Prove that $T$ is a scalar multiple of the identity operator Suppose $T \in \mathscr{L}(V)$ is such that every nonzero vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of identity operator.
My question is about why the certain statements we prove leads to the conclusion of the question.
After attempting the problem and not getting the correct solution. I found a solution online that stated the following:

To show $T$ is a scalar multiple of the identity, we must show that $a_v$ is independent of $v$ for $v\in V\setminus\{{0}\}$.
  Suppose $v,w \in V\setminus\{{0}\}$ We want to show $a_v = a_w$

When I attempted the problem I was under the impression that seeing that every vector in $V$ is an eigenvector of $T$ It would simply follow that $T$ is a scalar multiple of the identity since:  $$T(v) = \lambda v = \lambda I(v)$$
So why is it that these two conditions show that $T$ is a scalar multiple of the identity?
Note: I know how to do the mechanics of the problem already. It is the reasoning I am having problems with. 
 A: Everything being an eigenvector implies that there is only one eigenvalue:$av=Tv=T\sum v_i=\sum a_iv_i $ where $v_i$ are basis vectors and $a_i $ the corresponding eigenvalues.
So $av=av_1+\dots +av_n=a_1v_1+\dots +a_nv_n$, hence $a=a_i$ for all $i $.
A: If $\vec v \ne 0 \ne \vec w$ are linearly dependent, they are in fact collinear and there is some scalar $\alpha \ne 0$ with
$w = \alpha v; \tag 1$
letting $\mu_{\vec v}$ be the eigenvalue associated with vector $\vec v$, so that
$T\vec v = \mu_{\vec v} \vec v, \tag 2$
we have
$T\vec w = T(\alpha \vec v) = \alpha T \vec v = \alpha \mu_{\vec v} \vec v = \mu_{\vec v} (\alpha \vec v) = \mu_{\vec v} \vec w, \tag 3$
which shows that $\mu_{\vec v}$ is also an eigenvalue for $\vec w$, that is
$\mu_{\vec w} = \mu_{\vec v}; \tag 4$
the eigenvalues are the same in the case (1).
Now suppose $\vec v$ and $\vec w$ are not collinear, i.e, they are linearly independent, then
$\mu_{\vec v + \vec w} \vec v + \mu_{\vec v + \vec w} \vec w = \mu_{\vec v + \vec w} (\vec v + \vec w) = T(\vec v + \vec w) = T\vec v + T\vec w = \mu_{\vec v} \vec v + \mu_{\vec w} \vec w; \tag 5$
since $\vec v$ and $\vec w$ are linearly indendent this implies
$\mu_{\vec v + \vec w} = \mu_{\vec v} = \mu_{\vec w}, \tag 6$
which shows that any two linearly independent eigenvectors share the same eigenvalue; this there is but one eigenvalue $\mu$ shared by all non-zeero vectors, collinear or not; then we have
$T \vec v = \mu \vec v, \; \forall \vec v, \tag 7$
and $T = \mu I$ is thus a multiple of the identity mapping.
