Positive Semidefinite matrix A is a positive semidefinite

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Positive Semidefinite matrix A is a positive semidefinite (also called matrix nonnegative definite matrix)

Positive Semidefinite matrix A is a positive semidefinite (also called matrix nonnegative definite matrix)

Positive definite matrix A is a positive definite matrix

Positive definite matrix A is a positive definite matrix

Negative semidefinite matrix A is a negative semidefinite matrix

Negative semidefinite matrix A is a negative semidefinite matrix

Negative definite matrix A is a negative definite matrix

Negative definite matrix A is a negative definite matrix

Positive semidefinite matrix A is real symmetric matrix A is a positive semidefinite matrix

Positive semidefinite matrix A is real symmetric matrix A is a positive semidefinite matrix

Positive definite matrix A is real symmetric matrix A is a positive definite matrix

Positive definite matrix A is real symmetric matrix A is a positive definite matrix

Question Is It true that ? Yes

Question Is It true that ? Yes

Proof of Question ?

Proof of Question ?

Proof of Question ?

Proof of Question ?

Fact 1. 1. 6 n The eigenvalues of a Hermitian (resp. positive semidefinite ,

Fact 1. 1. 6 n The eigenvalues of a Hermitian (resp. positive semidefinite , positive definite) matrix are all real (resp. nonnegative, positive)

Proof of Fact 1. 1. 6

Proof of Fact 1. 1. 6

Exercise From this exercise we can redefinite: H is a positive semidefinite

Exercise From this exercise we can redefinite: H is a positive semidefinite

注意 A is symmetric

注意 A is symmetric

注意 之反例 is not symmetric

注意 之反例 is not symmetric

Proof of Exercise

Proof of Exercise

Remark n n Let A be an nxn real matrix. If λ is a

Remark n n Let A be an nxn real matrix. If λ is a real eigenvalue of A, then there must exist a corresponding real eigenvector. However, if λ is a nonreal eigenvalue of A, then it cannot have a real eigenvector.

Explain of Remark n p. 1 A, λ : real Az= λz, 0≠z (A-

Explain of Remark n p. 1 A, λ : real Az= λz, 0≠z (A- λI)z=0 By Gauss method, we obtain that z is a real vector.

Explain of Remark n p. 2 A: real, λ is non-real Az= λz, 0≠z

Explain of Remark n p. 2 A: real, λ is non-real Az= λz, 0≠z z is real, which is impossible

Elementary symmetric function kth elementary symmetric function

Elementary symmetric function kth elementary symmetric function

Kx. K Principal Minor kxk principal minor of A

Kx. K Principal Minor kxk principal minor of A

Lemma p. 1

Lemma p. 1

Lemma p. 2

Lemma p. 2

Explain Lemma

Explain Lemma

The Sum of Kx. K Principal Minors

The Sum of Kx. K Principal Minors

Theorem

Theorem

Proof of Theorem p. 1

Proof of Theorem p. 1

Proof of Theorem p. 2

Proof of Theorem p. 2

Rank P. 1 rank. A: =the maximun number of linear independent column vectors =the

Rank P. 1 rank. A: =the maximun number of linear independent column vectors =the dimension of the column space result = the maximun number of linear independent row vectors result =the dimension of the row space

Rank P. 2 rank. A: =the number of nonzero rows in a row-echelon (or

Rank P. 2 rank. A: =the number of nonzero rows in a row-echelon (or the reduced row echlon form of A)

Rank P. 3 rank. A: =the size of its largest nonvanishing minor (not necessary

Rank P. 3 rank. A: =the size of its largest nonvanishing minor (not necessary a principal minor) See next page =the order of its largest nonsigular submatrix.

Rank P. 4 1 x 1 minor Not principal minor rank. A=1

Rank P. 4 1 x 1 minor Not principal minor rank. A=1

Theorem Let A be an nxn sigular matrix. Let s be the algebraic multiple

Theorem Let A be an nxn sigular matrix. Let s be the algebraic multiple of eigenvalue 0 of A. Then A has at least one nonsingular (nonzero)principal submatrix(minor) of order n-s.

Proof of Theorem p. 1

Proof of Theorem p. 1

Geometric multiple Let A be a square matrix and λ be an eigenvalue of

Geometric multiple Let A be a square matrix and λ be an eigenvalue of A, then the geometric multiple of λ=dim. N(λI-A) the eigenspace of A corresponding to λ

Diagonalizable

Diagonalizable

Exercise A and have the same characteristic polynomial and moreover the geometric multiple and

Exercise A and have the same characteristic polynomial and moreover the geometric multiple and algebraic multiple are similarily invariants.

Proof of Exercise p. 1

Proof of Exercise p. 1

Proof of Exercise (2)Since A and p. 2 have the same characteristic polynomial, they

Proof of Exercise (2)Since A and p. 2 have the same characteristic polynomial, they have the same eigenvalues and the algebraic multiple of each eigenvalue is the same.

Proof of Exercise p. 3

Proof of Exercise p. 3

Explain: geom. mult=alge. mult in diagonal matrix

Explain: geom. mult=alge. mult in diagonal matrix

Fact For a diagonalizable(square) matrix, the algebraic multiple and the geometric multiple of each

Fact For a diagonalizable(square) matrix, the algebraic multiple and the geometric multiple of each of its eigenvalues are equal.

Corollary Let A be a diagonalizable(square) matrix and if r is the rank of

Corollary Let A be a diagonalizable(square) matrix and if r is the rank of A, then A has at least one nonsingular principal Submatrix of order r.

Proof of Corollary p. 1

Proof of Corollary p. 1