Hi ,

I would really appreciate if someone could help me out here.thanks.

Prove that if an nxn matrix A has n distinct eigenvalues ,then the set of n eigenvectors formed by

choosing one non zero vector from each eigenspace of A is a basis of R^n

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- Jun 9th 2013, 11:49 PMn22tests coming up-eigenvectors-prove
Hi ,

I would really appreciate if someone could help me out here.thanks.

Prove that if an nxn matrix A has n distinct eigenvalues ,then the set of n eigenvectors formed by

choosing one non zero vector from each eigenspace of A is a basis of R^n - Jun 10th 2013, 12:02 AMchiroRe: tests coming up-eigenvectors-prove
Hey n22.

Are you familiar with the diagonalization theorem?

Diagonalizable matrix - Wikipedia, the free encyclopedia - Jun 12th 2013, 09:36 PMn22Re: tests coming up-eigenvectors-prove
- Jun 13th 2013, 05:38 AMHallsofIvyRe: tests coming up-eigenvectors-prove
Not "any two vectors", one vector from each eigenspace. Do you know what an eigenspace is? You should have already proved that if v satisfies $\displaystyle Av= \lamba v$ then so does any multiple of it: $\displaystyle A(av)= a(Av)= a(\lambda v)= \lambda (av)$. From that it follows that the set of all eigenvectors, corresponding to a given eigenvalue (together with the 0 vector) is a subspace.

It is also easy to see that if two eigenvectors correspond to different eigenvalues, they are**independent**(one is not a multiple of the other). So if A has n different eigenvalues, I has n different eigenspaces.