Theorem about existence of ordered basis$ \beta $for diagonalizable linear operator

Question

Theorem :linear operator T on a finite-dimensional vector space V is diagonalizable if and only if there exists an ordered basis β for V consisting of eigenvectors of T [Why?] Furthermore, if T is diagonalizable, β = {$v_1 , v_2 , . . . , v_n$} is an ordered basis of eigenvectors of T, and $D = [T]_\beta$, then D is a diagonal matrix and $D_{jj}$ is the eigenvalue corresponding to $v_j$ for $1 ≤ j ≤ n$.

The book doesn't provide the proof, general sketch of doing this proof?

Answer 1

The linear operator T is diagonalisable if there is a basis of $V$ with respect to which the matrix of $T$ is diagonal. That is, $[T]_{\beta} = \text{diag}(\alpha_1, \alpha_2, \ldots, \alpha_n)$ for some ordered basis $\beta = \left\{v_1, v_2, \dots ,v_n \right\}$, and some $\alpha_1,\dots,\alpha_n \in \mathbb R.$ Let us show that this is equivalent to the existence of an ordered basis made up of eigenvectors of $T$. That is, we aim to show $$[T]_{\beta} = \text{diag}(\alpha_1, \alpha_2, \ldots, \alpha_n) \iff T(v_k) = \lambda_k v_k, \text{ for } k=1,\ldots,n,$$ where the $\lambda_k \in \mathbb R$ are the eigenvalues of $T$. Well, this follows from the definition of $[T]_{\beta};$ we have $T(v_k) = \lambda_k v_k$ if and only if the $k$th column of $[T]_{\beta}$ has a zero in every entry except the $k$th, where it has entry $\lambda_k$. So we see in fact that $[T]_{\beta} = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)$.