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Null Space and Eigenvalues/Eigenvectors

by psholtz
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psholtz
#1
Dec10-10, 04:01 PM
P: 136
Suppose I have a linear operator of dimension n, and suppose that this operator has a non-trivial null space. That is:

[tex]A \cdot x = 0[/tex]

Suppose the dimension of the null space is k < n, that is, I can find 0 < k linearly independent vectors, each of which yields the 0 vector when the linear operator A is applied to it.

Is it fair to say that this operator then has k eigenvalues, of value 0? and that the k eigenvectors corresponding to this eigenvalue=0 are linearly independent vectors of the null space?
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tiny-tim
#2
Dec10-10, 04:26 PM
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hi psholtz!
Quote Quote by psholtz View Post
Is it fair to say that this operator then has k eigenvalues, of value 0? and that the k eigenvectors corresponding to this eigenvalue=0 are linearly independent vectors of the null space?
yes

(what is worrying you about that? )
psholtz
#3
Dec10-10, 06:12 PM
P: 136
Nothing worrying me about that..

Just wanted to make sure I had it straight..

thanks!

HallsofIvy
#4
Dec11-10, 09:06 AM
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PF Gold
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Null Space and Eigenvalues/Eigenvectors

It would be more standard to say that 0 is an eigenvalue of geometric multiplicity k (which would imply that it had algebraic multiplicity greater than or equal to k: the characteristic equation has a factor [itex]x^n[/itex] for [itex]n\ge k[/itex]) rather than to talk about "k eigenvalues", all of value k, as if they were different eigenvalues that happened to have the same value. Its value is the only property an eigenvalue has!
psholtz
#5
Dec11-10, 11:22 AM
P: 136
Yes, thanks..

So algebraic multiplicity is the number of times the eigenvalue appears as a root in the characteristic equation.

Geometric multiplicity is the dimension of the subspace formed by the eigenvectors of that particular eigenvalue..

And the algebraic multiplicity is always going to be greater than or equal to the geometric multiplicity, correct?


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