Is this example of diagonalizable matrix in error?

Question

I am reading through Axler Linear Algebra Done Right and come across this example.

Define $T\in\mathcal{L}(\mathbb{R}^2)$ by $$T(x,y)=(41x+7y,-20x+74y).$$

The matrix of $T$ with respect to the standard basis of $\mathbb{R}^2$ is $$\begin{pmatrix}41&7\\-20&74\end{pmatrix},$$ which is not a diagonal matrix. However, $T$ is diagonalizable, because the matrix of $T$ with respect to the basis $(1,4),(7,5)$ is $$\begin{pmatrix}69&0\\0&46\end{pmatrix},$$ as you should verify.

Well, I am trying to verify it, but I keep getting the matrix of $T$ with respect to that latter basis as $$\begin{pmatrix}69&322\\276&230\end{pmatrix}.$$

Am I doing something wrong, or is the book in error? More generally, how can I find a diagonalization of a matrix whenever I want to?

Answer 1

Instead of putting to coefficients of $T(1,4)$ with respect to the basis $\mathcal{B} = ((1,4),(7,5))$ in the first column of your matrix, you have put $T(1,4)$ directly (or more precisely the coefficients of $T(1,4)$ with respect to the standard basis $((1,0),(0,1))$). You have

$$ T(1,4) = (41 + 28, -20 + 296) = (69, 276) = 69 \cdot (1,4) + 0 \cdot (7,5) $$

so the first column of $[T]_{\mathcal{B}}$ is $(69, 0)^t$ and not $(69, 276)^t$.