Let $A$ be an $n \times n$ matrix. If every non-zero vector $v$ is an evector of $A$, prove that $A$ is a diagonal matrix I'll start with the things i already know, I know that for a vector $v$ to be an evector of $A$, then the following must be true
$Av = \lambda v$  this is only true if and only if....
$(A - \lambda I)v = 0$
and i also know that a diagonal matrix is a matrix that has the following form..
$$ A = \begin{pmatrix} x_1 & 0 \\ 0 & x_2   \end{pmatrix}$$
($x_1$ and $x_2$ can be any number and can equal the same number as well)
i swear... i think the hardest part about these types of problem is knowing where to start... i think im too used to being told where to start by my professors.. which is a bad habit of mine...
any help will be appreciated

Hint:
For $k\in\mathbb{R}$,
$$kI_nv=kv$$

I'm only showing $n=2$. Use the same idea for higher $n$
Suppose $v=\binom{a}{b}$ and $A=\left(\begin{matrix}x_1&y_1\\y_2&x_2\end{matrix}\right)$.
We have 
\begin{align}
Av&=\lambda v\\
\left(\begin{matrix}x_1&y_1\\y_2&x_2\end{matrix}\right)\binom{a}{b}&=\lambda \binom{a}{b}\\
\binom{ax_1+by_1}{ay_2+bx_2}&=\binom{\lambda a}{\lambda b}
\end{align}
Hence, $$x_1=x_2=\lambda$$
and $$y_1=y_2=0$$
