Showing characteristic and minimum polynomial coincide for this particular matrix

Question

Given a matrix such as $$\pmatrix{0 & 0 & 2 \\ 1 & 0 & -1 \\0 & 1 & 1 \\ },$$ whose characteristic polynomial is $-X^3+X^2-X+2.$ $$$$How it could be deduced that it equals minus the minimum polynomial ?$$$$Thanks in advance.

Answer 1

It is easier to see that this is the minumum polynomial than that it is the characteristic polynomial. In fact this is a companion matrix, of which the final column always gives minus the non-leading coefficients of the minimum polynomial, which is also the characteristic polynomial, see this question or this answer.

Explicitly, if $e_i$ are the canonical basis vectors, then $n-1$ repeated multiplications by any companion matrix send $e_1\mapsto e_2\mapsto e_3\mapsto\cdots e_{n-1}\mapsto e_n$; one sees that no polynomial of degree $d$ with $0\leq d<n$ applied to $e_1$ gives $0$, and the minimal polynomial must be of degree at least $n$ (in fact exactly $n$ by Cayley-Hamilton, but I'm avoiding use of that much more complicated theorem). If the final column has successive entries $c_0,\ldots c_{n-1}$, this means that $$A\cdot e_n=c_0e_1+c_1e_2+\cdots+c_{n-1}e_n=c_0A^0\cdot e_1+c_1A^1\cdot e_1+\cdots+c_{n-1}A^{n-1}\cdot e_1=(c_0I+c_1A+\cdots+c_{n-1}A^{n-1})\cdot e_1,$$ so the indicated polynomial $P$ of $A$ annihilates $e_1$. Since it commutes with $A$, it also annihilates $A^i\cdot e_1=e_{i+1}$ for $i=1,\ldots,n-1$, in other words it annihilates as basis and therefore the whole space.

Answer 2

Think of this as the matrix of an endomorphism $u$ expressed in a suitable basis $(e_1,e_2,e_3)$: the three vectors $(e_1,u(e_1),u^2(e_1))$ form a basis (actually, $u(e_1)=e_2$ and $u^2(e_1)=e_3$). Now suppose $\mu$ is $u$'s minimal polynomial. What does the preceding remark tell us about the degree of $\mu$?

Well, suppose $Q$ is a non-zero polynomial of degree at most $2$, say $Q(X)=aX^2+bX+c$. Then $Q(u)\neq 0$, for if we apply the endomorphism $Q(u)$ to the vector $e_1$, the result is non zero: $$Q(u)\Big(e_1\Big)=au^2(e_1)+bu(e_1)+ce_1\neq 0.$$ That's because the family $(e_1,u(e_1),u^2(e_1))$ is linearly independent, and, by virtue of $Q\neq 0$, at least one of the three coefficients $a,b,c$ is non zero. Since $\mu(u)=0$, the preceding implies $\mathrm{deg}(\mu)\geq 3$. Also, by definition $\mu$ has leading coefficient $1$, and by the Cayley-Hamilton theorem it divides $P$ which has degree $3$ and leading coefficient $-1$. All of this together implies that $\mu=-P$.

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The same reasoning will show that if $x,u(x),\dots,u^{p-1}(x)$ are linearly independent where $x$ is some vector in a finite dimensional vector space $V$, and $u$ is an endomorphism of $V$, then $u$'s minimal polynomial has degree $\geq p$.