Let $A$ and $B$ be two rectangular matrices of same size, with more rows than columns. Let $B^+$ denote the pseudoinverse of $B$.
It seems it is common to solve $A=BM$ for $M$ by computing $M=B^+ A$; does this corresponds to minimizing $\|A-BM\|$? According to simple numerical computations, no.
Is there a simple way to minimize $\|A-BM\|$?
It all boils down to the choice of the matrix norm.
We know that the solution to the least-squares problem $\min \|Bx - a\|_2$ is given by $B^+a$. The idea is to interpret $\min \|BX-A\|$ as a series of vector least-squares problems.
Let me denote the columns of $X$ and $A$ by $x_i$ and $a_i$, respectively. To this end, let me use the Frobenius norm. Then $$ \|BX-A\|_F^2 = \sum_{i=1}^n \|Bx_i - a_i\|_2^2. $$ We know that the choice $x_i = B^+a_i$ minimizes the right-hand side. Hence the choice $X=B^+A$ minimizes the right-hand side.