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
 A: One needs to account for the possibility that there are different eigenvalues for different vectors.  We may do so as follows:
Note the following argument is quite general and applies to any linear operator on any vector space over any field, whether finite dimensional or not:
Since every non-zero vector $v$ is an eigenvector, each has an associated eigenvalue $\mu(v)$:
$Av = \mu(v) v; \tag 1$
now if $w$ is a vector linearly dependent upon $v$, 
$w = av \tag 2$
for some scalar $a$, then 
$\mu(w) w = Aw = A(av) = aAv = a\mu(v) v = \mu(v) (av) = \mu(v)w, \tag 3$
or
$(\mu(w) - \mu(v))w = 0; \tag 4$
since $w \ne 0$ this implies
$\mu(w) = \mu(v); \tag 5$
if, on the other hand, we choose $w \ne 0$ linearly independent from $v$, then
$\mu(v + w) v + \mu(v + w)w = \mu(v + w)(v + w)$
$= A(v + w) = Av + Aw = \mu(v) v + \mu(w) w; \tag 6$
comparing coefficients of $v$ and $w$ on each side yields
$\mu(v + w) = \mu(v) = \mu(w); \tag 7$
we thus see that in the event every non-zero vector is an eigenvector, all the eigenvalues are the same; for any $v$,
$Aw = \mu(v)w = \mu(v) I w, \tag 8$
which shows that
$A = \mu(v)I, \tag{9}$
a scalar multiple of the identity, a diagonal matrix.
A: 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 $$y_1=y_2=0$$
A: 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!

A: Matrix multiplication is a continuous function on $\mathbf{R}^n\to \mathbf{R}^n$.
If every $v$ is an eigen vector then the corresponding  eigenvalue  should vary continously with $v\in \mathbf{R}^n$, denote it by $\lambda_v$.
For simplicity, let us restrict to the subset $S=\{(x_1,x_2,\ldots,x_n)^T\in \mathbf{R}^n \mid \sum_ix_i^2=1\} $ which being closed and bounded a compact set, and is also connected.
So, $\lambda_v$ restricted to $v\in S$ is also a continuous real valued function, its image has to be an interval being the continuous image of a connected set, and has to be compact. So the image is a singleton, call it $\lambda.$
So all the eigenvalues are the same. That is given matrix is scalar multiplication by $\lambda$. In particular a diagonal matrix.
