How to prove that the eigenvalue(s) appears on the diagonal matrix $D=[T]_\beta$ are all the eigenvalue(s) of $T$?

Question

Let $T$ be a linear operator on $V$ that is diagonalizable with basis $\beta$, how to prove that the eigenvalue(s) appears on the diagonal matrix $D=[T]_\beta$ are all the eigenvalue(s) of $T$? I haven't come up with any idea.

Answer 1

The eigenvalues of $T$ are the roots of the characteristic polynomial of its matrix with respect to some basis, in particular with respect to $\beta$. But that matrix is $D$ and its eigenvalues are the entries of the main diagonal.

Answer 2

The $i$-th column vector of $[T]_{\beta}$ is $[T (b_i)]_{\beta}$, where $b_i$ is the $i$-th basis vector in $\beta$ (basis consisting of eigenvectors). Now $Tb_i=\lambda_i b_i$, so the column has $\lambda_i$ in the ith place and zeroes elsewhere. Starting with such a representation similarly one would show that these diagonal entries must correspond to eigenvalues.