Suppose that $V$ be a finite dimensional vector space, and $f:V \longrightarrow V$ be a non zero linear map. If the matrix of $f$ with respect to any basis of $V$ gives a diagonal matrix, why is $f= \lambda Id. $ where Id is an identity map ?
I am trying to show by contradiction. Let $f \neq \lambda Id.$ This implies that for any $v \in V$, $fv \neq \lambda Id v $ which is same as $fv \neq \lambda v. $ Thus, for all $v \in V$, $v$ and $fv$ are linearly independent. I don't know where this leads now.