Why can the trace of covariance matrix be the power?

Question

Why can the trace of covariance matrix be the power ? I mean let the $W_b$ is the covariance matrix of signal $s_b$,why can $tr(W_b)$ be equal to the power of $s_b$ ? Is there any proof about this?

Answer 1

Let $s_b = [s_b(1) \dots s_b(n)]^t \in \mathbb{C}^n$ be your random signal with the covariance matrix $W_b = \mathbb{E}[s_b s_b^H]$ its entries are $W_b[i,j] = \mathbb{E}[s_b(i) s_b(j)^H]$ . The power of the signal is $P_{s_b} = \mathbb{E}[s_b^H s_b] = \sum_{i=1}^n \mathbb{E}[|s_b(i)|^2]$ which is a summation over the diagonal elements of the matrix $W_b[i,i]$ , this is the main definition of the trace of a matrix. Thus:

$\mathrm{trace}(W_b) = \sum_{i=1}^n W_b[i,i] = \sum_{i=1}^n \mathbb{E}[|s_b(i)|^2] = P_{s_b}$

$\square$