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Unformatted text preview: = U ∗ AU U ∗ A∗ U
= U ∗ AA∗ U
= U ∗ A∗ AU
= U ∗ A∗ U U ∗ AU
= (U ∗ AU )∗ (U ∗ AU )
= T ∗T
Thus, T is normal and upper triangular, so it follows from the previous proposition that T must
be diagonal.
We can strengthen this theorem.
2.2.8 Theorem. (”Super” Spectral Theorem) Let A ∈ Mn . Then the following are equivalent:
(i) A is normal.
(ii) A is unitarily diagonalizable.
n
n
(iii)
 λj  2 =
ai,j 2 , where λ1 , λ2 , ..., λn are the eigenvalues of A (multiply counted).
j =1 i,j =1 Proof.
(i) ⇒ (ii) 5 Shown in the previous theorem.
(ii) ⇒ (i)
Suppose A = U DU ∗ for some diagonal D ∈ Mn and unitary U ∈ Mn .
Then:
A∗ A = (U DU ∗ )∗ (U DU...
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This note was uploaded on 01/16/2014 for the course MATH 6304 taught by Professor Bernhardbodmann during the Fall '12 term at University of Houston.
 Fall '12
 BernhardBodmann
 Matrices

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