# If the sum of eigenvectors is an eigenvector, then they all correspond to the same eigenvalue

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.

• If $\vec u,\vec v$ are dependent then one is a multiple of the other so of course they have the same eigenvalue.
– lulu
Commented Apr 28, 2017 at 23:46
• @lulu Sorry, but what do you mean by "$\mathbf{u}, \mathbf{v}$ are dependent"? Commented Apr 28, 2017 at 23:52
• Linear dependence...if the $\lambda_i$ aren't all the same then I can write those vectors as multiples of each other.
– lulu
Commented Apr 28, 2017 at 23:54
• 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. Commented Apr 29, 2017 at 0:01

$$(\lambda_1-\lambda_3)\vec u+(\lambda_2-\lambda_3)\vec v=\vec 0$$ 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.

• Excellent! That makes total sense! Commented Apr 29, 2017 at 0:31
• 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 $\lambda_1 = \lambda_2 = \lambda_3$. Commented Apr 29, 2017 at 0:34
• Ah yes I lost track of what needed to be proved haha. I have edited what you write in. Commented Apr 29, 2017 at 0:56
• @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? Commented Apr 29, 2017 at 3:20
• @srijan No problem, glad to help :) Commented Apr 29, 2017 at 12:05