Original question : Suppose that $T\in\mathcal{L}(V)$, where $V$ is a finite-dimensional vector space, is such that every non-zero vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator.
My proof: Since every non-zero arbitrary vector $v_j\in V$ is an eigenvector of $T$, we have $$Tv_j=\lambda_j v_j$$with $v_j\neq0$.
Suppose $v_1,\cdots,v_n$ is basis of $V$ (Clearly, there can't have zero vector) and suppose an arbitrary vector $v\in V$ with $v\neq 0.$ Then $$Tv=T(a_1v_1+\cdots+a_nv_n)=a_1\lambda_1v_1+\cdots+a_n\lambda_nv_n$$ Since we know by theorem that if $a$ and $b$ are eigenvector of $T$, then $a+b$ is also eigenvector of $T$ corresponding to the same eigenvalues. Thus we have $\lambda_1=\cdots=\lambda_n$. Therefore $$Tv=\lambda_1(a_1v_1+\cdots+a_nv_n)=\lambda_1v.$$This means no matter what vector we put, we have the same eigenvalues for all eigenvectors. Therefore the eigenvalues is independent of the eigenvectors and we can conclude that $$Tv=\lambda_1 v=\lambda_1 Iv=(\lambda_1 I)v$$for which $T$ is a scalar multiple of identity operator.
Is my proof valid?
Here's a hint. Let $e_1, e_2, ... e_n$ be a basis, $e_i \mapsto \lambda_i e_i, 1 \le i \le n.$ What can you say if $e_1 + e_2 \mapsto \lambda(e_1+e_2)?$