$A$ be $n \times n$ matrix over $\mathbb{C}$ such that every non-zero vector of ${\mathbb{C}}^n$ is an eigenvector of $A$ I am trying assignment questions in Linear Algebra and this question could not be solved by me. So, I thought of posting it here.

Let A be an $n\times n$ matrix over $\mathbb{C}$ such that every non-zero vector of ${\mathbb{C}}^n$ is an eigenvalue of A. Then which of the following are true:

*

*All eigenvalues of A are equal

*All eigenvalues of A are distinct

*$A=\lambda I $ for some $\lambda \in \mathbb{C}$ , where I is the n times n identity matrix

*The minimal and characterstic polynomial for A are equal.


I tried by taking eigenvectors to be distinct and the hypothesis but couldn't solve any options. Can you please give hints how to start the problem!
Thank you!
 A: I assume our OP Avenger means "every non-zero vector of ${\mathbb{C}}^n$ is an eigenvector of A" instead of
"every non-zero vector of ${\mathbb{C}}^n$ is an eigenvalue of A"; otherwise, the problem makes no sense.
If every non-zero vector $w$ is an eigenvector, each with eigenvalue $\mu_w$,
$Aw = \mu_w w \tag 1$
by choosing two linearly independent vectors $u$ and $v$ we have
$\mu_u u + \mu_v v  = Au + Av = A(u + v)$
$= \mu_{u + v}(u + v) =  \mu_{u + v}u + \mu_{u + v}v; \tag 2$
linear independence of $u$ and $v$ then forces
$\mu_u = \mu_{u + v} = \mu_v; \tag 3$
thus all eigenvalues are equal to some
$\mu \in \Bbb C, \tag 4$
that is, for any vector $w$,
$Aw = \mu w, \tag 5$
or
$A = \mu I; \tag 6$
$A$ is a scalar multiple of $I$.
The characteristic polynomial of $A$ is
$\det(A - xI) = \det (\mu I - xI)$
$= \det((\mu - x)I) = (\mu - x)^n; \tag 7$
however, $A = \mu I$ also satisfies its minimal polynomial
$m(x) = \mu - x, \tag 8$
and
$m(x) \ne (\mu - x)^n \tag 9$
unless $n = 1$.
So we have items (1) and (3) in the OP's question are true, but (2) and (4) are false.
A: Hint: if $u$ is an eigenvector for eigenvalue $\alpha$ and $v$ an eigenvector for eigenvalue $\beta \ne \alpha$, can $u+v$ be an eigenvector?
