Simple exercise on a linear operator $T$

Question

I'm given the following linear operator $$T(a,b)=(-2a+3b,-10a+9b)$$ on the vector space $V=\mathbb{R}^2$. I have to find the eigenvalues of $T$ and an ordered basis $\beta$ for $V$ such that $[T]_\beta$ is a diagonal matrix.

I've tried using the standard basis $\beta=\{(1,0), (0,1)\}$ and getting $$[T]_\beta= \begin{bmatrix} -2 & 3\\ -10 & 9 \end{bmatrix} $$ which is not a diagonal matrix.

When I deal with this type of exercises, if I find a basis such that $[T]_\beta$ is diagonal, the diagonal elements are the eigenvalues and I don't need to write the characteristic polynomial anymore, right?

In the case $[T]_\beta$ is not a diagonal matrix, can I still use it to write the characteristic polynomial $p(\lambda)=\det([T]_\beta-\lambda I_2)$ and then find the eigenvalues and write the corresponding diagonal matrix $D$? Is this procedure valid for any chosen basis $\beta$ and the corresponding $[T]_\beta$?

Answer 1

Yes, it is valid for any choosen basis $\beta$ and the corresponding $[T]_{\beta} $. And this is the general rule for finding characteristics polynomial and for finding the eigenvalue,$\det([T]_\beta-\lambda I_2) = 0 $

And here one thing, you can't write a $[T]_{\beta} $ as a diagonal matrix, if it is not diagonal with respect to $\beta $. In stead, you can find diagonal matrix similar to $[T]_{\beta} $ , whenever $[T]_{\beta} $ is diagonalizable.

Answer 2

To get a diagonal matrix you need to choose as your basis vectors eigenvectors.

You can use the matrix rel $\beta$ to write the characteristic polynomial, since it is invariant (that is, the same rel different bases).

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So, the characteristic polynomial is $p(x)=(-2-x)(9-x)+30=-18-7x+x^2+30=x^2-7x+12=(x-4)(x-3)$. Since there are two eigenvalues, $4$ and $3$, the matrix is diagonalizable. The diagonal matrix will be $\begin{pmatrix} 4&0\\0&3\end{pmatrix}$.