left[\begin{array}{ccc}
3 & 3-4 i & 5+2 i \\
3+4 i & 5 & -2+i \\
5-2 i & -2-i & 7
\end{array}\right]
\end{gathered}
$
here, A is Hermitian matrix as $\mathrm{A}=\mathrm{A}^\theta$
Note :
For any square matrix say A, with complex number entries,
$\mathrm{A}+\mathrm{A}^\theta$ is a Hermitian matrix
$
\left[\because\left(\mathrm{A}+\mathrm{A}^\theta\right)^\theta=\mathrm{A}^\theta+\left(\mathrm{A}^\theta\right)^\theta=\mathrm{A}^\theta+\mathrm{A}\right]
$
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