what is the solution to ${\partial{\|A-RY\|^2_F}\over{\partial{Y}}}= 0$?

Question

1. what is the solution to ${\partial{\|A-RY\|^2_F}\over\partial{Y}}= 0$? $A$ is $m$ by $n$, $R$ is $m$ by $s$ and $Y$ is $s$ by $n$

2. How to solve a quadratic equation?

$AXA^T+B^TXB + X=C$, where X is variable matrix, and $A,B,C$ are constant matrices

Thanks

Answer 1

1) For any matrix $\mathbf{W}$, the square of the Frobenius norm is simply the scalar product, i.e.

$$\mathbf{\|W\|^2_F \equiv W:W}$$

The differential of which is $\mathbf{d(W:W) = 2\,W:dW}$

In your case $\mathbf{W = R.Y-A}$, so $$\mathbf{dW = R.dY}$$

Substituting into the differential yields $$\mathbf{2(R.Y-A):R.dY = 2R'.(R.Y-A):dY}$$

Since this must equate to zero for any $\mathbf{dY}$, we're left with $$\mathbf{R'.(R.Y-A) = 0}$$

Solve for $\mathbf{Y}$ in a least-squares sense to obtain

$$\mathbf{Y = P.A}$$

where $\mathbf{P}$ is the pseudo-inverse of $\mathbf{R}$.

2) AFAIK there is no closed-form solution for the matrix quadratic, but the equation you've written is not quadratic in $\mathbf{X}$ so it can be solved with a "kronecker-vec" transformation.

Answer 2

I suggest a slight correction of the answer proposed by @lynne.

The matrix equation: $$ (RX-A)R' = 0 $$ If the matrix equation is consistent, i.e., it indeed has at least one solution. The general solution has the form: $$ X = R^{\dagger} (A R') R'^{\dagger} + G - R^{\dagger} R \cdot G \cdot R' R'^{\dagger} $$ where $ G $ is an arbitrary matrix $ \in\mathbb{R}^{s \times n} $.

(?) I'm not sure, if the other arbitrary matrix $F$ is necessary, which offers more "Degrees of Freedom" in the general solution.

The solution is unique, if and only if all columns of $R$ are linearly independent. And thus $$ X = R^{\dagger} A $$

btw, I think: $$ 2(R.Y−A):R.dY = 2R′.(R.Y−A):dY $$