DIAGONALIZATION OF HERMITIAN MATRICES 25 U u 1 u 2 u N is unitary In the real

# Diagonalization of hermitian matrices 25 u u 1 u 2 u

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DIAGONALIZATION OF HERMITIAN MATRICES 25 U = [ u 1 , u 2 , . . . , u N ] is unitary”. In the real case, the same equivalence is also true of orthogonal matrices. Properties: 1. If U is unitary, then U U = I implies U x , U y = U ( U x ), y = x , y for all x , y V . This says that when a unitary U acts as a linear operator on the inner product space, all inner products are left unchanged. In particular, all vector norms and orthogonality relationships remain unchanged. 2. A similar property holds for orthogonal operators O on real vector spaces: O x , O y = x , y for all x , y V . 3. When A is Hermitian, x , A y = A x , y (1.15) for all x , y V . 1.10 Diagonalization of Hermitian Matrices Consider a linear operator H that is Hermitian with respect to an inner product. Let u 1 , u 2 be two eigenvectors of H , with eigenvalues λ 1 , λ 2 respectively. Consider the following argument: ( λ 2 - ¯ λ 1 ) u 1 , u 2 = u 1 , λ 2 u 2 - λ 1 u 1 , u 2 = u 1 , H u 2 - H u 1 , u 2 = u 1 , H u 2 - u 1 , H u 2 = 0 Two important conclusions: The identity ( λ 2 - ¯ λ 1 ) u 1 , u 2 = 0 implies 1. If u 1 = u 2 (non-zero so u 1 , u 2 = | u 1 | 2 > 0), then 0 = λ 2 - ¯ λ 1 = λ 1 - ¯ λ 1 . In other words, all eigenvalues of a Hermitian operator H must be real . 2. If λ 1 = λ 2 , we must have u 1 , u 2 = 0. In other words, two eigenvectors with distinct eigenvalues of a Hermitian operator H must be orthogonal . When V = C N and the eigenvalues of a Hermitian matrix H are all single roots of the characteristic polynomial (so there are exactly n distinct eigenvalues, all of which are real), it follows that there are N mutually orthogonal eigenvectors of H , which of course form a basis. If these are normalized, we get an orthonormal basis of eigenvectors { u i } N i =1 for V . With more effort, one can show this result is true for any Hermitian matrix H . An orthonormal basis of eigenvectors corresponds to a matrix of eigenvectors U = [ u 1 , u 2 , . . . , u N ] which is unitary , U - 1 = U . Therefore, H will be unitarily diagonalizable: H = UDU Subscribe to view the full document.

26 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES) where D = diag { λ 1 , λ 2 , . . . , λ N } . This is an important result. Theorem 5. Let H be a size [ N , N ] Hermitian matrix. 1. H has N real eigenvalues λ i , i = 1 : N counting multiplicity, and an orthonormal basis of eigenvectors u i , i = 1 : N . Thus it is diagonalized, H = UDU , with U = [ u 1 , u 2 , . . . , u N ] unitary and D = diag { λ 1 , λ 2 , . . . , λ N } . 2. If H is real and hence symmetric, then the eigenvectors may be chosen to be real, and the diagonalizing matrix U is orthogonal. • Winter '10
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