# 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?
• Jan 30th 2010, 01:41 PM
girdav
If $E_{\lambda}$ is the eigenspace corresponding to the eigenvalue $\lambda$ then we can write $
E = E_{\lambda}\oplus F$
.
What can you say if $v$ and $w$ 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: