Singular Value Decomposition - Proof With singular value decomposition we can write the following:
\begin{equation}
A = U \Sigma V^{T}
\end{equation}
\begin{equation}
U^{T}AV=U^{T}U\Sigma V^{T} V
\end{equation}
Since $U,V$ orthogonal, the above equation leads to the following:
\begin{equation}
\Sigma =U^{T}AV 
\end{equation}
I've seen a proof that says the following
\begin{equation}
\Sigma^{-1}=V^TA^{-1}U
\end{equation}
Can someone help with to understand how we ended up to the latter equation. 

If all the matrices involved are square and invertible, we have $U^T = U^{-1}$ and $V^{T} = V^{-1}$, so
$$
\Sigma^{-1} = (U^{-1}AV)^{-1} = V^{-1}A^{-1}U = V^TA^{-1}U
$$
as desired.
