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!
EDIT
$$
v = c_1e_1 + c_2e_2 + \dots + c_ne_n \\
T(v) = \mu v = \lambda_1c_1e_1 + \lambda_2c_2e_2 + \dots + \lambda_nc_ne_n \\
$$
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?
Best Answer
Since all $v\in V$ are eigenvectors, we can choose $e_i$, the $i$th unit vector. Then by assumption we have $T e_i = \lambda_i e_i$ for some $\lambda_i$. It follows that $T$ is diagonal, with elements $\lambda_1,...,\lambda_n$ on the diagonal.
Now choose $v=e_1+...+e_n$, again for some $\lambda$, we have $Tv=\lambda v$, so we have $$T v = T(e_1+...+e_n) = \lambda_1 e_n +... + \lambda_n e_n = \lambda (e_1+...+e_n).$$ Since the $e_i$ are linearly independent, it follows that $\lambda = \lambda_1 = ... = \lambda_n$. Hence $Tx = \lambda x$, $\forall x$.