Calculate $\mathbb{E}\{ H X H^*\}$

Question

Let $X$ be an $n \times n$ matrix. Also, let $H$ be an $m \times n$ matrix with i.i.d. elements.
My aim is to calculate the expectation $$\mathbb{E}\{ H X H^*\},$$ where $H^*$ denotes the the conjugate transpose of $H$. When I was reading a paper, I saw the following relation $\mathbb{E}\{ H X H^*\}=$trace$(X) \mathbb{E}\{H H^* \}$. Is this correct ? if yes, how to prove it?

Answer 1

Here is the case when $H$ is just a vector and the i.i.d. elements have mean $0$ \begin{align} \mathbb{E}[H^*XH] &= \sum_{i,j=1}^n \mathbb{E}[\bar{H_i} X_{i,j} H_j] \\ &= \sum_{i=1}^n X_{i,i} \mathbb{E}[ \bar{H_i} H_i] + \sum_{i=1}^n \sum_{j\neq i} X_{i,j} \mathbb{E}[\bar{H_i} H_j ] \\ &= \sum_{i=1}^n X_{i,i} \mathbb{E}[\bar{H}_1 H_1] + \sum_{i =1}^n \sum_{j \neq i} X_{i,j} \mathbb{E}[\bar{H}_i] \mathbb{E}[H_j] \\ &=\textrm{trace}(X) \mathbb{E}[H^*H] \end{align}

If the mean of the i.i.d. elements is not zero, then consider $X = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)$, which will give, $H$ is still a vector, $$ \mathbb{E}[H^* XH]=\mathbb{E}[\bar{H}_2] \mathbb{E}[H_1] \neq 0 = \textrm{trace}(X) \mathbb{E}[H^*H].$$

So I think you need i.i.d. elements with mean $0$.