# 607 - Definition. For M Rnn, if M = MT and vTMv > 0 for all...

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Definition. For M R n × n , if M = M T and v T M v > 0 for all nonzero v R n , then M is said to be positive definite. If M = M T and v T M v 0 for all v R n , then M is said to be positive semi-definite. 2. For all N R m × n , N T N and NN T are positive semi-definite. Proof v T N T N v = w T w = || w || 2 0 v R n and w = N v R m . Also, NN T = ( N T ) T ( N T ). Properties: 1. For all M = M T R n × n , M is positive (semi-) definite if and only if all eigenvalues of M are positive (non-negative) . Proof an orthogonal P and a diagonal D = diag[ λ 1 λ 2 " λ n ] such that M = PDP T . All eigenvalues of M are positive (non-negative) . ⇔λ i > ( ) 0 i = 1, 2, , n . v T M v = v T PDP T v = z T D z = λ 1 z 1 + λ 2 z 2 + " + λ n z n > ( ) 0 v R n and z = P T v =[ z 1 z 2 " z n ] T R n . 2 22 Lemma: A R m × n with rank A = k . rank A T A = k . A T A has k positive eigenvalues and n k zero eigenvalues. Proof A T A x = 0 x T A T A x = 0 ⇒| | A x || 2 = 0 A x = 0 x R n . A x = 0 A T A x = 0 x R n . rank A T A = n Nullity A T A = n Nullity A = k . Suppose A = BDB 1 R n × n , where B = [ b 1 b 2 " b n ] is invertible and D = diag[ λ 1 λ 2 " λ n ]. Let B = { b 1 , b 2 , , b n } be an eigenvector basis . For T A (the matrix transformation induced by A ), [ T A ] B = B 1 AB = B 1 BDB 1 B = D . How about non-diagonalizable matrices? How about non-square matrices? Lemma: For M = M T R n × n , rank M = number of nonzero eigenvalues. Proof

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## This note was uploaded on 10/16/2010 for the course EE 155 taught by Professor Fong during the Spring '09 term at National Taiwan University.

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607 - Definition. For M Rnn, if M = MT and vTMv > 0 for all...

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