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

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?

• Just write down the definitions! What does mean that an operator is diagonalizable? How do you compute $[T]_\beta$ when $\beta$ is basis of eigenvectors? What do you get? This is inmediate given the apropiate defintions. Mar 14, 2020 at 5:25

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 $$[T]_{\beta} = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)$$.