Right Inverse from Singular Values

Question

Let $A$ be a matrix with SVD $A = U \Sigma V^*$. Suppose that $$ \Sigma = \begin{bmatrix}1&0&0&0\\ 0&2&0&0\\0&0&3&0\\0&0&0&0\end{bmatrix}.$$ The right-inverse of $A$ is the matrix $B$ such that $AB = I$. Determine if the right inverse is unique, and if not, find two such right inverses.

This question seems to ask a lot with very limited information. My analysis so far is as follows: Consider the square matrix $AA^*$, which is clearly invertible. Then the matrix $B = A^*(AA^*)^{-1}$ is a right inverse of $A$, and since $A$ is a $3 \times 4$ matrix, $B$ is a $4 \times 3$ matrix here. I'm trying to conclude something about the dimensionality of the null space, but it appears that can't be done from this line of reasoning.

Moreover, if the right inverse were not unique, how would I go about determining two of them? From the singular values alone, it seems like there are an infinite number of potential matrices.

Answer 1

You know that $B$ is $4\times 3$. You want $$U\Sigma V^*B=I.$$ If you multiply by $U^*$ on the left and by $U$ on the right, you get $$ \Sigma V^*BU=I.$$ So it is enough to look for right inverses $R$ for $\Sigma$, and then $B=VRU^*$ will give you a right inverse for $A$.

An obvious right inverse for $\Sigma$ is $$ R_0=\begin{bmatrix} 1&0&0\\ 0&1/2&0\\0&0&1/3\\ 0&0&0\end{bmatrix} $$ (this one is $A^*(AA^*)^{-1}$, by the way). But since the last column of $\Sigma$ is zero, anything in the fourth row of the right inverse will have no effect when doing $\Sigma R_0$. That is, $$ \begin{bmatrix} 1&0&0\\ 0&1/2&0\\0&0&1/3\\ r&s&t\end{bmatrix}. $$ is a right inverse for $\Sigma$ for any choice of $r,s,t$. So now you have infinitely many right inverses for $\Sigma$, which will give you infinitely many right inverses for $A$. Note that because $U,V$ are unitaries, if $V^*R_1U=V^*R_2U$, then $R_1=R_2$; thus each choice of $r,s,t$ will produce a different right inverse.