Frobenius Norm confusion on Hermitian and Transpose on matrix form

Question

I am taking an ML class, in which I must minimize the function: $\|A-XB\|_{2}^F$. Using the mathematics cookbook for ML, we know that $\|A\|_2^F = \sqrt{\operatorname{Tr}\big(AA^H\big)}\,$. Since I am given that $X$ is a real numbered matrix, I thought that $A^H=A^T$, so I know get $\sqrt{\operatorname{Tr}\big((A-XB)(A-XB)^T\big)}\,$.

But my professor and some online sources suggested that it in fact the other way around, meaning it equals $\sqrt{\operatorname{Tr}\big(A^TA\big)} $, so $\sqrt{\operatorname{Tr}\big((A-XB)^T(A-XB)\big)}\,$.

I am confused. Which is the right one?

Answer 1

The trace has the important property of being cyclically invariant: $$\operatorname{Tr}(ZA)\:=\:\operatorname{Tr}(AZ)\quad\forall\,A,Z\in\mathcal M_n$$

Thus, Answer( "Which is the right one?" ) = "Both."