How to show $BB^\dagger$ is hermitian?

Question

I have B is a $m\times n$ matrix, with complex entries. Then I want to show that $BB^{\dagger}$ is Hermitian.

In order to satisfy Hermiticity, then $(BB^{\dagger})^{\dagger}$=$BB^{\dagger}$.

I set

$B$ = $\begin{vmatrix} i & a & c\\ b+i & i & d \end{vmatrix}$

$B^{\dagger}$ = $\begin{vmatrix} -i & b-i \\ a & -i \\ c & d \end{vmatrix}$

$B*B^{\dagger} = \begin{vmatrix} a2+c2−1 & ia+i(b−i)+cd \\ ia+i(b−i)+cd & d2+(b−i)(b+i)−1 \end{vmatrix}$

So then I do:

$(B*B^{\dagger})^{\dagger} = \begin{vmatrix} a2+c2−1 & -ia+i(b+i)+cd \\ -ia+i(b+i)+cd & d2+(b+i)(b-i)−1 \end{vmatrix}$

Is this all?

Answer 1

You must be misunderstood the defenition of Hermitian matrix. It is not $BB^\star=B^\star B$, but $B=B^\star$.

You can use this three properties, are valid for all matrices: $A^\star=\bar{A}^T, \bar{A}^T=\bar{A^T}, (AB)^\star=B^\star A^\star$. ($\star$ is Hermitian adjoint operator.)

Then...

$$(BB^\star)^\star=(B^\star)^\star B^\star=\bar{(\bar{B^T})}^T B^\star=(B^T)^T B^\star=BB^\star$$

So. $BB^\star$ is Hermitian matrix.