# tests coming up-eigenvectors-prove

• June 10th 2013, 12:49 AM
n22
tests 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
• June 10th 2013, 01:02 AM
chiro
Re: tests coming up-eigenvectors-prove
Hey n22.

Are you familiar with the diagonalization theorem?

Diagonalizable matrix - Wikipedia, the free encyclopedia
• June 12th 2013, 10:36 PM
n22
Re: tests coming up-eigenvectors-prove
Quote:

Originally Posted by chiro
Hey n22.

Are you familiar with the diagonalization theorem?

Diagonalizable matrix - Wikipedia, the free encyclopedia

Hello,
I am familiar with the theorem. but knowing how to prove something is difficult-especially when I dont understa"choosing one non zero vector from each eigenspace" is .So if I picked any two vectors then its a basis of R^n.?please expand.thanks.
• June 13th 2013, 06:38 AM
HallsofIvy
Re: tests coming up-eigenvectors-prove
Quote:

Originally Posted by n22
Hello,
I am familiar with the theorem. but knowing how to prove something is difficult-especially when I dont understa"choosing one non zero vector from each eigenspace" is .So if I picked any two vectors then its a basis of R^n.?please expand.thanks.

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 $Av= \lamba v$ then so does any multiple of it: $A(av)= a(Av)= a(\lambda v)= \lambda (av)$. 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.