# Showing that minimal polynomial has the same irreducible factors as characteristic polynomial

I'm trying to show that the minimal polynomial of a linear transformation $T:V \to V$ over some field $k$ has the same irreducible factors as the characteristic polynomial of $T$. So if $m = {f_1}^{m_1} ... {f_n}^{m_n}$ then $\chi = {f_1}^{d_1} ... {f_n}^{d_n}$ with $f_i$ irreducible and $m_i \le d_i$.

Now I've managed to prove this using the primary decomposition theorem and then restricting $T$ to $ker({f_i}^{m_i})$ and then using the fact that the minimal polynomial must divide the characteristic polynomial (Cayley-Hamilton) and then the irreducibility of $f_i$ gives us the result.

However I would like to be able to prove this directly using facts about polynomials/fields without relying on the primary decomposition theorem for vector spaces. Is this fact about polynomials true in general?

We know that $m$ divides $\chi$ and so certainly $\chi = {f_1}^{m_1} ... {f_n}^{m_n} \times g$ but then how do we show that $g$ must have only $f_i$ as it's factors? I'm guessing I need to use the fact that they share the same roots. And I'm also guessing that it depends on $k$, i.e. if $k$ is algebraically closed then it is easy because the polynomials split completely into linear factors.

Help is much appreciated, Thanks

• Do you know about characteristic subspaces? – Olivier Bégassat Jun 8 '14 at 14:36
• Hmmm, why primary decomposition? It is easy using decomposition into cyclic (not necessarily primary) submodules as per the invariant factor decomposition. Do you want to rule that out as well? – Marc van Leeuwen Jun 8 '14 at 14:42
• @OlivierBégassat I'm afraid I do not! – Wooster Jun 8 '14 at 14:48
• @MarcvanLeeuwen I'm aware of that decomposition as well, (is this rational canonical form for a general field?) but I would also like to rule that out. I just thought that there should be an elementary way using just polynomial rings and the fact that the two polynomials divide each other and share the same roots! Thanks – Wooster Jun 8 '14 at 14:49
• @Wooster They are also known as "generalized eigenspaces". Does that ring a bell? – Olivier Bégassat Jun 8 '14 at 15:01

## 4 Answers

One can reason using only some field theory, rather than linear algebra, if one prefers. I will assume the Cayley-Hamilton theorem, and the fact that all eigenvalues are roots of the minimal polynomial$~\mu$, which holds because an eigenvector for$~\lambda$ is killed by $P[T]$ (if and) only if $P$ is a multiple of $X-\lambda$.

First observe that the result is true when $\chi$ splits over $k$ (and hence so does$~\mu$, which divides$~\chi$ by the Cayley-Hamilton theorem), since the factors $X-\lambda$ of either polynomial are precisely those with $\lambda$ an eigenvalue. This in particular takes care of the case where $k$ is algebraically closed; the remainder deals with the case where $\chi$ does not split into linear factors in $k[X]$.

If $K/k$ is a field extension, one can extend scalars to obtain $\def\ext{\otimes_kK}T\ext: V\ext\to V\ext$. Since one gets identical matrices$~A$ for $T$ and for $T\ext$, using some $k$-basis of $V$ and the corresponding $K$-basis of $V\ext$, they have the same characteristic polynomials. They have the same minimal polynomials as well, since the minimal polynomial can be found from the unique solution of the linear system $x_0I+X_1A+\cdots+x_{d-1}A^{d-1}=A^d$ (one equation for each matrix coefficient) for the smallest $d$ for which this system has a solution at all; solving a linear system with coefficients in$~k$ over a larger field$~K$ will not change the existence of a solution, or the solution in case it is unique.

So over a sufficiently large field$~k$ (a splitting field of$~\chi$ will do), the polynomials $\mu$ and $\chi$ split, and give rise to the same set of linear factors $X-\lambda$ (though some may occur with larger multiplicity for$~\chi$ than for$~\mu$). In order to conclude that they also have the same set of irreducible factors in $k[X]$, I propose two arguments. One is that the presence of an irreducible factor$~f$ in $k[X]$ can be read off from the presence of any one of the factors into which $f$ splits in $K[X]$, because no two (monic) irreducible polynomials over$~k$ share a root in$~K$: if $\alpha\in K$ is a root of a monic irreducible $f\in k[X]$, then the minimal polynomial of$~\alpha$ over$~k$ is $f$ (this is actually the minimal polynomial of multiplication by$~\alpha$ viewed as $k$-linear map $K\to K$). The other argument is that, having the same set of (linear) factors, $\chi$ divides in $K[X]$ some power of $\mu$. But this division is valid in $k[X]$ as well (where both $\mu$ and $\chi$ live), which shows that the irreducible factors of$~\chi$ occur among those of$~\mu$ (and C-H gives the converse).

