Question

Suppose $A\in R^{n\times n}$, where $R$ is a commutative ring. Let $p_i \in R $ be the coefficients of the characteristic polynomial of $A$: $\mathop{\mathrm{det}}(A-xI) = p_0 + p_1x + \dots + p_n x^n$.

I am looking for a proof that: $-\mathop{\mathrm{adj}}(A) = p_1 I + p_2 A + \dots + p_n A^{n-1}$.
In the case where $\mathop{\mathrm{det}}(A)$ is a unit, $A$ is invertible, and the proof follows from the Cayley-Hamilton theorem. But what about the case where $A$ is not invertible?

Answer 1

Here is a direct proof along the lines of the standard proof of the Cayley–Hamilton theorem. [This works universally, i.e. over the commutative ring $R=\mathbb{Z}[a_{ij}]$ generated by the entries of a generic matrix $A$.]

The following lemma combining Abel's summation and Bezout's polynomial remainder theorem is immediate.

Lemma Let $A(\lambda)$ and $B(\lambda)$ be matrix polynomials over a (noncommutative) ring $S.$ Then $A(\lambda)B(\lambda)-A(0)B(0)=\lambda q(\lambda)$ for a polynomial $q(\lambda)\in S[\lambda]$ that can be expressed as

$$q(\lambda)=A(\lambda)\frac{B(\lambda)-B(0)}{\lambda}+\frac{A(\lambda)-A(0)}{\lambda}B(0)=A(\lambda)b(\lambda)+a(\lambda)B(0) \qquad (*)$$

with $a(\lambda),b(\lambda)\in S[\lambda].$

Let $A(\lambda)=A-\lambda I_n$ and $B(\lambda)=\operatorname{adj} A(\lambda)$ [viewed as elements of $S[\lambda]$ with $S=M_n(R)$], then

$$A(\lambda)B(\lambda)=\det A(\lambda)=p_A(\lambda)=p_0+p_1\lambda+\ldots+p_n\lambda^n$$ is the characteristic polynomial of $A$ and $$A(0)B(0)=p_0 \text{ and } q(\lambda)=p_1+\ldots+p_n\lambda^{n-1}$$

Applying $(*),$ we get

$$q(\lambda)=(A-\lambda I)b(\lambda)-\operatorname{adj} A \qquad (**) $$

for some matrix polynomial $b(\lambda)$ commuting with $A.$ Specializing $\lambda$ to $A$ in $(**),$ we conclude that

$$q(A)=-\operatorname{adj} A\qquad \square$$

Answer 2

This formula can be obtained during a proof of the Cayley-Hamilton theorem, as is indicated on its Wikipedia article. The essence of the argument is that Euclidean division by a monic polynomial (on the left, say), can be performed in the polynomial ring over any (unitary) ring, not necessarily commutative; this follows directly from consideration of what Euclidean division does, or by a simple inductive argument.

Since I care about polynomials being monic, I'l define the characteristic polynomial of a matrix $A$ to be $\chi_A=\det(I_nX-A)=\sum_{i=0}^nc_iX^i$ where $c_n=1$ (and $c_0=\det(-A)$), so the result to prove becomes $\mathrm{adj}(-A)=c_1I_n+c_2A+\cdots+c_{n-1}A^{n-2}+A^{n-1}=\sum_{i=1}^nc_iA^{i-1}$

Consider the noncommutative ring $M=\mathrm{Mat}_n(R)$, and using Euclidean division in $M[X]$ (in which $R[X]$ is embedded by mapping $r$ to $rI_n$) divide $\chi_A$ on the left by $X-A$; the quotient and remainder will be $\mathrm{adj}(X-A)$ and $0$ (by uniqueness). Writing the quotient $\mathrm{adj}(X-A)=\sum_{i=0}^{n-1}B_iX^i$, its coefficients $B_i\in M$ are determined in the division successively as $B_{n-1}=c_n=1$ and $B_{i-1}=c_i+AB_i$ for $i=n-1,\ldots,1$, which expands to $B_{i-1}=c_iA^0+c_{i+1}A^1+\cdots+c_nA^{n-i}$. In particular the constant coefficient of the quotient is $B_0=\sum_{i=1}^nc_iA^{i-1}$, but this is also $\mathrm{adj}(-A)$ (by substituting $X=0$ into $\mathrm{adj}(X-A)$).

To retrieve the Cayley-Hamilton theorem from the formula found, multiply on the left or right by $A$ and move the left hand side to the right.

