What is the derivative of $1\over 2 $$\| P^TS-I\|^2_F$?

Question

What is the derivative of ${1}\over{2}$${\| P^TS-I\|^2_F}$? over $P$?

$P$ is a $n$ by $s$ matrix, $S$ is of the same dimension,

thanks

Answer 1

The Frobenius norm of a matrix equals the norm of its transpose. And its square is the same as the scalar matrix product, i.e. $$\|X^T\|^2_F = \|X\|^2_F = X:X$$

The differential of the scalar product is just $$d(X:X) = 2X:dX$$

In your particular function $X^T = (P^T\cdot{S}-I)$ and there's a scale factor of $({{1}\over{2}})$ so its differential is

$$ \eqalign { (S^T\cdot{P}-I):S^T\cdot dP \cr S\cdot(S^T\cdot{P}-I):dP \cr } $$

The derivative wrt $P$ is $$S\cdot(S^T\cdot{P}-I)$$