# Eigenvalue/Eigenvector proof.

• Jan 30th 2010, 05:57 AM
nlews
Eigenvalue/Eigenvector proof.
Suppose that the 2x2 matrix A has only one eigenvalue λ with eigenvector v, and that w is a non zero vector which is not an eigenvector..show that:

a) v and w are linearly independent
b) the matrix with respect to the basis {v, w} is
(λ c
0 λ) for some c =not to 0
c) for a suitable choice of w, c = 1

I am stuck.
I know how to show that eigenvectors are linearly independent, but how do I show that these two vectors are linearly independent to eachother?
as for b and c i dont know where to start! Please help!
• Jan 30th 2010, 01:41 PM
girdav
If $\displaystyle E_{\lambda}$ is the eigenspace corresponding to the eigenvalue $\displaystyle \lambda$ then we can write $\displaystyle E = E_{\lambda}\oplus F$.
What can you say if $\displaystyle v$ and $\displaystyle w$ are linearly dependent?
• Jan 31st 2010, 12:09 PM
nlews
I'm not sure what that sign means =/
but if they are dependent then the coefficients in this equation αv + βw = 0 will not be equal to zero. Im not sure how that helps. Sorry!
• Jan 31st 2010, 12:52 PM
HallsofIvy
Quote:

Originally Posted by nlews
Suppose that the 2x2 matrix A has only one eigenvalue λ with eigenvector v, and that w is a non zero vector which is not an eigenvector..show that:

a) v and w are linearly independent
b) the matrix with respect to the basis {v, w} is
(λ c
0 λ) for some c =not to 0
c) for a suitable choice of w, c = 1

I am stuck.
I know how to show that the eigenvalues are linearly independent,

Unfortunately, I suspect you do NOT know that- it doesn't even makes sense. eigenvalues are numbers and "independent" is only defined for vectors.

Quote:

but how do I show that these two vectors are linearly independent to eachother?
Use the definition of "independent". Suppose av+ bw= 0. Applying A to both sides, a(Av)+ b(Aw)= 3av+ b(Aw)= 0. If a were 0, then either Aw= 0 which means that w is an eigenvector with eigenvalue 0, which by hypothesis is not true, or a= 0. If b were 0, that says 3av= 0 so that v= 0, not an eigenvector. If both are non-zero, we would have w= -(a/b)v so that Aw= -(a/b)Av= -(3a/b)v, again contradicting the fact that w is not an eigenvector.

Quote:

as for b and c i dont know where to start! Please help!
• Jan 31st 2010, 01:36 PM
nlews
sorry yes I obviously meant eigenvectors not eigenvalues. I just meant that I knew the standard proof that eigenvectors are linearly independent if they belong to distinct eigenvalues, and I couldn't see how knowing that would help me here.
However thank you for your help, that makes sense to me! I had sort of a similar thing but the logic wasn't quite there because I hadn't applied A so thank you! part a done!