# Minimal polynomial for an invertible matrix and its determinant

So here's one that I can't quite crack:

Let $A\in M_n(\mathbb{F})$ be an invertible matrix with integer eigenvalues.
Its minimal polynomial is $m_A(\lambda)=\lambda^k+b_{k-1}\lambda^{k-1}+...+b_0$.
Prove that $detA$ is divisible by $b_0$.

I'm afraid I don't quite see the connection between the minimal polynomial and the determinant.
All I know is that for an invertible matrix $b_0 \neq 0$, but I can't find a way to relate that to the determinant.
Any tips and guidance would be much appreciated!
Thanks in advance.

• Minimal polynomial is a factor of characteristic polynomial. – Yimin May 20 '15 at 16:48
• All roots of char. polynomial are roots of minimal poly. too. Just no. of time they occur in min. poly may be lesser. So.... – Sry May 21 '15 at 11:43
• This is a weird question, since divisibility over a field is a trivial relation on nonzero elements. If one is asking to show that $\det A=0$ if and only if $b_0=0$, then why not just say so? – Marc van Leeuwen May 24 '15 at 11:32
• As no answer to my previous comment is forthcoming, I'll suppose that divisibility in $\Bbb Z$ is intended, not in $\Bbb F$. It would have been clearer if it had said: prove that the coefficients $b_i$ are all integral, and that $\det(A)$ divides $b_0$ in$~\Bbb Z$. – Marc van Leeuwen Oct 7 '18 at 10:54

## 4 Answers

There exists a polynomial $p(t)$ such that $$\DeclareMathOperator{char}{char}\char_A(t)=p(t)m_A(t)$$ It follows that $$(-1)^n\det(A)=\char_A(0)=p(0)m_A(0)=p(0)b_0$$ That is, $$\det(A)=(-1)^np(0)b_0$$ Hence $b_0$ divides $\det(A)$.

• Ah, it clicks now.<br> Thank you! – Elad Avron May 20 '15 at 17:02

Hint:

For any square matrix of order $n$, the constant term of the characteristic polynomial $\chi_A$is $(-1)^n\det A$, and the minimal polynomial of $A$ divides $\chi_A$.

• I know that the minimal polynomial divides the characteristic polynomial, but does that guarantee that its constant divides the constant as well? – Elad Avron May 20 '15 at 16:51
• Of course. Btw, this is also true for the dominant coefficient. – Bernard May 20 '15 at 16:59

For any ring $R$, and any polynomials $f(x), d(x), q(x) \in R[x]$, we have

$f(x) = d(x) q(x) \Rightarrow f_0 = d_0 q_0, \tag{1}$

where $f_0$ is the constant term of $f(x)$ etc. This is easy to see by simply examining the basic rule for polynomial multiplication, with

$d(x) = \sum_0^{\deg d} d_i x^i, \tag{2}$

$q(x) = \sum_0^{\deg q} q_i x^i, \tag{3}$

$f(x) = \sum_0^{\deg d + \deg q} f_j x^j = d(x)q(x) = \sum_{j = 0}^{j = \deg d + \deg q} (\sum_0^j d_i q_{j - i})x^j; \tag{4}$

looking at the coefficient of $x^0$ in the inner sum, we see from (4) that (1) binds. Now if the eigenvalues $\lambda_i$, $1 \le i \le n$, of $A$ are integers, it follows that $\det A = \prod_1^n \lambda_i$ is itself an integer, and since we may write the characteristic polynomial $p_A(x)$ as

$p_A(x) = \prod_1^n (x- \lambda_i), \tag{5}$

it follows from the fact that the minimal polynomial $m_A(x)$ of $A$ must divide $p_A(x)$ that $m_A(x)$ may be written

$m_A(x) = \prod_{i \in S} (x - \lambda_i) \tag{6}$

for some subset $S \subset \{1, 2, \ldots, n \}$; thus we see that the degree-zero term of

$m_A(x) = \sum_0^{\vert S \vert} b_j x^j, \tag{7}$

where $\vert S \vert$ is the cardinality of $S$, is given by

$m_0 = \pm \prod_{j \in S} \lambda_j, \tag{8}$

so $m_0$ itself is also an integer. It then follows from (1) etc. above that $m_0 \mid \det A$, as integers, the desired conclusion.

Nota Bene: Assuming $\Bbb F$ is a field ($\Bbb F$ for field, right?), the fact that $m_A(x)$ may be factored in the form (6) follows from the fact that $\Bbb F[x]$ is a principal ideal domain, hence unique factorization into primes (which in a PID are essentially the same as irreducibles) applies; the $x - \lambda_i$, each being irreducible in $\Bbb F[x]$, are thus primes etc. etc. etc. End of Note.

All roots of the minimal polynomial are eigenvalues, so here that (monic) polynomial is of the form $$\prod_{i=1}^k(X-a_i)$$ for some sequence $$(a_1,\ldots,a_k)$$ of integers, all eigenvalues, with maybe some repetitions. Then $$b_0= \prod_{i=1}^k(-a_i)$$. For the characteristic polynomial the situation is the same, but maybe with more repetitions of certain eigenvalues. The constant term then is an integer multiple of $$b_0$$, and on the other hand the constant term of the characteristic polynomial is $$\det(-A)$$. The result follows.