Let $T: V \rightarrow V$ be a linear operator.
I need to demonstrate that if all nonzero vectors of $V$ are eigenvectors of $T$, then there is one specific $\lambda \in K$ such that $T(v) = \lambda v$, for all $v \in V$.
I understand that, if all nonzero vectors of $V$ are eigenvectors of $T$, then $T$ must be a scaling transformation. It just stretch or shrinks vectors, but doesn't change their directions.
So, the statement says that if it happens, then, there is a single $\lambda$ such that $T(v) = \lambda v$. In other words, if there is such transformation, then it scales all vectors by the same scalar $\lambda$.
Applying the transformation to our standard basis vectors, we have: $$ T(e_1) = \lambda_1 e_1 \\ T(e_2) = \lambda_2 e_2 \\ \vdots \\ T(e_n) = \lambda_n e_n $$
I understand I need to prove that $\lambda_1 = \lambda_2 = \dots = \lambda_n$, but I can't see how!
$$ v = c_1e_1 + c_2e_2 + \dots + c_ne_n \\ T(v) = \mu v = \lambda_1c_1e_1 + \lambda_2c_2e_2 + \dots + \lambda_nc_ne_n \\ $$
Since what's multiplying $v$ coordinates is $\lambda_i$, then all of them must be $\mu$. I'm not sure how to 'mathematize' this. Is this idea correct?
EDIT 2 Extending the left hand side of EDIT 1, we have: $$ \mu v = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu(c_1e_1 + \dots + c_ne_n) = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu c_1e_1 + \dots + \mu c_ne_n = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ $$
And since $e_i$ are linearly independent, $\mu = \lambda_1 = \lambda_2 = \dots = \lambda_n$. Is this proof correct?