# Minimal polynomial and cyclic subspace.

I'm lost with minimal polynomials. I have to prove that the degree of the minimal polynomial $$m_{x_0}(x)$$ such that cancels a vector $$x_0 \in V$$ is equal to the dimension of the cyclic subspace spanned by $$x_0$$. In addition, $$m_{x_0}(x)$$ divides other polynomial that cancels $$x_0$$. I think that is straightforward from definition but I don't know how.

• I can't make sense of the phrase "the degree of the minimal polynomial $m_{x_0}(x)$ such that cancels a vector $x_0 \in V$". Could you explain what $m_{x_0}(x)$ is supposed to mean a bit more clearly? – Omnomnomnom Nov 25 '18 at 4:46

### Demo

Given an operator $$\mathcal T$$, the cyclic space generated by $$v\in V$$ is $$C_v=\mathrm {span}\{v, \mathcal Tv, \dots, \mathcal T^k v, \dots\}.$$ Assume $$V$$ is finite dimensional, then there is some $$p\in \mathbb N^*$$ s.t. $$(\mathcal T^j v)_0^{p-1}$$ is a linearly independent set, but $$(\mathcal T^j v)_0^{p}$$ is not. So there exists a set of scalars $$(a_j)_0^{p-1}$$ s.t. $$\mathcal T^p v = \sum_0^{p-1} a_j \mathcal T^j v.$$ Hence the polynomial $$g(x ) = x^p - \sum_0^{p-1}a_j x^j$$ satisfies that $$g(\mathcal T)v = 0$$, thus $$g(x)$$ annihilates $$\mathcal T|_{C_v}$$. Let $$m(x)$$ be the desired minimal polynomial, then since $$g(\mathcal T)v =0$$, $$m|g$$. On the other hand, if some monic polynomial $$r(x)$$ with degree $$q satisfying $$r(\mathcal T)v =0$$, then $$(\mathcal T^j v)_0^q$$ is linear dependent, contradicting the definition of $$p$$. So $$\deg m \geqslant \deg g$$ where $$m$$ is an annihilating polynomial for $$v$$. Since $$m, g$$ are monic by definition, $$m = g$$, hence $$\deg m = \deg g$$. By the definition of $$p$$, we could conclude that $$\dim(C_v)=p$$, hence the desired conclusion.

### UPDATE

Suppose $$g(x)$$ annihilates $$v$$, and $$m(x)$$ is the minimal one. Then perform the division with remainder, we could get polynomials $$q(x), r(x)$$ s.t. $$g =mq + r$$ where $$\deg r < \deg m$$. If $$r \neq 0$$, then since $$m$$ is the minimal one, $$m$$ annihilates $$v$$. Then $$0=g(\mathcal T)v = m(\mathcal T)q(\mathcal T)v + r(\mathcal T) = 0 + r(\mathcal T)v.$$ If $$r \neq 0$$, then $$r$$ is a polynomial annihilating $$v$$ with $$\deg r <\deg m$$, contradicting the minimality of $$m$$. Hence $$g=mq$$, equivalently $$m | g$$.

1. Peter Petersen Linear Algebra

A second course for LA. I think the choice of proofs are special.

1. Sheldon Axler Linear Algebra Done Right

A second course with theory developed without introducing determinants. It is pretty geometric.

1. Kenneth Hoffman, Ray Kunze Linear Algebra

A classic with deeply developed theories. Should be revisited after the 1st or 2nd exposure to the subject.

The crucial part is to use the polynomials to study the structure of linear spaces, which is covered by these books.

• How can you conclude that? – Bayesian guy Nov 25 '18 at 18:13
• Which part is confusing? – xbh Nov 26 '18 at 3:49
• The last one. By the definition of p... – Bayesian guy Nov 26 '18 at 3:55
• @Bayesianguy $p$ is the maximal integer s.t. $(\mathcal T^jv)_0^{p-1}$ is linear independent. You could show this set is actually a basis of $C_v$. If necessary, i would add the proof. – xbh Nov 26 '18 at 4:01
• @Bayesianguy I updated. The proof is easy to find in standard linear algebra texts. – xbh Nov 26 '18 at 4:30