Can a matrix have more than one inverse (Singular Value Decomposition) Assume there's a matrix $A$ with SVD as below
$$
A = U
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 3 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
V^*
$$
There's a matrix $B$, for which $AB = I$.
Question, can $A$ have two inverse matrices? IF yes, what would they be? If no, why not?
I think there are, and here's my first matrix:
I assume $AA^{-1} = I$, so $C^+C = I$, where $A^{-1}$ is actually the matrix, and $A$ is the left inverse of $A^{-1}$. After a series of calculations, I will get 
$$
A^{-1} = V
\begin{bmatrix}
1/1 & 0 & 0 & 0 \\
0 & 0.5 & 0 & 0 \\
0 & 0 & 1/3 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
U^*
$$
However, I don't know how to find another one though. Maybe I'm wrong at the beginning?
 A: In a singular value decomposition, the matrices $U$ and $V$ are unitary, so invertible. Hence the rank of $A$ is the same as the rank of the singular value matrix, which is $3$. So $A$ is not invertible.
What you write as
$$
V
\begin{bmatrix}
1/1 & 0 & 0 & 0 \\
0 & 1/2 & 0 & 0 \\
0 & 0 & 1/3 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
U^*
$$
is the pseudoinverse $A^+$ of $A$, not the inverse. And $AA^{+}$, in this case, is not the identity, because the rank of $A$ is $3$.
A: *

*The pseudoinverse matrix is unique. 

*The singular values are unique. 

*The singular value decomposition is not unique.


Example:
$$
\left[
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
 a & 0 \\
 0 & b \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{rr}
 -1 & 0 \\
 0 & -1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
 a & 0 \\
 0 & b \\
\end{array}
\right]
%
\left[
\begin{array}{rr}
 0 & -1 \\
 -1 & 0 \\
\end{array}
\right]
$$
The target matrix you posit has rank 3:
$$
  \mathbf{A} \in \mathbb{C}^{4\times 4}_{3}
$$
The singular value decomposition for your matrix is
$$
  \mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
\left[ \begin{array}{cc}
  \color{blue}{\mathbf{U}_{\mathcal{R}\left( \mathbf{A}\right)}} &
  \color{red}{\mathbf{U}_{\mathcal{N}\left( \mathbf{A}^{*}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
  \mathbf{S}  & \mathbf{0} \\
  \mathbf{0} & 0
\end{array} \right]
%
\left[ \begin{array}{l}
  \color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*}\right)}}^{*} \\
  \color{red}{\mathbf{V}_{\mathcal{N}\left( \mathbf{A}\right)}}^{*}
\end{array} \right]
%
$$
where
$$
  \mathbf{S} =
\left[
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 2 & 0 \\
 0 & 0 & 3 \\
\end{array}
\right]
$$
The Moore-Penrose pseudoinverse matrix is
$$
  \mathbf{A}^{\dagger} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
\left[ \begin{array}{cc}
  \color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*}\right)}} &
  \color{red}{\mathbf{V}_{\mathcal{N}\left( \mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
  \mathbf{S}^{-1}  & \mathbf{0} \\
  \mathbf{0} & 0
\end{array} \right]
%
\left[ \begin{array}{l}
  \color{blue}{\mathbf{U}_{\mathcal{R}\left( \mathbf{A}\right)}}^{*} \\
  \color{red}{\mathbf{U}_{\mathcal{N}\left( \mathbf{A}^{*}\right)}}^{*}
\end{array} \right]
%
$$
We can play games with signs on the pseudoinverse also. Both the matrix and its pseudoinverse are unique, up to the signs. A virtue of the SVD is that is orients an alignment between the unitary decomposition of the row and column spaces.
