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Positive Semidefinite matrix
A is a positive semidefinite matrix (also called nonnegative definite matrix)
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Positive definite matrix
A is a positive definite matrix
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Negative semidefinite matrix
A is a negative semidefinite matrix
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Negative definite matrix
A is a negative definite matrix
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Positive semidefinite matrix
A is real symmetric matrix A is a positive semidefinite matrix
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Positive definite matrix
A is real symmetric matrix A is a positive definite matrix
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Question Is It true that ? Yes
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Proof of Question ?
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Proof of Question ?
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Fact 1.1.6 The eigenvalues of a Hermitian (resp.
positive semidefinite , positive definite) matrix are all real (resp. nonnegative, positive)
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Proof of Fact 1.1.6
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Exercise From this exercise we can redefinite:
H is a positive semidefinite
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注意 A is symmetric
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注意 之反例 is not symmetric
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Proof of Exercise
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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
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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.
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Explain of Remark p.2 A: real, λ is non-real Az= λz, 0≠z
z is real, which is impossible
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Elementary symmetric function
kth elementary symmetric function
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KxK Principal Minor kxk principal minor of A
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Lemma p.1
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Lemma p.2
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Explain Lemma
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The Sum of KxK Principal Minors
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Theorem
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Proof of Theorem p.1
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Proof of Theorem p.2
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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 result
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Rank P.2 rankA:=the number of nonzero rows in a row-echelon
(or the reduced row echlon form of A)
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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
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Rank P.4 1x1 minor Not principal minor rankA=1
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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.
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Proof of Theorem p.1
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the eigenspace of A corresponding to λ
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
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Diagonalizable
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Exercise A and have the same characteristic polynomial and moreover
the geometric multiple and algebraic multiple are similarily invariants.
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Proof of Exercise p.1
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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.
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Proof of Exercise p.3
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Explain: geom.mult=alge.mult in diagonal matrix
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Fact For a diagonalizable(square) matrix,
the algebraic multiple and the geometric multiple of each of its eigenvalues are equal.
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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.
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Proof of Corollary p.1
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