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I have a 2x2 matrix where I need to find the rational canonical form over the field of rational numbers and real numbers. The matrix given is

$$A=\begin{pmatrix}2 &-1 \\ 1 & -1\end{pmatrix}.$$

What I've done is found the characteristic and minimal polynomials which happen to be the same ($x^2-x-1$). Using the format $c_2x^2 - c_1x - c_0$, I took the coefficients of c_1 and c_0 to get the rational canonical form to be

$$\begin{pmatrix}0 &-c_0\\1& -c_1\end{pmatrix} = \begin{pmatrix}0 & 1 \\ 1 & 1\end{pmatrix}.$$

Am I on the right track or completely off. Also, wouldn't this be the RCF for both the rational and real fields? Any help is appreciated. Thanks

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  • $\begingroup$ Over the reals, the matrix is diagonalizable. $\endgroup$ – Sammy Black Nov 18 '14 at 21:38
  • $\begingroup$ Got it, thanks! Is my RCF correct for the rational field though? $\endgroup$ – Raptors1102 Nov 18 '14 at 21:52
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Hint: Over the real numbers, $x^2 - x - 1 = (x - \phi_+)(x - \phi_-)$, where $$ \phi_+ = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad \phi_- = \frac{1 - \sqrt{5}}{2}. $$

In the canonical form, you get a block of sized $d \times d$ for each irreducible factor of degree $d$. Hence, over the reals, you have two $1 \times 1$ blocks, i.e., a diagonal matrix whose diagonal entries are the eigenvalues: $$ \begin{bmatrix} \phi_+ & 0 \\ 0 & \phi_- \end{bmatrix}. $$

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