Given matrices $A, B \in \mathbb{R}^{m \times n}$, I want to solve the following least-squares problem
$$ \min_{X \in \mathbb{R}^{n \times n}} \| A - B X \|_F^2 $$
where $\| \cdot \|_F$ is the Frobenius norm. In this problem, $m \ge n$.
Does anyone know how to solve this? Or possibly convert it into an ordinary least squares problem?
Recall that for $A \in \mathbb{R}^{m \times n},$
\begin{align*} \| A \|^{2}_{F} = \textrm{trace}(A^{\top} A). \end{align*}
With this in mind, consider the following (very relevant) post:
Derivative of squared Frobenius norm of a matrix
Hence, you can differentiate your expression in the usual way (being mindful of 2nd-order conditions for optimality), and find a solution to your optimization problem.
Additionally, a useful reference for a variety of matrix identities (including matrix calculus) is "The Matrix Cookbook" by Kaare Brandt Petersen and Michael Syskind Pedersen. A version of the document can be found here:
For future readers, the post linked by the accepted answer leads to the following result, $$ X = (B^T B)^{-1} B^T A $$ This system could be solved with a Cholesky decomposition of $B^T B$, however a more numerically stable method is to use a $QR$ decomposition of $B$. \begin{align} \left|\left| A - B X \right|\right|_F^2 &= \left|\left| A - Q \pmatrix{R \\ 0} X \right|\right|_F^2 \\ &= \left|\left| Q^T A - \pmatrix{R \\ 0} X \right|\right|_F^2 \\ &= \left|\left| C_1 - R X \right|\right|_F^2 + || C_2 ||_F^2 \end{align} where $$ Q^T A = \pmatrix{C_1 \\ C_2} $$ and $C_1$ is the first $n$ rows of $Q^T A$. Then, we just need to minimize the term $|| C_1 - R X||_F^2$ which can be made zero by the choice, $$ X = R^{-1} C_1 $$ This assumes of course that $B$ is invertible.
