How do you prove $\|A-B\|_F^2 \geq \|\Sigma_A - \Sigma_B\|_F^2$?

Question

Given the following matrices $A,B\in \Bbb{R}^{m\times n}$ with $A=U \Sigma_A V^t$ and $B=Q \Sigma_B R^t$ (full SVD), how do you prove

$$\|A-B\|_F^2 \geq \|\Sigma_A - \Sigma_B\|_F^2$$

Answer 1

Not quite a proof, but here's a citation that will work. The following is from the 1997 edition of Bhatia's Matrix analysis.

Problem III.6.13: Let $A,B$ be any $n \times n$ matrices and let $\Phi$ be any symmetric gauge function on $\Bbb R^n$. Then $$ \Phi(s(A) - s(B)) \leq \Phi(s(A - B)) $$

Now, taking $\Phi(x_1,\dots,x_n) := \sqrt{x_1^2 + \cdots + x_n^2}$ gives us the desired result. In particular: $\Phi(s(A) - s(B)) = \|\Sigma_A - \Sigma_B\|_F$, and $\Phi(s(A-B)) = \|A-B\|_F$.