# Thread: Help with a proof - eigenvalues

1. ## Help with a proof - eigenvalues

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

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

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

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 $\displaystyle \lambda \ne \mu$ and $\displaystyle u$ and $\displaystyle v$ are linearly dependent.

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

In other words, either $\displaystyle \lambda = \mu$, or $\displaystyle u$ and $\displaystyle v$ 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$\displaystyle \lambda \neq \mu$.
it's only said that $\displaystyle u+v$ is a eigenvector.
are you saying that I need to assume that $\displaystyle \lambda \neq \mu$ 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$\displaystyle \lambda \neq \mu$.
it's only said that $\displaystyle u+v$ is a eigenvector.
are you saying that I need to assume that $\displaystyle \lambda \neq \mu$ 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 $\displaystyle u+v$ is an eigenvector of $\displaystyle \phi$, but $\displaystyle \lambda \ne \mu$ ...

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

Great.
Thank you.