# What are some meaningful connections between the minimal polynomial and other concepts in linear algebra?

I’ve found that the most effective way for me to deeply grasp mathematical concepts is to connect them to as many other concepts as I can. Unfortunately, I’m seeing neither the importance nor the relevance of the minimal polynomial at all. Are there any significant connections between the minimal polynomial and linear algebra? Particularly regarding the relationship between the minimal polynomial of a linear operator on a vector space and other properties of that operator?

I think the problem is partially due to my textbook's emphasis on crunching out as many theorems about a concept as possible, rather than explaining it on a deep level and demonstrating its importance. But part of it is definitely also that this concept is not clicking well for me. Thanks for any help.

• Perhaps an obvious one, but: Connection to Invertibility: Let $T : V \to V$ be a linear transformation of a finite dimensional vector space $V$ over a field $\mathbb{F}$. Then $T$ is invertible if and only if $\lambda$ does not divide the minimal polynomial $m_T(\lambda)$. – user137731 Jun 7 '17 at 0:18
• Connection to Diagonalizibility: Let $T:V\to V$ as above. Then $T$ is diagonalizable if and only if its minimal polynomial factors completely over $\Bbb F$ into distinct linear factors. – user137731 Jun 7 '17 at 0:21
• What in your eyes is exactly the difference between crunching out a theorem about a concept and demonstrating its importance on a deep level? Maybe you just want a bit more comments to what the theorems are really saying. – Marc van Leeuwen Jun 7 '17 at 4:50
• Marc Van Leeuwen, you make a good point in that by not being comfortable with the concept yet, I am less capable of distinguishing between whether or not a theorem regarding the concept is important. That being said, I was very comfortable with the previous chapters and most of their concepts, yet I still found that the author prioritized quantity of theorems over explaining and connecting theorems. For example, after stating and proving the highly important rank-nullity theorem, the author made a few sentence long remark, and then moved on. I strongly disagree with that teaching approach. – Calvin John Jun 7 '17 at 5:04

This answer is devoted to the connection between commutative groups and linear algebra. It especially focuses on the link between exponent and minimal polynomial of an endomorphism, as the original poster rightfully find useful to connect notions between them.

Definition. Let $$R$$ be a commutative ring with unity, then a $$R$$-module is a triple $$(M,+,\cdot)$$, where $$(M,+)$$ is a commutative group and $$\cdot\colon R\times M\rightarrow M$$ is such that the following properties hold: \begin{align}r\cdot(x+y)&=r\cdot x+r\cdot y\\(r+s)\cdot x&=r\cdot x+s\cdot x\\(rs)\cdot x&=r\cdot (s\cdot x)\\1\cdot x&=x\end{align}

Remark. Modules over a field are exactly vector spaces over this field so that the notion of module is a generalisation of vector spaces to rings.

Observation. The $$\mathbb{Z}$$-modules are exactly commutative groups.

Proof. The direct implication follows from the definition.

Let $$(G,+)$$ be a commutative group and let define $$\cdot\colon\mathbb{Z}\times G\rightarrow G$$ by: $$0\cdot g:=0_G,n\cdot g:=(n-1)\cdot g+g,(-n)\cdot g:=n\cdot(-g).$$ Then, $$(G,+,\cdot)$$ is a $$\mathbb{Z}$$-module. $$\Box$$

Observation. Let $$k$$ be a field, then $$k[T]$$-module are exactly the $$k$$-vector spaces endowed with an endomorphism.

Proof. Let $$(M,+,\cdot)$$ be a $$k[T]$$-module, then notice that by restriction, $$(M,+,\cdot_{\vert k\times M})$$ is a $$k$$-vector space. Furthermore, $$\varphi\colon M\rightarrow M$$ defined by: $$\varphi(m):=T\cdot m$$ is an endomorphism of $$M$$.

Conversely, let $$(E,+,\cdot)$$ be a vector space and $$\varphi\in\textrm{End}(E)$$, then let define $$\star\colon k[T]\times E\rightarrow E$$ by: $$f\star x:=f(\varphi)(x).$$ Then, $$(E,+,\star)$$ is a $$k[T]$$-vector space. $$\Box$$

Definition. Let $$M$$ be a $$R$$-module, then $$M$$ is finitely generated if and only if there exists finetely many elements $$x_1,\ldots,x_n$$ of $$M$$ such that: $$M=\bigoplus_{k=1}^nRx_k:=\left\{\sum_{k=1}^nr_kx_k;r_k\in R\right\}.$$

Remark. Respectively, this extends the notion of finitely generated commutative groups and finite-dimensional vector spaces.

