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The companion matrix of a monic polynomial in 1 variable over a field $F$ plays an important role in understanding the structure of finite dimensional $F[x]$-modules.

It is an important fact that the characteristic polynomial and the minimal polynomial of $C(f)$ are both equal to $f$. This can be seen quite easily by induction on the degree of $f$ .

Does anyone know a different proof of this fact? I would love to see a graph theoretic proof or a non inductive algebraic proof, but I would be happy with anything that makes it seem like more than a coincidence!

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4 Answers 4

up vote 6 down vote accepted

Suppose your matrix is over a field $\mathbb{F}$. Look at $G = F[x]/f$, where $f$ is your polynomial of degree $n$. Then $G$ is a vector space over $\mathbb{F}$, and $C(f)$ is the matrix (with respect to the basis $1,x,x^2,\ldots,x^{n-1}$) corresponding to the linear operator $g \mapsto x \cdot g$.

Since $f = 0$ in $G$, also $fx^i = 0$ in $G$, and so $f$ is a polynomial of degree $n$ such that $f(C(f)) = 0$; thus it must be the characteristic polynomial of $C(f)$. Moreover, any polynomial $g$ of smaller degree does not reduce to $0$ in $G$, so in particular $g(C(f))$ applied to the vector $1$ does not equal the zero vector. So $f$ is the minimal polynomial of $C(f)$.

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Shouldn't your «thus it must be the characteristic polynomial of $C(f)$» be «thus $f$ is divisible by the minimal polynomial of $C(f)$»? –  Mariano Suárez-Alvarez Nov 14 '10 at 23:12
    
@Mariano: both polynomials have the same degree. –  Yuval Filmus Nov 15 '10 at 3:07
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@YuvalFilmus: Mariano is right, it is strange to conclude that the polynomial annihilating $C(f)$ must be its characteristic polynomial, and later conclude that it must also be the minimal polynomial because that has degree $n$ too. The proper reasoning is: $f$ is the minimal polynomial because it annihilates and no lower degree polynomial does so; given this it follows by Cayley-Hamilton that it is also the characteristic polynomial. –  Marc van Leeuwen Jan 31 '13 at 8:55

This is essentially Yuval's answer expressed in a slightly different way. Let your companion matrix be $$C=\pmatrix{0&1&0&\cdots&0\\\\ 0&0&1&\cdots&0\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ 0&0&0&\cdots&1\\\\ -a_0&-a_1&-a_2&\cdots&-a_{n-1}}.$$ Then for the vector $v=(1\,\,0\,\,0\cdots 0)$, $$v\sum_{j=0}^{n-1} b_j C^j= \pmatrix{b_0&b_1&b_2&\cdots&b_{n-1}}$$ so that $g(C)\ne0$ for all nonzero polynomials $g$ of degree less than $n$. So the minimal polynomial has degree $n$, and equals the characteristic polynomial (via Cayley-Hamilton). But $vC^n=(-a_0\,\, {-a_1}\,\, {-a_2}\cdots{-a_{n-1}})$ and for $v(C^n+\sum_{j=0}^{n-1}b_j C^j)=0$ we need $a_j=b_j$. So the minimal and characteristic polynomials both equal $f$.

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Does Yuval's answer use Cayley-Hamilton? –  user1119 Nov 14 '10 at 12:05
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He says "thus it must be the characteristic polynomial". –  Robin Chapman Nov 14 '10 at 13:19

The fact that the minimal polynomial $C(f)$ is $f$ is obvious, as has been indicated above. The fact that its characteristic polynomial is also $f$ is a classical computation exercise. The computation is to be preferred over applying Cayley-Hamilton because this fact can be used as an ingredient to an elementary proof of that theorem (at least over fields) as has been said above. I will give a simpler argument below that requires no modules over a PID.

First the computation of the characteristic polynomial $$\left|\matrix{x&0&0&\ldots&a_0\\ -1&x&0&\ldots&a_1\\ 0&-1&x&\ldots&a_2\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ 0 & \cdots & 0 & -1 & x+a_{n-1}}\right| . $$ One way is to add the last row $x$ times to the previous row, then that row $x$ times to the one before and so on up to the first row, which results in a determinant of the form $$\left|\matrix{0&0&0&\ldots&f\\ -1&0&0&\ldots&*\\ 0&-1&0&\ldots&*\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ 0 & \cdots & 0 & -1 & *}~\right| = f $$ where the polynomial $f$ at the upper right is in fact obtained as in a Horner scheme $f=a_0+x(a_1+x(\cdots(a_{n-2}+x(a_{n-1}+x))\cdots))$.

Another method is to develop the matrix by the first row, and apply induction on the size. The minor that the $x$ is multiplied by is a matrix of the same kind but for $(f-a_0)/x=a_1+a_2x+\cdots+a_{n_1}x^{n-2}+x^{n-1}$, and the coefficient $a_0$ gets multiplied by $(-1)^{n-1}$ times an upper triangular matrix of size $n-1$ with all diagonal entries $-1$, which gives $a_0$; he starting case, the matrix of this type for the polynomial $a+x$, is a $1\times1$ matrix with $x+a$ as coefficient. Again the polynomial is found as in a Horner scheme.

