# Thread: Help with a proof - eigenvalues

1. ## Help with a proof - eigenvalues

let $\varphi :V\rightarrow V$ be a linear transformation.
let $u$ be eigenvector of $\varphi$ with the eigenvalue $\lambda$
let $v$ be eigenvector of $\varphi$ with the eigenvalue $\mu$
I need to prove that if $u+v$ is an eigenvector of $\varphi$, then $\lambda=\mu$

this is what I tried:
$\varphi(u+v)=a(u+v)=au+av$
on the other hand:
$\varphi(u)+\varphi(v)=\lambda u+\mu v$

but how can I conclude that $a=\lambda$ and $a=\mu$?

2. ## Re: Help with a proof - eigenvalues

φ(u+v) = λu + μv = λ’(u+v)
(λ- λ’)u + (μ- λ’)v = 0
u and v are linearly independent so λ = λ’ , μ = λ’

3. ## Re: Help with a proof - eigenvalues

Why are they linear independent?

4. ## Re: Help with a proof - eigenvalues

Originally Posted by Stormey
Why are they linear independent?
Suppose $\lambda \ne \mu$ and $u$ and $v$ are linearly dependent.

Then there is some $t \ne 0$ such that $v=tu$
In that case $\phi v = \phi (tu) = t (\phi u) = t (\lambda u) = \lambda (tu) = \lambda v$, which is a contradiction (why?).

In other words, either $\lambda = \mu$, or $u$ and $v$ are linearly independent.

5. ## Re: Help with a proof - eigenvalues

Originally Posted by Stormey
Why are they linear independent?
If λ =’μ (starting assumption) the eigen vectors are linearly independent. You'll have to look it up in a linear algebra book or on-line.

=’ means not equal

6. ## Re: Help with a proof - eigenvalues

Thanks for the help.

I know that vectors of different eigenvalues are linear independent, you don't need to convince me about that.
the thing is: it's not said that $\lambda \neq \mu$.
it's only said that $u+v$ is a eigenvector.
are you saying that I need to assume that $\lambda \neq \mu$ and show that it leads to a contradiction?
(just want to wrap my head around it)

7. ## Re: Help with a proof - eigenvalues

Originally Posted by Stormey
Thanks for the help.

I know that vectors of different eigenvalues are linear independent, you don't need to convince me about that.
the thing is: it's not said that $\lambda \neq \mu$.
it's only said that $u+v$ is a eigenvector.
are you saying that I need to assume that $\lambda \neq \mu$ and show that it leads to a contradiction?
(just want to wrap my head around it)
This is where a proof by contradiction comes in.

Suppose $u+v$ is an eigenvector of $\phi$, but $\lambda \ne \mu$ ...

If it leads to a contradiction, then we must conclude that if $u+v$ is an eigenvector of $\phi$, then $\lambda = \mu$.

Great.
Thank you.