If we let $E=u^* v$ which an outer product. Then is the frobenius norm of E
$||E||_F=||U||_F||V||_F$
How would I show this is true.
The frobenius norm is $E=(\sum_{i=1}^{m} \sum_{j=1}|e_{ij}^2|)^{1/2}$
But I am not sure how to proceed
Showing a property of the frobenius norm
1
$\begingroup$
matrices
norm
1 Answers
2
You have
$$ \| E \|_F^2 = \sum |u_i v_j|^2 = \left( \sum |u_i|^2 \right) \left( \sum |v_j|^2 \right) = \| U \|_{F}^2 \| V \|_{F}^2. $$