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Let $T$ be a linear operator on $R^3$ which is represented in the standard ordered basis by the matrix $\begin{pmatrix} 6 &-3 &-2 \\ 4 &- 1& -2 \\ 10 &- 5& -3 \\ \end{pmatrix}$

Express the minimal polynomial $p$ for $T$ in the form $p = p_1p_2$, where $p_1$ and $p_2$ are monic and irreducible over the field of real numbers.

The characteristic polynomial is: $(x-2)(x^2+1)$. In this case who will be the minimum polynomial? The characteristic polynomial itself? I can not decompose more than this because exercise demands that they $p_1$ and $p_2$ be irreducible in $R$.

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  • $\begingroup$ I think there's a transcription error - as written, I get a determinant of $18+60-40-60-36+20=-38$ and $2$ is not an eigenvalue. On the other hand, if the $2$ in the upper right were $-2$ instead, that determinant becomes $2$ and we easily see $2$ is an eigenvalue. $\endgroup$ – jmerry Feb 8 '19 at 22:56
  • $\begingroup$ In position 1,3 is -2 in place of 2 $\endgroup$ – Ricardo Freire Feb 9 '19 at 0:00
  • $\begingroup$ OK, everything fits now. $\endgroup$ – jmerry Feb 9 '19 at 0:28
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Well actually $X-2$ and $X^2+1$ are irreducible over $\mathbb R$.

As for the minimal polynomial, you know that it divides $(X-2)(X^2+1)$. Notice that $X-2$ and $X^2+1$ do not annihilate your matrix. Hence the minimial polynomial is the characteristic polynomial itself!

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The minimal polynomial has the same irreducible factors as the characteristic polynomial, so it is divisible by both factors. As it also divides the characteristic polynomial, the two are indeed the same.

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