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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 reinserting U x for y gives
λmin U x2 ≤< DU x, U x >=< Ax, x >≤ λmax U x2
=⇒ λmin x2 ≤< Dx, x >≤ λmax x2
because U is unitary, i.e. normpreserving. 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
x2
x2 x2 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 x2 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
x2 A similar argument applies f...
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 Fall '12
 BernhardBodmann
 Math, Matrices

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