3
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Let $B=\{v_1,v_2,v_3\}$, a basis of $V$ above $\mathbb{R}$. Let $$ [T]_B = \left(\begin{array}{cccc} 6&-3&-2\\4&-1&-2\\10&-5&-3 \end{array}\right)$$

The characteristic polynomial is $f_T(x) = (x-2)(x^2+1)$. Hence, $m(x) = (x-2)(x^2+1)$

We have $W_1 = \ker (T-2I) = \text{span}\{v_1,2v_3\}$ and $W_2 = \ker (T^2+I) = \text{span}\{v_1+v_2, v_3\}$ and $V = W_1 \oplus W_2$.

Now, we denote $$C_1 = \{v_1 + 2v_3\} \\ C_2 = \{ v_1+v_2,v_3 \}$$

I don't quite understand why

$$[T_{|W_1}]_{C_1} = (2) \\ [T_{|W_2}]_{C_2} = \left(\begin{array}{cccc} 3&-2\\5&-3 \end{array}\right)$$

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  • $\begingroup$ Notice that, $W_1$ = span{$v_1+2v_3$} $\endgroup$ – Sry Jul 24 '15 at 11:20
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$ \{v_1 + 2v_3\}=\left( \begin{array}{c} 2\\ 0\\ 4\\ \end{array} \right)\\ \{v_1+v_2,v_3\} =\left(\begin{array}{c} 3\\ 3\\ 5\\ \end{array}\right),\left(\begin{array}{c} -2\\ -2\\ -3\\ \end{array}\right)$

and $T\left( \begin{array}{c} 2\\ 0\\ 4\\ \end{array} \right)\ = 2\left( \begin{array}{c} 2\\ 0\\ 4\\ \end{array} \right)\\$

thus your first matrix comes out as $(2)$

Now, $T\left(\begin{array}{c} 3\\ 3\\ 5\\ \end{array}\right)=3\left(\begin{array}{c} 3\\ 3\\ 5\\ \end{array}\right)+5\left(\begin{array}{c} -2\\ -2\\ -3\\ \end{array}\right)$

$T\left(\begin{array}{c} -2\\ -2\\ -3\\ \end{array}\right)=-2\left(\begin{array}{c} 3\\ 3\\ 5\\ \end{array}\right) -3\left(\begin{array}{c} -2\\ -2\\ -3\\ \end{array}\right)$

hence second matrix comes out as $\left(\begin{array}{cccc} 3&-2\\5&-3 \end{array}\right)$

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  • $\begingroup$ How did you get those vectors? for example, $v_1+2v_3 = (2,0,4)^T$ $\endgroup$ – jmiller Jul 24 '15 at 11:58
  • $\begingroup$ $v_1$ and $v_3$ are first and third column vector of $T$. I am not sure, is this what you asked? $\endgroup$ – Sry Jul 24 '15 at 12:01
  • $\begingroup$ $[T]_B = (T(v_1)_B, T(v_2)_B, T(v_3)_B)$. Then we have $T(v_1)_B + 2T(v_2)_B = (2,0,4)$. $\endgroup$ – jmiller Jul 24 '15 at 16:19
  • $\begingroup$ but why is it true that $v_1 + 2v_2 = (2,0,4)$? $\endgroup$ – jmiller Jul 24 '15 at 16:19
  • $\begingroup$ $[T]_B = (T(v_1)_B, T(v_2)_B, T(v_3)_B)$ is not correct. Just think of $[T]_B(x_1,x_2,x_3)^t$ this will result into $x_1v_1+x_2v_2+x_3v_3$. So now you can think what are $v_1,v_2,v_3$ $\endgroup$ – Sry Jul 26 '15 at 6:45

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