I'm trying to put a matrix into RCF, and I keep running into problems. I've checked my work a few times, so I think I must be making a conceptual error. Here's what I've got: $$A=\left( \begin{matrix}2 & -2 & 14 \\ 0 & 3 & -7 \\ 0 & 0 & 2\end{matrix}\right)\quad \text{ so }\quad xI-A=\left( \begin{matrix}x-2 & 2 & -14 \\ 0 & x-3 & 7 \\ 0 & 0 & x-2\end{matrix}\right).$$Diagonalizing gives $$\overset{C_1\leftrightarrow C_2}{\longrightarrow}\left( \begin{matrix}2 & x-2 & -14 \\ x-3 & 0 & 7 \\0 & 0 & x-2\end{matrix}\right)\overset{R_2-(\frac{1}{2}x-\frac{3}{2})R_1\mapsto R_2}{\longrightarrow}\left( \begin{matrix}2 & x-2 & -14 \\ 0 & -\frac{1}{2}x^2+\frac{5}{2} x-3& 7x-14 \\0 & 0 & x-2\end{matrix}\right)$$$$\overset{C_2-(\frac{1}{2}x-1)C_1\mapsto C_2}{\longrightarrow}\left( \begin{matrix}2 & 0 & -14 \\ 0 & -\frac{1}{2}x^2+\frac{5}{2} x-3& 7x-14 \\0 & 0 & x-2\end{matrix}\right)\overset{C_3+7C_1\mapsto C_3}{\longrightarrow}\left(\begin{matrix}2 & 0 & 0 \\ 0 & -\frac{1}{2}x^2+\frac{5}{2} x-3& 7x-14 \\0 & 0 & x-2\end{matrix}\right)$$$$\overset{C_3 \leftrightarrow C_2}{\longrightarrow}\left( \begin{matrix}2 & 0 & 0 \\ 0 & 7x-14 & -\frac{1}{2}x^2+\frac{5}{2} x-3\\0 & x-2 & 0 \end{matrix}\right)\overset{R_2\leftrightarrow R_3}{\longrightarrow}\left( \begin{matrix}2 & 0 & 0 \\ 0 & x-2 & 0\\0 & 7x-14 & -\frac{1}{2}x^2+\frac{5}{2} x-3 \end{matrix}\right)$$$$\overset{R_3-7R_2\mapsto R_3}{\longrightarrow}\left( \begin{matrix}2 & 0 & 0 \\ 0 & x-2 & 0\\0 & 0 & -\frac{1}{2}x^2+\frac{5}{2} x-3 \end{matrix}\right) \overset{ \frac{1}{2}C_1 }{ \longrightarrow } \left( \begin{matrix}1 & 0 & 0 \\ 0 & x-2 & 0\\0 & 0 & -\frac{1}{2}x^2+\frac{5}{2} x-3 \end{matrix}\right)$$$$\overset{-2C_3}{\longrightarrow}\left( \begin{matrix}1 & 0 & 0 \\ 0 & x-2 & 0\\0 & 0 & x^2-5x+6 \end{matrix}\right).$$The row operations we used were $$R_2-(\frac{1}{2}x-\frac{3}{2})R_1\mapsto R_2, R_2\leftrightarrow R_3, R_3-7R_2\mapsto R_3,$$ giving operations on the basis elements $$\{\beta_1, \beta_2, \beta_3\}$$ as follows: $$\beta_1+(\frac{1}{2}x - \frac{3}{2})\beta_2\mapsto \beta_1,$$ $$\beta_2\leftrightarrow \beta_1, \beta_2+7\beta_3\mapsto \beta_2.$$ So if we start with the standard basis for $$\mathbb{Q}^3 \quad \varepsilon_1=(1,0,0)\quad \varepsilon_2=(0,1,0)\quad \varepsilon_3=(0,0,1),$$ they should change to $$(\varepsilon_1, \varepsilon_2, \varepsilon_3)\to (\varepsilon_1+(\frac{1}{2}x-\frac{3}{2})\varepsilon_2, \varepsilon_2, \varepsilon_3)\to (\varepsilon_1+(\frac{1}{2}x-\frac{3}{2})\varepsilon_2, \varepsilon_3, \varepsilon_2)$$$$\to (\varepsilon_1+(\frac{1}{2}x-\frac{3}{2})\varepsilon_2, \varepsilon_3+7\varepsilon_2, \varepsilon_2).$$(adsbygoogle = window.adsbygoogle || []).push({});

I calculate that first entry is zero, as it should be since we have a unit in the first entry of the matrix. Then I get that the other two are $$(1,0,0)+7(0,1,0)=(0,7,1)$$ and $$(0,1,0).$$ This would give a conjugation matrix $$P=\left( \begin{matrix}0 & 0 & -2 \\ 7 & 1 & 3\\1 & 0 & 0 \end{matrix}\right).$$But this is not the answer in the text. What am I doing wrong? This is driving me crazy.

Here's a link to the way they do it in the text, using a different way to row reduce.

http://imgur.com/8BLgD

http://imgur.com/1OJuU

http://imgur.com/N7riK

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# Putting a matrix into Rational Canonical Form

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