Prove that for for an arbitrary linear operator all eigenvalues lie in a circular ring.
{z $\in$ $С$ | $\sigma_n(A)$ $\leq$ $|z|$ $\leq \sigma_1(A)$}
Prove that for for an arbitrary linear operator all eigenvalues lie in a circular ring.
{z $\in$ $С$ | $\sigma_n(A)$ $\leq$ $|z|$ $\leq \sigma_1(A)$}
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Hint: Use the fact that $$ \sigma_1(A) = \max_{\|x\| = 1} \|Ax\|, \quad \sigma_n(A) = \min_{\|x\| = 1} \|Ax\|. $$