Minimizing Frobenius matrix norm

Question

I am looking for a way to determine a complex matrix ${\bf C} \in \mathbb{C}^{n\times m}$, $m\leq n$ such that

$$ \| {\bf A} {\bf C} - {\bf I}_m \|_{\rm F}^2 $$

gets minimized, where $\mathbf{I}_m$ is the $m$-dimensional identity matrix, $\| \cdot \|_{\rm F}$ is the Frobenius norm and ${\bf A} \in \mathbb{C}^{m \times n}$ is known. In principle, there are no further assumptions made to the matrices $\bf A$ and $\bf C$. I have already figured out that a SVD and the low-rank approximation may be some clues for solving that problem, but I did not manage to apply this to my specific problem. Would it be helpful to choose $\bf C$ to be the pseudo-inverse of $\bf A$? I'd be grateful for all helpful advices or even solutions!

Answer 1

Considering the definition of the pseudoinverse I think you have already answered your question.

Answer 2

Given non-thin $\mathrm A \in \mathbb C^{m \times n}$, we have the unconstrained quadratic program (QP) in $\mathrm X \in \mathbb C^{n \times m}$

$$\text{minimize} \quad \| \mathrm A \mathrm X - \mathrm I_m \|_{\text{F}}^2$$

If $\mathrm A$ is square ($m=n$) and has full rank, then the minimizer is the inverse of $\mathrm A$

$$\hat{\mathrm X} := \mathrm A^{-1}$$

If $\mathrm A$ is fat ($n > m$) and has full row rank, then the minimizer is the right inverse of $\mathrm A$

$$\hat{\mathrm X} := \mathrm A^* (\mathrm A \mathrm A^*)^{-1}$$

Let us now consider the general case. Note that

$$\| \mathrm A \mathrm X - \mathrm I_m \|_{\text{F}}^2 = \mbox{tr} \left( (\mathrm A \mathrm X - \mathrm I_m)^* (\mathrm A \mathrm X - \mathrm I_m) \right) = \mbox{tr} \left( \mathrm X^* \mathrm A^* \mathrm A \mathrm X \right) - \langle \mathrm X, \mathrm A^* \rangle - \langle \mathrm A^*, \mathrm X \rangle + m$$

Taking the derivative with respect to $\mathrm X$ and finding where it vanishes, we obtain the following matrix equation

$$\boxed{\mathrm A^* \mathrm A \mathrm X = \mathrm A^*}$$

which always has a solution. Note that if $\mathrm A^{-1}$ or $\mathrm A^* (\mathrm A \mathrm A^*)^{-1}$ exist, they are solutions to the matrix equation above. Note that if $\mathrm A$ is fat, then the matrix equation above has infinitely many solutions.

Answer 3

Yes, the Moore-Penrose pseudoinverse will get the product matrix $\mathbf{A}\mathbf{A}^{+}$ as close to the identity as possible in the $2-$norm.