# Let $A$ be an $n \times n$ matrix, prove that a linear combination, $cv + dw$ with $c \not=d$ of two eigenvectors

Let A be an $n\times n$ matrix. Prove that a linear combination, $cv + dw$ with $c \not= d$ of two eigenvectors corresponding to different eigenvalues is never an eigenvector.

I think I get the general idea here.

1. Let $v$ and $w$ be eigenvectors of $A$ with distinct eigenvalues $\lambda_1 \not= \lambda_2$.
2. If we apply the characteristic equations to $v$ and $w$, we get $$Av = \lambda_1 v,\ \ \ Aw = \lambda_2 w.$$

3. Now if we form a linear combination of $v$ and $w$, we get $$cv + dw = x.$$

4. Now we just prove that $x$ is not an eigenvector of A with the following characteristic equation: $$Ax \not= \lambda x.$$

5. Here is my actual proof: $$Ax = A(cv + dw) = A(cv) +A(dw) = c(Av) + d(Aw) = c(\lambda_1v) + d(\lambda_2w).$$

6. It is clear to see that $c(\lambda_1v) + d(\lambda_2w) \not= \lambda x$

7. Therefore, we conclude that $x$ can never be an eigenvector.

Is this the right way to prove this? Did I make a mistake somewhere?

• This claim, as stated, isn't true. You must also require that $c\ne0$ and $d\ne0$. – zipirovich Jan 3 '18 at 0:37

"It is clear" is something that sould be used with utmost care. In this particular case, you are using "it is clear" precisely in the key step of the proof, number 6 in your list, where you provide no details.

Note that, as stated the claim is false: you need both $c,d$ nonzero, as step 6 shows.

To see 6, if you had $$\lambda cv+\lambda dw=\lambda x = Ax=\lambda_1 cv+\lambda_2 d w,$$ you get $$c(\lambda-\lambda_1)v+d(\lambda-\lambda_2)w=0.$$ Now it's the moment to use that $\lambda_1\ne\lambda_2$ (which implies that $v$ and $w$ are linearly independent), and that $c,d$ are not zero.

• I think something like the line in Doyun's answer would make the last step here a little easier to understand: "Because v and w are eigenvectors of different eigenvalues, {v,w} are linearly independent" – muzzlator Jan 3 '18 at 0:27
• Actually, we must require that both $c$ and $d$ are nonzero, not just one of them. – zipirovich Jan 3 '18 at 0:39
• @zipirovich: indeed. Edited in. – Martin Argerami Jan 3 '18 at 0:54
• @muzzlator: good point. I have edited that. – Martin Argerami Jan 3 '18 at 0:55

I think your proof is almost right and great! In my opinion, we have to add an assumption that both $c$ and $d$ are not zero.

I think it is good to add the following sentence to prove step 6) more precisely:

"Because $v$ and $w$ are eigenvectors of different eigenvalues, $\{v,w\}$ are linearly independent."