# question about representation of the minimum polynomial

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 $$R$$.

• 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. – jmerry Feb 8 '19 at 22:56
• In position 1,3 is -2 in place of 2 – Ricardo Freire Feb 9 '19 at 0:00
• OK, everything fits now. – jmerry Feb 9 '19 at 0:28

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!