Eigenvalue of transition matrix Background: This is an exam problem which I was not able to solve entirely. After the exam, I discussed with other students, almost in vain. 
The problem:
Let $V$ be a $n$-dimensional linear space and $(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n})$ and $(\beta_{1}, \beta_{2}, \ldots, \beta_{n})$ be its two different basis. Let $C$ be the transition matrix from the basis of $(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n})$ to that of $(\beta_{1}, \beta_{2}, \ldots, \beta_{n})$. Given an element $v \in V$, its coordinates under the basis of $(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n})$ and $(\beta_{1}, \beta_{2}, \ldots, \beta_{n})$ are $x = (x_1, \ldots, x_n)^{\top}$ and $y = (y_1, \ldots, y_n)^{\top}$, respectively. Please prove that $x = ky$ for some constant $k$ if and only if $k$ is an eigenvalue of the transition matrix $C$.     
My Partial Solution:
For the direction of "$x = ky \Rightarrow k$ is an eigenvalue of $C$":
From some "change of basis"-like lemma, we have $x = Cy$. Together with $x = ky$, we get $Cy = ky$. Therefore, $k$ is an eigenvalue of $C$ and furthermore, $y$ is the corresponding eigenvector.
For the direction of "$k$ is an eigenvalue of $C \Rightarrow x = ky$ ":
I failed here. I can only get that there exists some $y' \in V$ such that $Cy' = ky'$. However, 

How should I relate $Cy' = ky'$ with $x = Cy$ to obtain $x = ky$?


You can not. The whole base change construction is only dressing up the problem. You have a matrix $C$ and two column vectors related by $y=Cx$. More information is not given, and even if $k$ is an eigenvalue, $x$ need not be any eigenvector.
$y=kx$ if and only if $k$ is an eigenvalue and $x$ its eigenvector
would work, but would also be rather trivial.
