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.
- Let $v$ and $w$ be eigenvectors of $A$ with distinct eigenvalues $\lambda_1 \not= \lambda_2$.
If we apply the characteristic equations to $v$ and $w$, we get $$Av = \lambda_1 v,\ \ \ Aw = \lambda_2 w.$$
Now if we form a linear combination of $v$ and $w$, we get $$cv + dw = x.$$
Now we just prove that $x$ is not an eigenvector of A with the following characteristic equation: $$Ax \not= \lambda x.$$
Here is my actual proof: $$Ax = A(cv + dw) = A(cv) +A(dw) = c(Av) + d(Aw) = c(\lambda_1v) + d(\lambda_2w).$$
It is clear to see that $c(\lambda_1v) + d(\lambda_2w) \not= \lambda x$
Therefore, we conclude that $x$ can never be an eigenvector.
Is this the right way to prove this? Did I make a mistake somewhere?