I need help with this proof
"T is a linear operator on a fin-dim vector space V
Beta is an ordered basis for V
Prove that Lambda is an eigenvalue of T iff Lambda is an eigenvalue of [T](subBeta)"
Let $\displaystyle T:V\to V$ be a linear transformation with $\displaystyle k$ and eigenvalue, so $\displaystyle Tv = kv$ for some $\displaystyle v\in V^{\times}$.
Let $\displaystyle A = [T]_{\beta}$ be the corresponding matrix with respect to this ordered basis $\displaystyle \beta$.
Remember that $\displaystyle (Tv)_{\beta} = A(v)_{\beta}$ where $\displaystyle ( ~ )_{\beta}$ is the coordinate with respect to $\displaystyle \beta$.
Since $\displaystyle Tv = kv \text{ iff } (Tv)_{\beta} = (kv)_{\beta} \text{ iff }A(v)_{\beta} = k(v)_{\beta}$.
Thus, $\displaystyle k$ is an eigenvalue of $\displaystyle A$ since $\displaystyle (v)_{\beta}$ is non-zero.