Determine if a basis consists of eigenvectors So, this might be a silly question, but here it is. I am doing a couple of problems computing $[T]_\beta$, and determining whether $\beta$ is a basis consisting of eigenvectors of $T$. My problem is this: I understand that if $[T]_\beta$ is a diagonal matrix, then clearly $\beta$ is a basis consisting of eigenvectors of $T$, since we've written each vector in $\beta$ as a scalar times itself. However, I can't seem to reason to myself that if $[T]_\beta$ is not a diagonal matrix, then $\beta$ is not a basis consisting of eigenvectors of $T$. 
If $[T]_\beta$ is not diagonal, then for some $\beta_1 \in \beta$, $\beta_1 = \sum^\infty_{i=1}a_i\beta_i$, but how do we know for sure that means that $\beta_1 \neq \lambda\beta_1$, for some $\lambda$?

Prove that contra-positive:
If $\beta = \{u_1, \ldots, u_n\}$ consists of eigenvectors of $T$, then there are $\lambda_i$'s for which:
$$T(u_1) = \lambda_1 u_1 \\ \ldots \\ T(u_n) = \lambda_n u_n$$
Then $[T]_{\beta} = (b_{ij})$, where $b_{ii} = \lambda_i $ and $b_{ij} = 0$ $\forall \ i\neq j$.
