In Is this rank-$1$ matrix semidefinite?, I have seen that $X = xx^T$ is PSD when $x$ is real. What about the case when $X$ is Hermitian?
I know that it is PSD but I'm not exactly sure how to prove it. One final question is about the rank of $X$ if $x$ is a vector, it should be one. Is that right?
As you have already computed, with $X = xx^*$, we have $$ v^*Av = |x_1v_1|^2 + \bar x_1 v_1 x_2 \bar v_2 + x_1 \bar v_1 \bar x_2 v_2 + |x_2v_2|^2. $$ Verify that this is the same as $$ |x^*v|^2 = x^*v \overline{x^*v} = (x_1v_1 + x_2v_2)(\overline{x_1 v_1 + x_2 v_2}). $$ Alternatively, simply note that $$ v^*Xv = v^*xx^*v = (v^*x)(v^*x)^* = |v^*x|^2. $$ In either case, we note that $v^*Xv$ is the magnitude of a complex number and is therefore non-negative.