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. 
 A: 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.
A: Singular value decomposition
Begin with a nonzero matrix $\mathbf{A}\in\mathbb{C}^{m\times n}_{\rho}$ such that the matrix rank $\rho<m$ and $\rho<n$. The singular value decomposition, guaranteed to exist, is
$$
\mathbf{A}
= 
\mathbf{U} \, 
\Sigma \, 
\mathbf{V}^{*} 
=
\left[
  \begin{array}{cc}
    \color{blue}{\mathbf{U}_{\mathcal{R}}} &
    \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array}
\right]
%
\left[
  \begin{array}{c}
    \mathbf{S} & \mathbf{0} \\
    \mathbf{0} & \mathbf{0}
  \end{array}
\right]
%
\left[
  \begin{array}{c}
    \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\
    \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array}
\right].
$$
The SVD provides an orthnormal basis for the four fundamental subspaces:
$$
\begin{array}{ll}
%
 matrix & subspace \\\hline
%
  \color{blue}{\mathbf{U}_{\mathcal{R}}}\in\mathbb{C}^{m\times\rho} & 
     \color{blue}{\mathcal{R}\left(\mathbf{A}\right)} \\
%
  \color{blue}{\mathbf{V}_{\mathcal{R}}}\in\mathbb{C}^{n\times\rho} &
     \color{blue}{\mathcal{R}\left(\mathbf{A}^{*}\right)} \\
%
  \color{red}{\mathbf{U}_{\mathcal{N}}}\in\mathbb{C}^{m\times m-\rho} &
     \color{red}{\mathcal{N}\left(\mathbf{A^{*}}\right)} \\
%
  \color{red}{\mathbf{V}_{\mathcal{N}}}\in\mathbb{C}^{n\times n-\rho} &
     \color{red}{\mathcal{N}\left(\mathbf{A}\right)}
%
\end{array}
$$
There are $\rho$ singular values which are ordered and real:
$$
  \sigma_{1} \ge \sigma_{2} \ge \dots \ge \sigma_{\rho}>0.
$$
These singular values for the diagonal matrix of singular values
$$
\mathbf{S} = \text{diagonal} (\sigma_{1},\sigma_{1},\dots,\sigma_{\rho}) \in\mathbb{R}^{\rho\times\rho}.
$$
The $\mathbf{S}$ matrix is embedded in the sabot matrix $\Sigma\in\mathbb{R}^{m\times n}$ whose shape insures conformability.
$$
 \Sigma = 
\left[
  \begin{array}{c}
    \mathbf{S} & \mathbf{0} \\
    \mathbf{0} & \mathbf{0}
  \end{array}
\right]
=
% Sigma
  \left[ \begin{array}{cccc|cc}
     \sigma_{1} & 0 & \dots &  0&   0& \dots &  0 \\
     0 & \sigma_{2}  &&&&&  \\
     \vdots && \ddots &&\vdots&& \vdots\\
      0 & & & \sigma_{m} & 0 & \dots & 0 \\\hline
      0 & \dots && 0 & 0 & \dots & 0 \\
      \vdots &   && \vdots & \vdots & \ddots & \vdots \\
      0 & \dots && 0 & 0 & \dots & 0 \\
  \end{array} \right]
$$
Moore-Penrose pseudoinverse
The components of the SVD can be used to construct the Moore-Penrose pseudoinverse:
$$
\mathbf{A}^{\dagger}
= 
\mathbf{V} \, 
\Sigma^{\dagger} \, 
\mathbf{U}^{*} 
=
\left[
  \begin{array}{cc}
    \color{blue}{\mathbf{V}_{\mathcal{R}}} &
    \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array}
\right]
%
\left[
  \begin{array}{cc}
    \mathbf{S}^{-1} & \mathbf{0} \\
    \mathbf{0} & \mathbf{0}
  \end{array}
\right]
%
\left[
  \begin{array}{c}
    \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\
    \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array}
\right].
$$
A common source of confusion comes from $\Sigma$ manipulations, so the main forms are shown here. Note that $\mathbf{S}^{\mathrm{T}} = \mathbf{S}$. Pay close attention to the shapes:
$$
\begin{align}
%
  \Sigma_{m\times n} &=
\left[
  \begin{array}{cc}
    \mathbf{S} & \mathbf{0} \\
    \mathbf{0} & \mathbf{0}
  \end{array}
\right] , \quad
% transpose
  \Sigma^{\mathrm{T}}_{n\times m} =
\left[
  \begin{array}{cc}
    \mathbf{S} & \mathbf{0} \\
    \mathbf{0} & \mathbf{0}
  \end{array}
\right] , \quad
% inverse
  \Sigma^{\dagger}_{n\times m} =
\left[
  \begin{array}{cc}
    \mathbf{S}^{-1} & \mathbf{0} \\
    \mathbf{0} & \mathbf{0}
  \end{array}
\right]
%
\end{align}
$$
This provides a crisp construction of $\Sigma^{\dagger}$.
Derivation
To isolate the term $\Sigma^{\dagger}$, start with the definition of the psuedoinverse.
$$
\begin{align}
  \mathbf{A}^{\dagger}
&= 
\mathbf{V} \, 
\Sigma^{\dagger} \, 
\mathbf{U}^{*}  \\
%
  \mathbf{V}^{*}\, \mathbf{A}^{\dagger}
&= 
\mathbf{V}^{*}\, \mathbf{V} \, 
\Sigma^{\dagger} \, 
\mathbf{U}^{*}  \\
%
  \mathbf{V}^{*}\mathbf{A}^{\dagger}\,  \mathbf{U}
&= 
\Sigma^{\dagger} \, 
\mathbf{U}^{*}\, \mathbf{U}  \\
%
  \mathbf{V}^{*}\mathbf{A}^{\dagger}\,  \mathbf{U}
&= 
\Sigma^{\dagger}
%
\end{align}
$$
