
#1
Dec1010, 04:01 PM

P: 136

Suppose I have a linear operator of dimension n, and suppose that this operator has a nontrivial 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? 



#2
Dec1010, 04:26 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi psholtz!
(what is worrying you about that? ) 



#3
Dec1010, 06:12 PM

P: 136

Nothing worrying me about that..
Just wanted to make sure I had it straight.. thanks! 



#4
Dec1110, 09:06 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

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!




#5
Dec1110, 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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