Frobenius Norm Inequality with SVD

Question

Let $A\in \mathbb{R}^{m\times n}$ and $x\in \mathbb{R}^n$ a column vector. I want to prove that $$||Ax||_2 \leq||A||_F||x||_2$$ using SVD where $||\cdot||_2$ is the euclidean norm and $||\cdot||_F$ is the Frobenius norm.

Answer 1

Let $A=USV^T$ be a SVD and let $y=V^Tx$. Since Frobenius norm and Euclidean norm are unitarily invariant, the inequality in question can be rewritten as $\|Sy\|_2\le\|S\|_F\|y\|_2$, which should be easy to prove.

Answer 2

We may assume wlog that $\|x\|_2=1$. First note that $\|Ax\|_2 \leq \sigma_{\mathrm{max}}$. This follows from $\|Ax\|^2_2= \langle Ax,Ax\rangle = \langle x, A^T A x\rangle $ and the fact that $\sigma_{\mathrm{max}}^2$ the biggest eigenvalue of $A^T A$ is. Then note that $$ \|A\|_F^2 = \mathrm{trace}(A^T A)= \sum_{i=1}^{\min\{m,n\}} \sigma_i^2 \geq \sigma_{\mathrm{max}}^2. $$