If you believe in invariant factors, this is easily seen as follows. The minimal polynomial is the last of the invariant factors (arranged so that each divides the next) and the characteristic polynomial is their product. It follows that the minimal polynomial divides the characteristic one (Cayley-Hamilton) and that the characteristic polynomial divides the minimal polynomial to the power the number of invariant factors. With each one dividing a power of the other, they must have the same irreducible factors.

Or if you don't (want to) believe in invariant factors, you can see the non-C-H direction by induction on the dimension, as follows. Leaving the $0$-dimensional base case as exercise, choose any nonzero vector$~v$, let $W$ be the cyclic submodule it generates and let $P\in k[X]$ be the minimal degree monic polynomial such that $P[T](v)=0$. Then $P$ is the characteristic polynomial of the restriction to$~W$ of$~T$ (a well known fact about cyclic modules; it also follows from C-H and dimension consideration). Then the full characteristic polynomial$~\chi$ of$~T$ is the product of $P$ and the characteristic polynomial$~\chi'$ of the linear operator $T_{/W}$ that $T$ induces in the quotient module $V/W$ (this follows by considering the block triangular matrix of$~T$ on a basis of$~V$ that extends one of$~W$). Now let $f$ be an irreducible factor of$~\chi$. If $f$ divides$~P$ we are done, since the minimal polynomial$~\mu$ of$~T$ is a multiple of$~P$ (since $\mu(T)(v)=0$). So suppose the contrary, then $f$ must divide$~\chi'$. By the inductive hypothesis $f$ divides the minimal polynomial$~\mu'$ of $T_{/W}$, which is the minimal degree monic polynomial such that $\mu'[T](V)\subseteq W$. But $\mu$ satisfies $\mu[T](V)=\{0\}\subseteq W$, so $\mu$ is a multiple of $\mu'$, and we conclude $f\mid\mu'\mid\mu$. QED

• Thank you for this answer. I was trying to avoid decomposition, however just out of interest, I can see why the minimal polynomial is the last of the invariant factors, but why is the characteristic polynomial their product? – Wooster Jun 8 '14 at 14:52
• The characteristic polynomial of a direct sum of modules (here cyclic) is the product of their characteristic polynomials. One does not even need a direct sum; for a submodule and the quotient by the submodule one has a product decomposition for the characteristic polynomial as well, as is used in the addition to my answer. – Marc van Leeuwen Jun 8 '14 at 18:41
• Thank you for the detailed and well-explained answer. However, I am surprised there isn't an answer to this question with just an elementary polynomial argument, if you have two polynomials, one dividing the other, and they share the same roots, then it seems that you should be able to deduce that they have the same irreducible factors? I thought that even using the primary decomposition theorem was too much. – Wooster Jun 8 '14 at 18:47
• Well, that is only true over algebraically closed fields, since otherwise the roots might just not be there. It is true that if two monic irreducible polynomials have a common root in the algebraic closure, then they must be equal. This requires a bit of field theory. But I was under the impression that you wanted to avoid extending scalars to the algebraic closure. – Marc van Leeuwen Jun 8 '14 at 18:56
• Ah okay, I was in fact looking for a more field theory/ring theoretic answer, I should mention that in the question. Although thank you for your direct argument! – Wooster Jun 8 '14 at 18:58

There is another proof using the properties of the determinant.

Claim: $\chi$ divides $m^n$ in $k[X]$.

To see this, fix a basis of $V$ and consider the matrix $M$ of $T$ in said basis. We see $\mathcal{M}_n(k),\mathcal{M}_n(k[X])$ as a subrings of $\mathcal{M}_n(k(X))$ given the natural embeddings $k\rightarrow k[X] \rightarrow k(X)$. In this context $\chi = \det(X.I_n-M)$ is a regular determinant of $X.I_n-M\in \mathcal{M}_n(k(X))$.

Write $m=\sum \limits_{k=0}^d a_kX^k$. We have: $m(X.I_n)=m(X.I_n) - m(M)=\sum \limits_{k=0}^d a_k(X^k.I_n-M^k)= \sum \limits_{k=1}^d a_k(X^k.I_n-M^k)=(X.I_n-M)\sum \limits_{k=1}^da_k\sum \limits_{p=0}^{k-1}X^p.M^{k-1-p}$.

Let $B:=\sum \limits_{k=1}^da_k\sum \limits_{p=0}^{k-1}X^p.M^{k-1-p}$. Since $B\in \mathcal{M}_n(k[X])$ where $k[X]$ is a ring, we have $\det(B) \in k[X]$.