Yet another way is to write the determinant as $$ x^n+\left|\matrix{x&0&0&\ldots&a_0\\ -1&x&0&\ldots&a_1\\ 0&-1&x&\ldots&a_2\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ 0 & \cdots & 0 & -1 & a_{n-1}}\right| $$ and develop by the last column, observing that the cofactor by which the entry $a_k$ is multiplied is $(-1)^{n-1-k}$ times a minor that has a block decomposition $M=\left|{L\atop0}~{0\atop{U}}\right|$ where $L$ is lower triangular matrix of size $k$ with entries $x$ on the diagonal, and $U$ is an upper triangular matrix of size $n-1-k$ with entries $-1$ on the diagonal, making the cofactor $x^k$, and the characteristic polynomial $f$.

Now the elementary proof of the Cayley-Hamilton theorem. Proceed by induction on $n$, the case $n=0$ being trivial. For $n>0$ take a nonzero vector $v$, and let $V$ be the subspace generated by its repeated images under the linear transformation $\phi$, which has a basis $v,\phi(v),\ldots,\phi^{d-1}(v)$ where $d=\dim(V)>0$ is the degree of the minimal polynomial $P$ that annihilates $v$ when acting by $\phi$. Extend to a basis of the whole space, in which basis $\phi$ has a matrix of the form $M=\left({A\atop0}~{{*}\atop{B}}\right)$, where $A$ is the companion matrix of $P$.

One has $\chi_M=\chi_A\chi_B$, where $\chi_A=P$, by the computation above. Now one gets null matrices when evaluating $P$ in $A$ (because $P$ is the minimal polynomial) and (by induction) when evaluating $\chi_B$ in $B$. Thus evaluating $\chi_M=P.\chi_B$ in $M$ gives a matrix product that in block form is $\left({0\atop0}~{{*}\atop{*}}\right)\cdot\left({{*}\atop0}~{{*}\atop0}\right) =\left({0\atop0}~{0\atop0}\right)$. (Note that one cannot use the induction hypothesis for $A$, since one might have $d=n$, which in fact will be the case "generically".)

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I have been thinking about the the problem a bit today. What Robin and Yuval have both shown is that if the Cayley-Hamilton theorem is true, then the characteristic and minimal polynomial of $C(f)$ are both equal to $f$ .

Conversely, assume that for all $f \in F[x]$, the the characteristic and minimal polynomial of $C(f)$ are both equal to $f$ .

Let $V$ be a finite dimensional $F$-vector space and $T : V \to V$ a linear transformation. we know from the classification theorem of modules over PIDs that there exists a basis $B$ of $V$ and $f_1, \dots, f_s \in F[x]$ such that $f_1 \mid \dots \mid f_s$ and

$$ [T]_B = \begin{pmatrix} C(f_1) & & \\ & \ddots & \\ & & C(f_s) \\ \end{pmatrix} := M$$

It is clear that the characteristic polynomial of $M$ is the product of the characteristic polynomials of all of the $C(f_i)$s and the minimal polynomial of $M$ is the least common multiple of the minimal polynomials of all the $C(f_i)$s. We see form the assumption that the characteristic polynomial of $T$ is $f_1 f_2 \dots f_s$ and the minimal polynomial of $T$ is $f_s$. This proves the Cayley-Hamilton theorem.

This shows that the Cayley-Hamilton theorem is equivalent to the fact "for all $f \in F[x]$, the the characteristic and minimal polynomial of $C(f)$ are both equal to $f$".

Proving the Cayley-Hamilton theorem without assuming knowledge of modules over PIDs or companion matrices is quite delicate (from what I remember from first year university).

This seems to support the idea that in order prove to the Cayley-Hamilton theorem (or the fact about companion matrices), you need to at some point get your hands dirty (whether it be directly computing the minimal and characteristic polynomials of a companion matrix or going through a delicate proof of the Cayley-Hamilton theorem).

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There are several nice ways to prove Cayley-Hamilton without doing anything delicate. Since it's a polynomial identity, it suffices to prove it over some field of characteristic zero. Over C it's obvious because it's obvious for diagonalizable matrices, and those are dense. Alternately, one can work "universally," i.e. in Z[x_{ij}]. This method is demonstrated here: mathoverflow.net/questions/32133/… –  Qiaochu Yuan Nov 14 '10 at 22:51
    
I still have high hopes for someone giving us a combinatorial proof though :D –  DBr Nov 14 '10 at 22:52
    
@Qiaochu: I suppose my last comment was a bit much! Thanks for the argument and link –  DBr Nov 14 '10 at 23:01
    
@DBr: I am skeptical about a combinatorial proof. Combinatorics is good at proving identities but to prove that the minimal polynomial is f one has to prove a non-identity (that is, that no smaller polynomial works) which seems much less amenable to combinatorial techniques. Of course, I could be wrong. –  Qiaochu Yuan Nov 14 '10 at 23:19
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@Qiaochu: Let's renumber the basis $e_0,\ldots$. Then for $k < n$, $A^k e_0 = e_k$. So the only polynomial $P$ of degree less than $n$ such that $P(A)e_0 = 0$ is $P = 0$. –  Yuval Filmus Nov 15 '10 at 20:04

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