# Right-inverse approximation in Frobenius Norm

Let $$A \in \mathbb{R}^{m\times n}$$, with $$m \geq n$$, be a matrix of rank $$r$$, and suppose we have a SVD decomposition $$A = U\Sigma V^t$$.

We define the pseudo-inverse of $$A$$ as $$A^{\dagger} := V\Sigma^{\dagger}U^t$$, where $$\Sigma^{\dagger}_{i, j} = 0$$, if $$\Sigma_{i, j} = 0$$, and $$\Sigma^{\dagger}_{i, j} = 1/\sigma_i$$, if $$\Sigma_{i, j} = \sigma_i$$.

The question is to prove that $$A^{\dagger}$$ is the matrix with lower Frobenius norm that solves the problem

$$\min_{B \in \mathbb{R}^{n \times m}} ||AB - I_m ||_F$$

• Frobenius norm is unitarily invariant. So, you may assume that $A=\Sigma$. This special case should be easy. – user1551 Aug 19 '19 at 15:22

If $$B=[b_1,\ldots,b_m]$$ where $$b_i$$ are the columns of $$B$$ and $$e_i$$ are the columns of the identity $$I_m$$, you can write $$\|AB-I_m\|_F^2=\sum_{i=1}^m\|Ab_i-e_i\|_2^2,$$ so the Frobenius norm minimization can be turned into $$m$$ independent minimizations $$\min_{B}\|AB-I_m\|_F^2=\sum_{i=1}^m\min_{b_i}\|Ab_i-e_i\|_2^2.$$ Now we have $$m$$ linear least squares problems on the right-hand side and if you know that the solution to a linear LS is the pseudo-inverse solution, we have $$b_i=A^\dagger e_i$$, that is, the $$i$$th column of $$B$$ is the $$i$$th column of the MP pseudo-inverse and hence $$B=A^\dagger$$.