Can a matrix have more than one inverse (Singular Value Decomposition)

Question

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?

Answer 1

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 $3$.

Answer 2

1. The pseudoinverse matrix is unique.
2. The singular values are unique.
3. 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.