# Eigenvector proof

1. Jun 19, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
If v and w are eigenvectors with different (nonzero) eigenvalues, prove that they are
linearly independent.

2. Relevant equations

3. The attempt at a solution
Define an operator A such that a is an nxn matrix, and Av=cIv with
c an eigenvalue and v and eigenvector. Define a basis
<v1...vn> in that v=vi and w=vk 1<=k<=n and 1<=i<=n,
and let ci,i, be an element in A. I is the identity matrix.

Consider I*v, a 1xn column matrix with its lonely nonzero (1) at position 1,i.
Let the value at ci,i=c.Multiplying I*v by A gives c*Iv . If I*w (another 1xn column
matrix) had 1 at position 1,i, it would
correspond with ci,i on A and we would get c*Iw. But we assume w has a different
eigenvalue. Therefore I*w must have its 1 at a different position to correspond with
a different value on A (call it k). Since I*v must have 1 at a row different from I*w,
let c*Ia1v+k*Ia2w=0, and since 1 is at different rows, and c and I and k are not zero,
a1 and a2 must be 0, so we have c*I*0*v+I*k*0*w=0*v+0*w=0, thus
a1 and a2 are trivial so v and w are linearly independent.

I kind of have a gut feeling that this may be too wordy.

Last edited: Jun 19, 2009
2. Jun 20, 2009

### CompuChip

What you have looks correct, but - as you said - perhaps you are overdoing a little :)

I would just start with stating that v and w are eigenvectors:
(1a) A v = c v
(1b) A w = d w
for some numbers c and d. We know that c is not equal to d and neither is equal to 0.

Now not being linearly independent means that there does not exist a number k such that w = k v. This is not pleasant to work with, so a proof by contradiction suggests itself. Suppose that there does exist a k such that
(2) w = k v.

Now can you derive a contradiction?
(Note: the rest of the proof is rather straightforward, because all you have to work with are equations (1a), (1b) and (2)).

3. Jun 20, 2009

### Dick

This is a special case of the last problem you posted. You don't need a basis and you don't need a matrix. If Av=av and Au=bu (a not equal b) you want to show that if cv+du=0 then both c and d are zero. Hit that equation with (A-aI).