Question: for each linear operator T on V, find the eigenvalues ot T and an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix.
$V=R^2$ and $T(a,b)=(-2a+3b,-10a+9b)$
The question I want to ask are two definitions below:
Definition. Let T be a linear operator on an n-dimensional vector space V with order basis $\beta$. Define the charaacteristic polynomial f(t) of T to be $A=[T]_\beta$, that is $f(t)=det(A-t I_n)$
Definition: A linear operator T on a finite-dimensional vector space V is diagonalizable if there is an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix. In here, $[T]_\beta=\begin{pmatrix} \lambda_1 & 0 & \dots & 0 \\ 0 & \lambda_2 & \dots & 0\\ \vdots & \vdots & & \vdots \\ 0 & 0 & \dots &\lambda_n \end{pmatrix}$
But in this question, $[T]_\beta$=$\begin{pmatrix} -2 & 3 \\ -10 & 9 \end{pmatrix}$. My question is $[T]_\beta$ has two meanings in here, why?