We then compute the determinant of those matrices: $m^n=m^n.\det(I_n)=\det(m(X.I_n))=\det((X.I_n-M)B)=\det(X.I_n-M)\det(B)=\chi\det(B)$, hence the result.

### Motivation

$$\newcommand{\proofstart}{\mathbf{Proof.}\blacktriangleleft} \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\lcm}{lcm} \newcommand{\bm}{\boldsymbol} \newcommand {\F}{\Bbb F}$$ I was doing a problem and I need the claim in the OP, so i tried to prove this. My reference used the fact that $$\F$$ always have a field extension which is algebraic closed, but I have not learned field theory systematically, so I try to use the tools only from basic undergrad math and here is my proof only within the range of linear algebra and the theory of univariate polynomial ring over a field $$\F$$.

### The proof

$$\proofstart$$ We first prove that if $$m = p^r$$ where $$p\in \F[x]$$ is irreducible over $$\F$$, then $$\chi= p^n$$ for some $$n\geqslant r$$. By Cayley-Hamilton theorem, $$m \mid \chi$$. Thus $$\chi = p^n g$$ for some $$n\geqslant r$$. If $$\deg(g)>0$$ and $$p \nmid g$$ then $$V = \Ker(\chi(\mathcal T)) = \Ker(p(\mathcal T)^n) \oplus \Ker(g(\mathcal T))$$ by the fact #2. Let these subspaces be $$V_1,V_2$$ respectively, then $$\mathcal T|_{V_1}$$ is annihilated by $$p^n$$, $$\mathcal T|_{V_2}$$ is annihilated by $$g$$. Thus the minimal polynomial of $$\mathcal T|_{V_1}$$ is $$p^k$$ for some $$k\leqslant n$$, and that of $$\mathcal T|_{V_2}$$ is some $$d$$ that divides $$g$$. Thus $$m = p^k d$$ by the fact #4. Now by the fact #1, $$m=p^n$$ is the uniqueness decomposition, so we have $$p \mid d$$, then $$p\mid g$$ which is a contradiction. Therefore $$\deg(g) =0$$. Since $$m, \chi$$ are both monic, $$g = 1$$, then $$\chi = p^n$$.

Now for the general case, suppose $$\chi =\prod _1^s p_j^{n_j}$$ where $$p_j$$ are distinct irreducible monic polynomials. Then by the C-H theorem and the fact #3 $$V = \Ker(\chi(\mathcal T)) = \bigoplus_1^s \Ker(p_j(\mathcal T)^{n_j}).$$ Let $$V_j = \Ker(p_j(\mathcal T)^{n_j})$$. Then $$V_j$$ is invariant under $$\mathcal T$$. Then $$m_j$$, the minimal polynomial of $$\mathcal T|_{V_j}$$ is $$p_j^{r_j}$$ for some $$r_j \leqslant n_j$$, since $$p_j$$ is irreducible and $$m_j \mid p_j^{n_j}$$. Thus by #4, the minimal polynomial $$m$$ of $$\mathcal T$$ is $$m = \lcm(p_j^{r_j})_1^s = \prod_1^s p_j^{r_j}$$.

For the converse, suppose $$m = \prod_1^s p_j^{r_j}$$, then by the fact #3 $$V = \Ker(m(\mathcal T)) = \bigoplus_1^s \Ker(p_j(\mathcal T)^{r_j}).$$ Let $$W_j = \Ker(p_j(\mathcal T)^{r_j})$$, then $$W_j$$ is invariant under $$\mathcal T$$. Then the minimal polynomial of $$\mathcal T|_{W_j}$$ is $$\mu_j = p_j^{k_j}$$ for some $$k_j \leqslant r_j$$. Now apply the claim at the very beginning to $$\mathcal T|_{W_j}$$ and $$\mu_j$$, we have that the characteristic polynomial $$\chi_j$$ of $$\mathcal T|_{W_j}$$ is $$\chi_j = p_j^{n_j}$$ for some $$n_j \geqslant k_j$$. Noe use the fact #5, we have $$\chi =\prod_1^s p_j^{n_j}$$.

Conclusively, $$\chi$$ and $$m$$ always share the same irreducible factors. $$\blacktriangleright$$

### Appendices: the fact I used

First we should admit the following fact:

1.[Uniqueness Decomposition of Polynomials] In $$\Bbb F [x]$$ where $$\Bbb F$$ is a field, every $$f \in \Bbb F[x]$$ of degree $$\geqslant 1$$ could be uniquely decomposed as a product of finitely many irreducible polynomials in $$\F[x]$$.

Now we need a decomposition theorem.

