Exercise 7.D.20

Answers

Since $r$ is a singular value of $S + T$ , for some $v \in V$ and some isometry $C$ ,

\begin{array}{l} (S + T) v & = rCv \\ ⟹ | | (S + T) v | |^{2} & = | | rCv | |^{2} \\ ⟹ | | Sv | |^{2} + | | Tv | |^{2} + ⟨ Sv, Tv ⟩ + ⟨ Tv, Sv ⟩ & = r^{2} | | v | |^{2} \\ ⟹ \frac{| | Sv | |^{2} + | | Tv | |^{2} + ⟨ Sv, Tv ⟩ + ⟨ Tv, Sv ⟩}{| | v | |^{2}} & = r^{2} \end{array}

where the third line is true because $C$ is an isometry.

Since $s$ and $t$ are the largest singular values of $S$ and $T$ , $\frac{| | Sv | |}{| | v | |} \leq s$ and $\frac{| | Tv | |}{| | v | |} \leq t$ .

By Cauchy-Schwartz Inequality, $| ⟨ Sv, Tv ⟩ | \leq | | Sv | | | | Tv | |$ and $| ⟨ Tv, Sv ⟩ | \leq | | Sv | | | | Tv | |$ .

Thus,

\begin{array}{l} \frac{⟨ Sv, Tv ⟩ + ⟨ Tv, Sv ⟩}{| | v | |^{2}} & \leq \frac{2 | | Sv | | | | Tv | |}{| | v | |^{2}} \\ \leq 2 st . \end{array}

Altogether, this means

r^{2} = \frac{| | Sv | |^{2} + | | Tv | |^{2} + ⟨ Sv, Tv ⟩ + ⟨ Tv, Sv ⟩}{| | v | |^{2}} \leq s^{2} + t^{2} + 2 st = {(s + t)}^{2}

which implies $r \leq s + t$ since $r, s, t$ are all nonnegative.

mqc

2024-06-20 01:13

Could you also do triangle inequality and use the result from 18 (a)? It reduces the work a little bit

tfbray1 • 2024-06-30