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 $\lambda$?

Of course, we have
$$
T \beta_j = \sum_{i=1}^n \alpha_i \beta_i
$$
For some scalars $\alpha_i$. In particular, $\alpha_i$ is the $i$th entry of the column of $[T]_\beta$ corresponding to $\beta_j$ (the $j$th column, in particular).
If $[T]_\beta$ is not diagonal, then for some $i \neq j$, we have $\alpha_j \neq 0$.
Because $\beta_i$ form a basis, every vector is uniquely expressible as a linear combination of the $\beta_i$.
This necessarily implies that $\sum_{i=1}^n \alpha_i \beta_i$ is not a multiple of $\beta_j$, which means that $\beta_j$ is not an eigenvector of $T$.