Answer 3

EDIT OF AUG. 31, 2010. The proof of the Cayley-Hamilton Theorem I like best (among the ones I know) is on page 21 (proof of Proposition 2.4) of Introduction to Commutative Algebra by Atiyah and MacDonald. The argument can be phrased as follows.

Let $K$ be a commutative ring; let $n$ be a positive integer; let $A=(a_{ij})\in M_n(K)$ be an $n$ by $n$ matrix with entries in $K$; let $\chi$ be its characteristic polynomial; define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}\,A-a_{ij}$; let $(e_i)$ be the canonical basis of $K^n$; observe $$\sum_i\ \ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Applying (a trivial case of) Fubini's Theorem to the double sum $\sum_{i,j}\ c_{jk}\ b_{ij}\ e_i$, we get $\chi(A)=0$.

Thank you very much to darij grinberg! [I'm leaving the previous edits "for the record".] END OF EDIT OF AUG. 31, 2010.

EDIT OF DEC. 11, 2010. For a nice application of the Cayley-Hamilton Theorem, see this answer by Balazs Strenner.

PREVIOUS EDITS:

Here is a proof of the Cayley-Hamilton Theorem.

Let $K$ be a commutative ring, let $n$ be a positive integer, let $X$ be an indeterminate, let $A\in M_n(K)$ be an $n$ by $n$ matrix with coefficients in $K$, and let $\chi:=\det(X-A)$ be the characteristic polynomial. Equip $K^n$ with the $K[X]$-module structure induced by $A$. We must check $\chi K^n=0$. Form the right $M_n(K[X])$-module $$H:=\mathrm{Hom}_{K[X]}(K[X]^n,K^n).$$ Let $e\in H$ be the evaluation at $A$ (note $K[X]^n=K^n[X]$). As $e$ is surjective, it suffices to show $e\chi=0$. As $X-A$ divides $\chi$ on the left, it suffices to show $e(X-A)=0$. But this is obvious.

EDIT OF AUG. 1, 2010. Here is a diagrammatic rewriting of the argument.

EDIT OF AUG. 30, 2010. Here is a coordinate version of the above argument. [Compare with the proof of Propositon 3 page 81 of Weil's Basic Number Theory, and with the proof of Propositon 2.4 page 21 of Introduction to Commutative Algebra by Atiyah and MacDonald].

Weil's formulation. Put $$B(X)=(b_{ij}(X)):=X-A\in M_n(K[X]),$$ and let $C(X)=(c_{ij}(X))$ be the adjugate of $B(X)$. We have $$\sum_j\ c_{jk}(X)\ b_{ij}(X)=\delta_{ik}\ \chi(X)\in K[X].$$ Replacing $X$ with $A$, evaluating on $e_i$ (the $i$-th vector of the canonical basis of $K^n$), and summing over $i$ gives $$\sum_j\ c_{jk}(A)\ \sum_i\ b_{ij}(A)\ e_i=\chi(A)\ e_k\in K^n.$$ But the second sum is 0 by definition of $b_{ij}(X)$.

Atiyah-MacDonald's formulation. Put $A=(a_{ij})$ and define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}A-a_{ij}$; observe $$\sum_i\ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Computing $\sum_{i,j}\,c_{jk}\,b_{ij}\,e_i$ in two ways we get $\chi(A)=0$.

Answer 4

As an arithmetic geometer, I have no choice but to use topological methods hand in hand with algebraic methods. Very likely necessity has been the mother of aesthetics here, but I find proofs of linear algebra facts using genericity arguments to be beautiful and insightful.

Qiaochu has shown how to answer the OP's question using these methods [he uses the "analytic" -- i.e., usual -- topology on $\mathbb{C}^n$, but close enough] assuming the Cayley-Hamilton theorem. Here I want to show that one can also prove the Cayley-Hamilton theorem quickly by these methods.

Step 1: To prove C-H as a polynomial identity, it is enough to prove that it holds for all $n \times n$ matrices over $\mathbb{C}$.

Proof: Indeed, to say C-H holds as a polynomial identity means that it holds for the generic matrix $A = \{a_{ij}\}_{1 \leq i,j \leq n}$ whose entries are independent indeterminates over the ring $R = \mathbb{Z}[a_{ij}]$. But this ring embeds into $\mathbb{C}$ -- indeed into any field of characteristic zero and infinite absolute transcendence degree -- and two polynomials with coefficients in a domain $R$ are equal iff they are equal in some extension domain $S$.

