If the matrix $A$ is perturbed by a symmetric matrix $E$, then its eigenvalues do not move by more the $||E||$

Question

"Matrix Computations" by Golub and Van Loan, 4th ed, Chapter 2, Corollary 2.4.4 says:

If $A\in R^{m\times n} and E\in R^{m\times n}$, then $$\sigma_{min} (A+E)\geq \sigma_{min}(A)-||E||_2$$ $\sigma$ are singular values.

The proof says the results above follow from
$$\sigma_{min}(A)\cdot ||x||_2\leq||Ax||_2 \leq\sigma_{max}(A)\cdot ||x||_2.$$

I tried the following: $$\sigma_{min}(A)||x||_2\leq||Ax||_2\leq||A||_2||x||_2$$ for some $x$ s.t. $||x||_2=1$, so $$\sigma_{min}(A)\leq||A||_2\leq||(A+E)||_2+||E||_2.$$

But it does not seem to work well.

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Edit1: sorry I mistyped the sign in the equation I was intended to prove.

Answer 1

$$\sigma_{min}(A+E)=\min_{\|x\|_2=1}\|Ax+Ex\|_2\leq \min_{\|x\|_2=1}\|Ax\|_2+\max_{\|x\|_2=1}\|Ex\|_2=\sigma_{min}(A)+\|E\|_2.$$

The inequality follows by setting $x$ to be a value minimizing $\|Ax\|_2$ subject to $\|x\|_2=1.$