Positive Semidefinite matrix A is a p ositive semidefinite matrix (also called n onnegative definite matrix)

Positive definite matrix A is a p ositive definite matrix

Negative semidefinite matrix A is a n egative semidefinite matrix

Negative definite matrix A is a n egative definite matrix

Positive semidefinite matrix A is a p ositive semidefinite matrix A is r eal symmetric matrix

Positive definite matrix A is a p ositive definite matrix A is r eal symmetric matrix

Question Yes Is It true that ?

Proof of Question ?

?

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

Proof of Fact 1.1.6

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

注意 A is symmetric

注意 之反例 is not symmetric

Proof of Exercise

Remark 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 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 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

KxK Principal Minor kxk principal minor of A

Lemma p.1

Lemma p.2

Explain Lemma

The Sum of KxK Principal Minors

Theorem

Proof of Theorem p.1

Proof of Theorem p.2

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

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

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

Rank P.4 1x1 minor Not principal minor rankA=1

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

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

Diagonalizable

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.2 (2)Since A and 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

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

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 A, then A has at least one nonsingular principal Submatrix of order r.

Proof of Corollary p.1