Lecture #13 Notes

# yn then dy y n j 1 so min n j 1 j yj

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: such an A, U, and D, we have the following: &lt; Ax, x &gt;= &lt; U ∗ DU X, x &gt; =&lt; DU x, (U ∗ )∗ x &gt; =&lt; DU x, U x &gt; 1 Now set y1 Ux = y = . . . yn Then &lt; Dy, y &gt;= n ￿ j =1 So λmin n ￿ j =1 λj ||yj ||2 2 ||y || ≤&lt; Dy, y &gt;≤ λmax n ￿ j =1 ||yj ||2 =⇒ λmin ||y ||2 ≤&lt; Dy, y &gt;≤ λmax ||y ||2 Now re-inserting U x for y gives λmin ||U x||2 ≤&lt; DU x, U x &gt;=&lt; Ax, x &gt;≤ λmax ||U x||2 =⇒ λmin ||x||2 ≤&lt; Dx, x &gt;≤ λmax ||x||2 because U is unitary, i.e. norm-preserving. Thus we have obtained the desired result of 1. 2. From the preceding part, we conclude that x x &lt; Ax, x &gt; = inf &lt; A( ), &gt; x￿=0 x￿=0 ||x||2 ||x||2 ||x||2 λmin ≤ inf = inf &lt; Au, u &gt; ||u||=1 However this inﬁmum is realized, because we can take x ∈ Cn \ {0} with Ax = λmin x , then &lt; Ax, x &gt;= λmin ||x||2 and &lt; Au, u &gt;= λmin for u = x/￿x￿. Thus we obtain the desired result that &lt; Ax, x &gt; λmin = min &lt; Ax, x &gt;= min x￿=0 ||x=1|| ||x||2 A similar argument applies f...
View Full Document

Ask a homework question - tutors are online