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I am asked to find the minimal polynomial of the matrix
\begin{bmatrix} 4&-2&2\\ 6&-3&4\\ 3&-2&3\end{bmatrix}
I've calculated the characteristic polynomial is $\Delta(x)=(x-2)(x-1)^2$.
As we know that the minimal polynomial $m(x)$ must divide $\Delta(x)$, so the possibilities of being $m(x)$ are $(x-2),(x-1),(x-1)^2,(x-2)(x-1),(x-2)(x-1)^2$.
Among these, by Cayley-Hamilton theorem, for which least degree factor $\Delta(A)=O$ is satisfied will be the minimal polynomial.

But for some reason, the first three above mentioned factors were ignored in my book. What can be the possible reason for this?

Manjoy Das
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  • If $v\neq0$ is an eigenvector of $T$ for the eigenvalue $r$, and $E=(v)$ is the subspace generated by $v$, then the minimal polynomial of $T|_E$ divides the minimal polynomial of $T$. This is because all polynomials $p$ such that $p(T)=0$ will also make $p(T_E)=0$. Since $T_E=rI$, the minimal polynomial of $T_E$ is $x-r$. Therefore, $x-r$ divides the minimal polynomial of $T$. – plop Aug 16 '22 at 19:23

1 Answers1

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Because every root of the characteristic polynomial is always a root of the minimal polynomial. In the case, for instance, of $x-1$, $2$ is not a root, but it is a root of that characteristic polynomial.

José Carlos Santos
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