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

I just like to highlight the hint from steven gregory and Andrés E. Caicedo:
If every vector is an eigenvector of $A$, so $e_1,\ldots,e_n$ are, where
$$
e_i = (0,0,\ldots,0,\underbrace{1}_{i.th~position},0,\ldots,0)^T
$$
for $i=1,\ldots,n$.
So there exists $\lambda_1,\ldots,\lambda_n$such that $Ae_i=\lambda_ie_i$. And now compute directly what $Ae_i$ is and you will can deduce the form of $A$.
A little bit more precisely:

 If $$A=\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\a_{21} & a_{22} & \dots & a_{2n}\\\vdots & \vdots & \ddots & \vdots\\a_{n1} & a_{n2} & \dots & a_{nn}\end{pmatrix}$$ then $$Ae_1=\begin{pmatrix}a_{11}\\a_{21}\\\vdots\\a_{n1}\end{pmatrix}$$. So if $Ae_1=\lambda_1e_1$, then $$\begin{pmatrix}a_{11}\\a_{21}\\\vdots\\a_{n1}\end{pmatrix}=\begin{pmatrix}\lambda_1\\0\\\vdots\\0\end{pmatrix}.$$ So $a_{11}, a_{21},\ldots,a_{n1}$ are determined by this equation!