2.Suppose $$f=gh$$ where $$f ,g,h \in \F[x]$$ and $$g,h$$ are coprime and of degrees both $$>0$$, then given an $$\F$$-linear space $$V$$, for all $$\mathcal T\in \mathcal L(V) [=\mathrm {End}(V)]$$, $$\Ker(f(\mathcal T)) = \Ker(g(\mathcal T)) \bigoplus \Ker(h(\mathcal T)).$$

$$\proofstart$$ Clearly $$\Ker(g(\mathcal T )), \Ker(h(\mathcal T)) \subseteq \Ker(f(\mathcal T ))$$, hence $$\Ker(g(\mathcal T)) + \Ker(h(\mathcal T)) \subseteq \Ker(f(\mathcal T))$$. For the converse, since $$g,h$$ are coprime, there exists $$u,v\in \F[x]$$ s.t. $$ug+vh =1$$. Thus $$u(\mathcal T)g(\mathcal T) + v(\mathcal T)h(\mathcal T) = \mathcal I$$. Then for every $$\alpha \in \Ker(f(\mathcal T))$$, $$\alpha = u(\mathcal T)g(\mathcal T)\alpha + v(\mathcal T)h(\mathcal T)\alpha.$$ Since $$u(\mathcal T) g (\mathcal T)\alpha \in \Ker (h(\mathcal T))$$, $$v(\mathcal T)h(\mathcal T)\alpha \in \Ker(g(\mathcal T))$$, we have $$\Ker(f(\mathcal T)) = \Ker(g(\mathcal T)) + \Ker(h(\mathcal T)).$$ To prove the sum is direct, suppose $$g(\mathcal T)\beta = h(\mathcal T)\beta = 0$$, then $$0 = u(\mathcal T) g(\mathcal T) \beta + v(\mathcal T)h(\mathcal T)\beta = \beta,$$ thus the sum is direct. $$\blacktriangleright$$

This could be generalized as:

3.Suppose $$f\in \F[x]$$ has a decomposition $$f = \prod_1^s f_j$$ where $$(f_j)$$ are pairwise coprime, then for $$\mathcal T\in \mathcal L(V)$$, $$\Ker(f(\mathcal T)) = \bigoplus_1^s \Ker(f_j (\mathcal T)).$$

This could be proved by induction on $$s$$.

Now for the minimal polynomial, we have

4.Suppose $$V$$ is an $$\F$$-linear space, $$\mathcal T \in \mathcal L(V)$$. If $$V$$ could be decomposed as $$V = \bigoplus_1^s V_j$$ where every $$V_j$$ is invariant under $$\mathcal T$$, then the minimal polynomial $$m\in \F[x]$$ of $$\mathcal T$$ and the minimal polynomials $$m_j$$ of $$\mathcal T|_{V_j}$$ satisfy $$m = \mathrm {lcm}(m_1, \dots, m_s).$$

$$\proofstart$$ Suppose $$p(\mathcal T) = \mathcal O$$ for $$p \in \F[x]$$. Then $$p(\mathcal T|_{V_j}) = \mathcal O$$ as well. Therefore $$m_j\mid p$$ for all $$j$$, thus $$g = \lcm(m_j)_1^s$$ satisfies $$g \mid p$$. Specifically $$m(\mathcal T) = \mathcal O$$, so $$g \mid m$$. Let $$g = m_j d_j$$ for $$j =1, \dots, s$$. By the direct sum expression, each $$\alpha$$ is actually $$\alpha = \sum_1^s \alpha_j$$ for $$\alpha_j \in V_j$$. Thus $$g(\mathcal T)\alpha = \sum_1^s(d_j \cdot m_j)(\mathcal T) \alpha_j = \sum_1^s (d_j\cdot m_j) (\mathcal T|_{V_j}) \alpha_j =0,$$ i.e. $$g(\mathcal T) = \mathcal O$$. Thus $$m \mid g$$ by the minimal property of $$m$$. Since $$m,g$$ are monic, $$m =g.\blacktriangleright$$

5.Under the same assumption of #4, suppose $$\chi_j$$ is the characteristic polynomial of $$\mathcal T|_{V_j}$$, then $$\chi = \prod_1^s \chi_j$$.

$$\proofstart$$ By the decomposition of $$V$$, we could find a basis of $$V$$ under which $$\mathcal T$$ has a block diagonal matrix $$\bm T =\mathrm {diag}(\bm T_1, \dots, \bm T_s)$$, where $$\bm T_j$$ is a $$\dim(V_j) \times \dim(V_j)$$ matrix over $$\F$$. Then we actually have $$\chi(x) = \det (x\bm I - \bm T) = \prod_1^s (x \bm I - \bm T_j) = \prod_1^s \chi_j(x),$$ as desired. $$\blacktriangleright$$