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

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