Pseudoinverse matrix and SVD

Question

I'm trying to solve an homework question but I got stuck.

Let A be a m x n matrix with the SVD $A = U \Sigma V^*$ and $A^+ = (A^* A)^{-1} A^*$ its pseudoinverse.

I'm trying to get $A^+ = V \Sigma^{-1} U^*$, but I'm missing something.

Can anyone help me with this please?

Thanks!

Answer 1

$$ \begin{align} A^+ &= (A^*A)^{-1}A^* \\ &=(V\Sigma U^*U\Sigma V^*)^{-1} V\Sigma U^* \\ &=(V\Sigma^2 V^*)^{-1} V\Sigma U^* \\ &=(V^*)^{-1} \Sigma^{-2} V^{-1} V\Sigma U^* \\ &= V \Sigma^{-2}\Sigma U^* \\ &= V\Sigma^{-1}U^* \end{align} $$ using the properties of the matrices $U,V,\Sigma$ in the Singular_value_decomposition.

Answer 2

First you need to assume that the matrix $A^*A$ is invertible. For which you need $n \leq m$ and rank($A$) is $n$.

So when $n \leq m$ and when rank($A$) is $n$, then the reduced SVD of $A$ is $A = U \Sigma V^*$ where $U \in \mathbb{R}^{m \times n}$, $\Sigma \in \mathbb{R}^{n \times n}$ and $V \in \mathbb{R}^{n \times n}$ such that $U^* U = I_{n \times n}$, $V^* V = I_{n \times n}$, $V V^* = I_{n \times n}$ and $\Sigma$ is a square diagonal matrix and has only (positive) real entries.

Note that $V^{-1}=V^*$.

Also note that $A^* = V \Sigma^* U^* = V \Sigma U^*$ since $\Sigma^* = \Sigma$.

Further note that if $M_1,M_2 \text{and} M_3$ are invertible matrices then $(M_1 M_2 M_3)^{-1} = M_3^{-1} M_2^{-1} M_1^{-1}$.

Use these to get the final answer.