How to show that following relation is true for matrix multiplication?

Question

I have one complex matrix $\mathbb{A}$ (of size $L\times L$) and one complex column vector $\mathbb{b}$ (of size $L$). How can I show that $$\mathbb{A^Hbb^HA}=\sum_{i=1}^{L}\mathbb{c_i^Hb}\mathbb{c_i}^T\mathbb{b^*}$$ where $\mathbb{c_i^H}$ is the $i$-th row of $\mathbb{A^H}$. Any help in this regard will be much appreciated. Thanks in advance. Please note that superscript $\mathbb{H}$ denotes the hermitian operator and superscript $T$ denotes the transpose operator. Further the superscript $*$ denotes the conjugate.

Answer 1

Let us first calculate $A^H b$ where $H$ is the hermitian operator. That is, $$ (A^H)_{ij} = A_{ji}^*$$ where $*$ represents the complex conjugate.

\begin{align} A^H b b^H A & = (A^H b) (b^H A) \\ & = (A^H b) (A ^H b )^H \\ & = y y^H \text{ where } y = A^H b \end{align}

Now calculate $y $.

$$ y = A^H b = \begin{bmatrix} c_1^H b \\ \vdots \\ c_L^H b \end{bmatrix}$$ where $c_i^H$ is the $i$th row of $A^H$.

Now note that $$y ^H = \begin{bmatrix} c_1^H b \\ \vdots \\ c_L^H b \end{bmatrix} ^ H = \begin{bmatrix} (c_1^H b)^* & \dots & (c_L^H b)^* \end{bmatrix} = \begin{bmatrix} c_1^T b^* & \dots & c_L^T b^* \end{bmatrix}(\text{ do you know why?})$$ Thats it,

Now its just a matrix multiplication. $$ y y^H = \sum_{i = 1} ^L c_i ^ H b c_i^T b^*$$