Question about SVD and orthogonal matrices

Question

Let $X$ be a $m \times n$ matrix. By SVD, I obtain $X = UDV^T$, where $U$ and $V$ are both orthogonal matrices, and $D$ is a diagonal matrix.

I think the following is true (but not sure why):

$(VDV^T + aI_n)^{-1} = V(D + aI_n)^{-1}V^T$

where $a\in \mathbb{R}$ is some scalar, and $I_n$ is a $n\times n$ identity matrix. Can anyone explain to me why one is able to move $V$ to the left of the matrix inverse and $V^T$ to the right?

Answer 1

I don't know what the SVD has to do with it but yes, your manipulation is correct.

We are given that $D$ is diagonal and $V$ is orthogonal. Note that $V^{-1} = V^T$ and $(V^T)^{-1} = V$. So, we have $$ \begin{align} (VDV^T + aI_n)^{-1} &= (VDV^T + V(a I_n)V^T)^{-1} \\ & = (V(D + aI_n)V^T)^{-1} \\ & = (V^T)^{-1}(D + aI_n)^{-1}V^{-1} = V(D + aI_n)^{-1}V^T. \end{align} $$

Answer 2

If $V$ is invertible,

$$(VMV^{-1}+aI)^{-1}=(VMV^{-1}+aVV^{-1})^{-1}=(V(M+aI)V^{-1})^{-1}=V(M+aI)^{-1}V^{-1}.$$

If, in addition, $V$ is orthogonal, this is

$$V(M+aI)^{-1}V^T.$$

Nothing else is required.