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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?

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  • $\begingroup$ What did you try to reach $\left(\begin{array}{}69&322\\276&230\end{array}\right)$? $\endgroup$ – Element118 Oct 20 '15 at 12:10
  • $\begingroup$ Calculated $T(1,4)=(41(1)+7(4),-20(1)+74(4))=(69,276)$ and $T(7,5)=(41(7)+7(5),-20(7)+74(5))=(322,230)$. This is how I have always calculated a matrix of $T$ with respect to bases of $V$. $\endgroup$ – nonremovable Oct 20 '15 at 12:13
  • $\begingroup$ This is not the definition of the matrix of $T$ with respect to a basis of $V$. See my answer. $\endgroup$ – levap Oct 20 '15 at 12:14
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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 $(69, 276)^t$.

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    $\begingroup$ Aha. Thank you for that correction. I see my mistake clearly now. $\endgroup$ – nonremovable Oct 20 '15 at 12:15

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