Answer by Elliot Herrington for Theorem about existence of ordered basis$...
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...
View ArticleTheorem about existence of ordered basis$ \beta $for diagonalizable linear...
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...
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