Diagonal Associated Matrix

Topics: Algebra - Linear Transformation - Eigenvalues and Eigenvectors

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(theorem)

Let $T:V\to V$ with $V$ a vector space on $\mathbb{K}$ with finite dimension. Then, $T$ is diagonalisable if and only if there exists a basis $\beta =\{x_1,\dots ,x_n\}$ for $V$ and scalars ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ (not necessarily distinct) such that:

$$T(x_j)={\lambda }_j{x}_j$$

…with $1\le j\le n$.

Under these circumstances, $[T{]}_{\beta }$ is diagonal:

$[T{]}_{\beta }=\left(\begin{array}{cccc}{a}_1& & & \\ & a_2& & \\ & & \ddots & \\ & & & a_3\end{array}\right)$

The vectors in $\beta$ are the eigenvectors of $T$, while the lambdas are its eigenvalues.