How to prove $\mathbf V \mathbf \Sigma^H \mathbf \Sigma \mathbf V^H$ has a inverse matrix

Question

I want to prove this matrix has inverse matrix

$\mathbf H_{AB}$ is a $2$ by $7$ matrix,now i do SVD to $\mathbf H_{AB}$,that is $\mathbf H_{AB}=\mathbf U \mathbf \Sigma \mathbf V^H$

Now,i want to prove that $\mathbf V \mathbf \Sigma^H \mathbf \Sigma \mathbf V^H$ has a inverse matrix ,i mean,if now $\mathbf B=\mathbf V \mathbf \Sigma^H \mathbf \Sigma \mathbf V^H$,then $\mathbf B^{-1}$ exists, and $\mathbf B\mathbf B^{-1}=\mathbf B^{-1}\mathbf B=\mathbf I$

Answer 1

The inverse pf $\mathbf{B}$ exists only if the diagonal square matrix $\mathbf \Sigma^H \mathbf \Sigma$ has non zero diagonal entries. So your $\mathbf{\Sigma}$ which looks like this $$\mathbf{\Sigma} = \begin{bmatrix} \sigma_1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 &\sigma_2 & 0 & 0 & 0 & 0 & 0 \\\end{bmatrix}$$ (or it's transpose) should maintain $\sigma_{1,2} \neq 0$