Theorem. Let $$R$$ be a principal ideal domain and $$M$$ be a finitely generated $$R$$-module, then there exists $$d_1\vert\cdots\vert d_n$$ elements of $$R\setminus R^\times$$ such that: $$M=\bigoplus_{k=1}^nM/(d_k).$$ Furthermore, the $$d_k$$ are unique up to multiplication by a unit of $$R$$.

Proof. See the corresponding chapter in Basic Algebra I by N. Jacobson. $$\Box$$

Remark. This theorem gives the structure of finitely generated commutative group as a direct sum of cyclic groups and the Frobenius decomposition.

Finely, here is the connection I claimed:

In the case of $$R=\mathbb{Z}$$, the least common multiple of the $$d_k$$ in the theorem leads to the notion of the exponent of a commutative group and for $$R=k[T]$$ to the minimal polynomial of an endomorphism.

In a sense, every theorem on the exponent of a commutative group can be transposed in the linear algebra world through the minimal polynomial. Here are a few examples, in particular, you will see how the minimal polynomial offers informations on the linear transformation:

Proposition. Let $$G$$ be a finite abelian group, then $$G$$ is cyclic if and only if its order equals its exponent.

Proposition. Let $$E$$ be a $$n$$-dimensional vector space and $$\varphi\in\textrm{End}(E)$$, then there exists $$x\in E$$ such that $$\left\{x,\varphi(x),\cdots,\varphi^{n-1}(x)\right\}$$ is a basis of $$E$$ if and only if the minimal polynomial of $$\varphi$$ equals its characteristic polynomial.

Remark. In the case $$R=\mathbb{Z}$$, the product of the $$d_k$$ in the theorem is equal to the order of the group and for $$R=k[T]$$ to the characteristic polynomial.

Proposition. Let $$G$$ be a group of prime order, then $$G$$ is a simple group.

Proposition. Let $$E$$ be a $$n$$-dimensional vector space and $$\varphi\in\textrm{End}(E)$$, if the minimal polynomial of $$\varphi$$ is irreducible, then $$\{0\}$$ and $$E$$ are the only $$\varphi$$-invariant subvector spaces of $$E$$.

• The downvote hurts a bit, I did my best giving a short introduction to module-theory that leads straight to the connection between group-theory and linear algebra. Even though, it might be over the head of the original poster it can probably give some insight to more advanced students. – C. Falcon Jun 7 '17 at 1:19
• You can't take it personally. Downvotes happen. Sometimes without any obvious reason why. – user137731 Jun 7 '17 at 1:42
• I'm not the downvoter, nor am I claiming that this is a bad answer, but presumably the downvote may have been because the op was complaining about long lists of theorems, and then you wrote a bunch of theorems:) – shalop Jun 7 '17 at 7:09
• @CalvinJohn I am glad that my answer provided you some deep insight! I am also grateful that you put some time to understand modules, it is a good investment. – C. Falcon Jun 7 '17 at 12:23

Let $V$ be a finite dimensional vector space over a field $F$.

If a polynomial $f \in F[X]$ is the product $g_1 \cdots g_m$ of coprime polynomials, then $$\ker f(T) = \ker g_1(T) \oplus \cdots \oplus \ker g_m(T)$$ When $f(T)=0$, we have $\ker f(T)=V$ and a decomposition of $V$ into invariant subspaces. This an important tool for understanding $T$.

It's natural to consider the simplest $f$ such that $f(T)=0$. This is the minimal polynomial.

By factoring $f$ into irreducible polynomials, we get the primary decomposition theorem.

When $f$ splits into linear factors (which happens for instance when $F$ is algebraically closed), we get a decomposition into generalized eigenspaces, which leads to the Jordan canonical form.

Thus, there is a close connection between polynomials and linear transformations. Somewhat surprisingly, this connection depends on the arithmetic properties of the field $F$, which for the most part of linear algebra does not really play a part, until you reach the decomposition theorems.

Here is one. The following two conditions are equivalent:

(I) the characteristic polynomial of square ($n$ by $n$) matrix $A$ and the minimal polynomial are the same

(II) the only matrices that commute with $A$ are of the form $$a_0 I + a_1 A + a_2 A^2 + \cdots + a_{n-1} A^{n-1}$$

The set of matrices in (II) make up a vector space of dimension $n.$ The set of all matrices (such as commute with $I$) is of dimension $n^2,$ much bigger

• Good; just a minor remark: I'd prefer to state (I) as "the minimal polynomial of (the $n\times n$ matrix) $A$ has degree $n$"; the characteristic polynomial really plays no role in this equivalence. Also maybe for (II) "every matrix that commutes with $A$ can be written as polynomial in $A$", as in this part the bound $n$ appears a bit of an afterthought. See also questions math.stackexchange.com/q/57308 and math.stackexchange.com/q/81467. – Marc van Leeuwen Jun 7 '17 at 5:05