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So I have this question here which says:

Let $T:\Bbb R^3\to\Bbb R^3$ defined by

$$T\left(\begin{matrix}x_1 \\ x_2 \\ x_3 \end{matrix}\right)=\left(\begin{matrix}-x_1+7x_2-x_3 \\ x_2 \\ 15x_2-2x_3 \end{matrix}\right).$$

  • (a) Let $\mathcal E=\{e_1,e_2,e_3\}$, where $e1=\left(\begin{matrix}1 \\ 0 \\ 0 \end{matrix}\right), e2=\left(\begin{matrix}0 \\ 1 \\ 0 \end{matrix}\right), e3=\left(\begin{matrix}0 \\ 0 \\ 1 \end{matrix}\right)$, be the standard basis of $\Bbb R^3$. Find $[T]_\mathcal E$.
  • (b) Find a basis $B=\{b_1,b_2,b_3\}$ of $\Bbb R^3$ such that $[T]_B$ is a diagonal matrix.
  • (c) What is $[T]_B$?

I already solved part $(a)$. The answer is:

$[T]_{\mathcal E} =\left( \begin{array}{ccc} -1 & 7 & -1\\ 0 & 1 & 0 \\ 0 & 15 & -2\\ \end{array} \right) $

However, I'm not sure what the difference between part $(b)$ and part $(c)$ is. If I find a basis, isn't that automatically giving me $[T]_{B}$?

Second question. How do I actually do part $(b)$? As far as I can tell, I just have to diagonalize $[T]_{B}$ by finding the eigenvalues and the eigenvectors since that will give me a basis for the eigenspace which in turn, is a diagonal matrix. Is that all though? Is there more that I have to do?

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    $\begingroup$ You are right on track. $\endgroup$ – xbh Dec 11 '18 at 6:55
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Yes, all you have to do do solve part (b) is to find eigenvalues and eigenvectors for $T$. And then the answer to (c) is the diagonal matrix such tha the entries of the main diagonal are the eigenvalues that you got while solving (b).

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