Can anyone help me with a relation:
I want to represent Frobenius norm of matrix $X$ in terms of l2 norm of constituent vectors $b$ and $a$
$\Vert X\Vert_F^2=\Vert ba \Vert_F^2$ where $b$ is a column vector and $a$ is a sparse row vector containing some zero elements.
something like this: $\Vert X\Vert_F^2 \le\ge = \Vert b\Vert_2\Vert a\Vert_2 $
Best Answer
For any matrix $A$, $\|A\|_F^2=\sum_{i=1}^r \sigma_i^2$, where $\sigma_i$ is the $i$th singular value, and $r$ is the rank of $A$. For a rank-$1$ matrix $A=ba^T$, the singular value decomposition becomes trivial: $$ A = \left( \frac{b}{\| b\|_2}\right)\left( \|b\|_2\|a^T\|_2\right)\left( \frac{a^T}{\| a^T\|_2}\right) $$
So clearly $\| A\|_F^2 = \left( \|b\|_2\|a^T\|_2\right)^2 = \|b\|_2^2\|a\|_2^2$.