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I believe I am very close to finishing this proof, but I cannot figure out the last part. If anybody could check my work and maybe give me a little hint, it would be greatly appreciated!

Let $V$ be a finite-dimensional vector space and $T \in \mathcal{L}(V)$, and let $\mathbf{u,v} \in V$ be eigenvectors of $T$.

Claim: If $\mathbf{u} + \mathbf{v}$ is an eigenvector of $T$, then $\mathbf{u}, \mathbf{v}$, and $\mathbf{u+v}$ all correspond to the same eigenvalue.

Proof (So far!): Suppose $T(\mathbf{u}) = \lambda_1 \mathbf{u}$ and $T(\mathbf{v}) = \lambda_2 \mathbf{v}$ with $\mathbf{u}, \mathbf{v} \neq \mathbf{0}$.

Now suppose $T(\mathbf{u+v}) = \lambda_3(\mathbf{u+v})$ with $\mathbf{u+v}\neq \mathbf{0}$.

Then, $$ T(\mathbf{u}) + T(\mathbf{v}) = \lambda_3 \mathbf{u} + \lambda_3 \mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} = \lambda_3 \mathbf{u} + \lambda_3\mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} - \lambda_3 \mathbf{u} -\lambda_3\mathbf{v} = \mathbf{0}\\ (\lambda_1 - \lambda_3) \mathbf{u} + (\lambda_2 - \lambda_3)\mathbf{v} = \mathbf{0} $$

Now I know in order to show that $\lambda_1 = \lambda_2 = \lambda_3$, I must show that the only solution to the last line is the trivial one. This would imply that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent which I am unconvinced of!

The only information I have to my advantage I haven't used yet is the fact that $\mathbf{u}, \mathbf{v}, \mathbf{u+v} \neq \mathbf{0}$. I really cannot see how this information can help me though.

Perhaps I am going about this wrong, but that's why I want to ask! Thanks for your help.

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    $\begingroup$ If $\vec u,\vec v$ are dependent then one is a multiple of the other so of course they have the same eigenvalue. $\endgroup$
    – lulu
    Apr 28, 2017 at 23:46
  • $\begingroup$ @lulu Sorry, but what do you mean by "$\mathbf{u}, \mathbf{v}$ are dependent"? $\endgroup$ Apr 28, 2017 at 23:52
  • $\begingroup$ Linear dependence...if the $\lambda_i$ aren't all the same then I can write those vectors as multiples of each other. $\endgroup$
    – lulu
    Apr 28, 2017 at 23:54
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    $\begingroup$ Your work is correct, but your conclusions need to be written correctly. From the last equation, it is clear thar all $\lambda_i $s are equal if $u,v $ are linearly independent (so this case is done). If $u,v $ are linearly dependent, there is nothing to prove since then all of your three vectors are multiples of each other and therefore correspond to the same eigenvalue. $\endgroup$
    – AnyAD
    Apr 29, 2017 at 0:01

1 Answer 1

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$$(\lambda_1-\lambda_3)\vec u+(\lambda_2-\lambda_3)\vec v=\vec 0$$ Case 1:

$\vec u$ and $\vec v$ are linearly independent. Then as you say, $\lambda_1-\lambda_3=\lambda_2-\lambda_3=0$, so done.

Case 2:

$\vec u$ and $\vec v$ are linearly dependent. Then $\vec v=k\vec u$ for a scalar $k$. Then $T(\vec v)=\lambda_2 \vec v$, but also $T(\vec v)=T(k\vec u)=kT(\vec u)=k\lambda_1\vec u=\lambda_1\vec v$. Since $\vec v$ is an eigenvector, it is non zero, and so $\lambda_1=\lambda_2$. Then you can go back to the previous equation $$\lambda_1 (\vec u+\vec v)=\lambda_3(\vec u+\vec v)$$and again using the fact that $\vec u+\vec v$ is an eigenvector so non-zero, it can be concluded that all the $\lambda_i$'s are equal.

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  • $\begingroup$ Excellent! That makes total sense! $\endgroup$ Apr 29, 2017 at 0:31
  • $\begingroup$ Although I do believe case 2 needs a bit more work to actually be done. So we know $\lambda_1 = \lambda_2$. Then we can go back to my second last line and say $\lambda_1 \mathbf{u} + \lambda_1 \mathbf{v} = \lambda_3 \mathbf{u} + \lambda_3 \mathbf{v} \implies \lambda_1 (\mathbf{u} + \mathbf{v}) = \lambda_3 (\mathbf{u} + \mathbf{v})$ and since $\mathbf{u} + \mathbf{v} \neq \mathbf{0}$, then $\lambda_1 = \lambda_2 = \lambda_3$. $\endgroup$ Apr 29, 2017 at 0:34
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    $\begingroup$ Ah yes I lost track of what needed to be proved haha. I have edited what you write in. $\endgroup$
    – John Doe
    Apr 29, 2017 at 0:56
  • $\begingroup$ @johnDoe +1 for the nice answer. Can we do by the method of contradiction? For example, if I suppose that $\lambda$ is an eigenvalue of $A$ with respect to eigenvector $u+v$ but not with respect to eigenvector $u$, then we may get contradiction? $\endgroup$
    – Srijan
    Apr 29, 2017 at 3:20
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    $\begingroup$ @srijan No problem, glad to help :) $\endgroup$
    – John Doe
    Apr 29, 2017 at 12:05

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