The Frobenius norm of a $m \times n$ matrix $F$ is defined as
$$\| F \|_F^2 := \sum_{i=1}^m\sum_{j=1}^n |f_{i,j}|^2$$
If I have $FG$, where $G$ is a $n \times p$ matrix, can we say the following?
$$\| F G \|_F^2 = \|F\|_F^2 \|G\|_F^2$$
Also, what does Frobenius norm mean? Is it analogous to the magnitude of a vector, but for matrix?
Best Answer
Actually there is $$||FG||^2_F \leqslant||F||^2_F||G||^2_F$$ The proof is as follows. \begin{align} \|FG\|^2_F&=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{p}\left|\sum\limits_{k=1}^nf_{ik}g_{kj}\right|^2 \\ &\leqslant\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{p}\left(\sum\limits_{k=1}^n|f_{ik}|^2\sum\limits_{k=1}^n|g_{kj}|^2\right)\tag{Cauchy-Schwarz} \\ &=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{p}\left(\sum\limits_{k,l=1}^n|f_{ik}|^2|g_{lj}|^2\right) \\ &=\sum\limits_{i=1}^{m}\sum\limits_{k=1}^{n}|f_{ik}|^2\sum\limits_{l=1}^{n}\sum\limits_{j=1}^{p}|g_{lj}|^2 \\ &=\|F\|^2_F\|G\|^2_F \end{align} Frobenius norm is like vector norm and similar to $l_2$.