# 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

## 2 Answers

"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."