Question

Show that the following matrices are Hermitian whether Ais Hermitian or not:$\mathit{A}{\mathit{A}}^{\mathbf{t}}\mathbf{,}\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}\mathbf{,}\mathit{i}\left(A-A^t\right)$ .

Short answer

The $i\left(A-A^T\right)$is the Hermitian matrix.

Step-by-step solution

Given information.

The given matrices are:

_{$A{A}^T,A+A^T,i\left(A-A^T\right)$}

Hermitian matrix and condition for a matrix to be Hermitian.

A Hermitian matrix is one in which the conjugate transpose of a complex square matrix is equal to itself.

Any square matrix Ais said to be Hermitian. If,
$A^{*}=A^T$

Where Ais the complex conjugate and $A^T$is the transpose of A.

Show that the first and second equations are Hermitian or not.

Solve for the first case.

$\left({\left(A{A}^T\right)}^T={\left(A^T\right)}^T{A}^T\right)$

Since,

$$\begin{gathered} {\left(AB\right)}^T=B^T{A}^T \\ =A{A}^T \\ \left(A{A}^T\right)=A{A}^T \end{gathered}$$

Hence, $A{A}^T$is the Hermitian matrix

Now solve for the second case:

$\left(A+A^T\right)=A^T+{\left(A^T\right)}^T$

Since,

${\left(A+B\right)}^T=A^T+B^T$

Then,

$A^T+A=A^T$

Hence,

$\left(A+A^T\right)=A+A^T$

Therefore, the $A+A^T$is the Hermitian matrix.

Show that the third equation is Hermitian or not.

Now solve the third equation:

${\left(i{\left(A-A\right)}^T\right)}^T=i{A}^T-{\left(A^T\right)}^T$

Since,

$\left(\because {\left(A-B\right)}^T=A^T-B^T\right)$

Then,

$$\begin{gathered} i\left(A^T-{\left(A^T\right)}^T\right)=-i\left(A^T-A\right)\left(\because \overline{i}=-i\right)\overline{i} \\ {A}^T-\left(A^T\right)=i\left(A-A^T\right) \\ {\left(i{\left(A-A\right)}^T\right)}^T=i\left(A-A^T\right) \end{gathered}$$

Therefore, the $i\left(A-A^T\right)$is the Hermitian matrix.