Step 2: C-H is easy to prove for complex matrices $A$ with $n$ distinct eigenvalues $\lambda_1,\ldots,\lambda_n$.

Proof: The characteristic polynomial evaluated at $A$ is $P(A) = \prod_{i=1}^n(A-\lambda_i I_n)$. Let $e_1,\ldots,e_n$ be a basis of $\mathbb{C}^n$ such that each $e_i$ is an eigenvector for $A$ with eigenvalue $\lambda_i$. Then -- using the fact that the matrices $A - \lambda_i I_n$ all commute with each other -- we have that for all $e_i$,

$P(A)e_i = \left(\prod_{j \neq i} (A-\lambda_j I_n)\right) (A-\lambda_i I_n) e_i = 0.$

Since $P(A)$ kills each basis element, it is in fact identically zero.

Step 3: The set of complex matrices with $n$ distinct eigenvalues is a Zariski-open subset of $\mathbb{C}^n$: indeed this is the locus of nonvanishing of the discriminant of the characteristic polynomial. Since we can write down diagonal matrices with distinct entries, it is certainly nonempty. Therefore it is Zariski dense, and any polynomial identity which holds on a Zariski dense subset of $\mathbb{C}^{n^2}$ holds on all of $\mathbb{C}^{n^2}$.

Answer 5

I guess it is worth giving a fuller answer, and then Victor can tell me more precisely where I am missing some subtlety. As I said, the definition I know of the adjugate is that it is a matrix whose entries are polynomials in the entries $a_{ij}$ of $A$ and which satisfies $A \text{ adj}(A) = I \det A$ identically, e.g. over $\mathbb{Z}[a_{ij}]$. Assuming Cayley-Hamilton, we know that $p_0 I + p_1 A + ... + p_n A^n = 0$ identically and that $p_0 = \det A$, where $p_k \in \mathbb{Z}[a_{ij}]$ as well.

Specializing now to $a_{ij} \in \mathbb{C}$ and supposing that $A$ is invertible, we conclude that

$$A \text{ adj}(A) = - p_1 A - p_2 A^2 - ... - p_n A^n$$

implies

$$\text{adj}(A) = - p_1 I - p_2 A - ... - p_n A^{n-1},$$

as you say.

Lemma: The invertible $n \times n$ matrices are dense in the $n \times n$ matrices with the operator norm topology.

Proof. Let $A$ be a non-invertible $n \times n$ matrix, hence $\det A = 0$. The polynomial $\det(A - xI)$ has leading term $(-1)^n x^n$, hence cannot be identically zero, so in any neighborhood of $A$ there exists $x$ such that $A - xI$ is invertible.

But everything in sight is continuous in the operator norm topology, so the conclusion follows identically over $\mathbb{C}$ and hence identically.

(I should mention that this is not even my preferred method of proving matrix identities. Whenever possible, I try to prove them combinatorially by interpreting $A$ as the adjacency matrix of some graph. For example - confession time! - this is how I think about Cayley-Hamilton. This is far from the cleanest or the shortest way to do things, but my combinatorial intuition is better than my algebraic intuition and I think it's good to have as many different proofs of the basics as possible.)

Answer 6

HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d\: B = d\: C$ for $\rm d = det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d\: b_{i,j} = d\: c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}\:]$ where $\;\rm d = det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\;\rm b_{i,j} = c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.

Notice that the crucial insight is that $\;\rm b_{i,j}\:, \; c_{i,j}\:,\; d\;$ have polynomial form in $\;\rm a_{i,j}\:$, i.e. they are elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a domain, enjoys cancelation of elts $\ne 0$. Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$

Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint. As a result, many students cannot easily resist the obvious topological temptations and instead derive hairier proofs employing density arguments (e.g see elswhere in this thread).

Analogously, the same generic method of proof works for many other polynomial identities, e.g.

$\rm\quad\; det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$

$\rm\quad\quad det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$

Now, for our pièce de résistance of topology, we derive the polynomial derivative purely formally.

For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$ and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition since it is linear and it takes the same value on the basis monomials $\rm x^n$. Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule

$$ \rm f(x)g(x) - f(y)g(y)\; = \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$$

$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $

by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$, i.e. specialize the difference "quotient" from the product rule for differences. Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\;\rm det \:A\;$ in all of the examples given above.