Gilbert Strang Frobenius question

Question

I wish to show that $$\|AB\|_F \leq (\Sigma_{j=1}^{n}\|a_j\|\cdot\|b_j^T\|),$$ where $a_j$ is the j:th colon and $b_j^T$ the j:th row.

My idea of a solution is to consider the SVD of $AB =U\Sigma V^T$. Setting $A' = U^T A$ and $B' = BV^T$, the Frobenius norm of A'B' is the same as that of AB, as U,V are orthonormal. Moreover, since $A'B' = \Sigma$ , $$(\Sigma\|{a'}_j \|\cdot \| {b'}_j^T \|) \geq (\Sigma \|{a'}_j\cdot {b'}_j^T \|) = \sigma_1 +...\sigma_n.$$ Applying Since U,V are orthogonal matrices, $$(\Sigma\|a_j\|\cdot\|b_j^T) = (\Sigma\|{a'}_j \|\cdot \|{b'}_j^T \|)$$ This implies the inequality.

Am I correct here?

Answer 1

No need for SVD.

Note that $AB = \sum_k a_k b_k^T$ and so $\|AB\|_F \le \sum_k \|a_k b_k^T\|_F = \sum_k \|a_k\|_2 \|b_k\|_2$.