Let $h_{i} \in \mathbb{C}^{1\times M}$ and $M$ is a postive constant. $H \in \mathbb{C}^{K\times M}$ is a matrix including $K$ vectors, i.e. $H = [h_{1}^{T},...,h_{K}^{T}]^{T}$, $K<M$. Let $W$ represent the right inverse matrix of $H$ and is formulated as $H^*(HH^*)^{-1}$. $*$ indicates the conjugate transpose. Colunm vectors are used to experss $W$ as $[w_{1},w_{2},...,w_{K}]$.
Question: Determining the two norm of column vector in $W$, i.e. $ \left\|w_{i}\right\|^{2} $.
I calculated that as:
$ \left\|w_{i}\right\|^{2} $
$= w_{i}*w_{i}^{*}$
$= [(H^*(HH^*)^{-1})(H^*(HH^*)^{-1})^*]_{i,i}$
$= [(H^*(H^*)^{-1}H^{-1})(((HH^*)^{-1})^{*}H)]_{i,i}$
$= [H^{-1}((HH^*)^{-1})H)]_{i,i}$
$= [H^{-1}((H^*)^{-1}H^{-1}H)]_{i,i}$
$= [H^{-1}(H^*)^{-1}]_{i,i}$
$= [(H^*H)^{-1}]_{i,i}$
However, the answer is $[(HH^*)^{-1}]_{i,i}$. I think there must be something wrong with the calculation, which I can't find. I want to know what is the correct calculation process.
You've missed a $*$ between lines 4 and 5:
$= [(H^*(H^*)^{-1}H^{-1})(((HH^*)^{-1})^{*}H)]_{i,i}$
$= [H^{-1}((HH^*)^{-1})^*H)]_{i,i}$
Expand the inverse first, and then transpose.
You also need to use $(H^{-1})^*=(H^*)^{-1}$.