How many ways are there to prove Cayley-Hamilton Theorem?

Question

I see many proofs for the Cayley-Hamilton Theorem in textbooks and net, so I want to know how many proofs are there for this important and applicable theorem?

Answer 1

My favorite : let $k$ be your ground field, and let $A = k[X_{ij}]_{1\leqslant i,j\leqslant n}$ be the ring of polynomials in $n^2$ indeterminates over $k$, and $K = Frac(A)$.

Then put $M = (X_{ij})_{ij}\in M_n(A)$ the "generic matrix".

For any $N=(a_{ij})_{ij}\in M_n(k)$, there is a unique $k$-algebra morphism $\varphi_N:A\to k$ defined by $\varphi(X_{ij}) = a_{ij}$ that satisfies $\varphi(M)=N$.

Then the characteristic polynomial of $M$ is separable (ie $M$ has $n$ distinct eingenvalues in an algebraic closure $\widetilde{K}$ of $K$). Indeed, otherwise its resultant $Res(\chi_M)$ is zero, so for any $N\in M_n(k)$, $Res(\chi_N) = Res(\chi_{\varphi_N(M)})= \varphi_N(Res(\chi_M)) = 0$, so no matrix $N\in M_n(k)$ would have distinct eigenvalues (but obviously some do, just take a diagonal matrix).

It's easy to show that matrices with separable characteristic polynomial satisfy Cayley-Hamilton (because they are diagonalizable in an algebraic closure), so $M$ satisfies Cayley-Hamilton.

Now for any $N\in M_n(k)$, $\chi_N(N) = \varphi_N(\chi_M(M)) = \varphi_N(0) = 0$.

Answer 2

Here is a neat proof from Qiaochu Yuan's answer to this question:

If $L$ is diagonalizable with eigenvalues $\lambda_1, \dots \lambda_n$, then it's clear that $(L - \lambda_1) \dots (L - \lambda_n) = 0$, which is the Cayley-Hamilton theorem for $L$. But the Cayley-Hamilton theorem is a "continuous" fact: for an $n \times n$ matrix it asserts that $n^2$ polynomial functions of the $n^2$ entries of $L$ vanish. And the diagonalizable matrices are dense (over $\mathbb{C}$). Hence we get Cayley-Hamilton in general.

Answer 3

One can prove this theorem by use of the fact that the matrix representation of all linear map on a complex vector space, is Triangularisable with respect to a basis $\{v_1,...,v_n\}$.

So if $T$ be a linear map there are $\{\lambda_1,...,\lambda_n\}$ s.t
$$T(v_1)=\lambda_1 v_1 $$ $$T(v_2)=a_{11} v_1+\lambda_2 v_2 $$ $$.$$ $$.$$ $$.$$ $$T(v_n)=a_{n1}v_1+a_{n2}v_2+...+\lambda_n v_n $$

And by computation we can find that the matrix $S=(T-\lambda_1)(T-\lambda_2)...(T-\lambda_n)$ vanishes all $v_i$, and so $S\equiv 0$.
For more details you can see Herstein's Topics in Algebra.

Answer 4

Here is a proof which doesn't assume that our field is algebraically closed (or involve passing to a field extension). For any square matrix $M$, let $c_M(t)$ be the characteristic polynomial of $M$. What we want to show is that $c_M(M)=0$.

Lemma 1: Let $A$ be an $n \times n$ matrix and suppose that there is a vector $v$ such that $v$, $Av$, $A^2 v$, ..., $A^{n-1} v$ is a basis of $k^n$. Then $c_A(A)=0$.

Proof: Since $v$, $Av$, $A^2 v$, ..., $A^{n-1} v$ is a basis, we have some linear relationship $A^n v + \sum_{j=0}^{n-1} f_j A^j v=0$. We set $f_n=1$ so we can write $$\sum_{j=0}^n f_j A^j v = 0 \quad (\ast).$$ Then, in the basis $v$, $Av$, $A^2 v$, ..., $A^{n-1} v$, the linear operator $A$ has matrix $$ \begin{bmatrix} 0&0&0&\cdots&0&-f_n \\ 1&0&0&\cdots&0&-f_{n-1} \\ 0&1&0&\cdots&0&-f_{n-2} \\ 0&0&1&\cdots&0&-f_{n-3} \\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots \\ 0&0&0&\cdots&1&-f_1 \\ \end{bmatrix}. $$ Using this basis, we compute that $c_A(t) = \sum f_j t^j$. Now, multipltying $(\ast)$ by $A^k$ on the left, we deduce that $$A^k \left( \sum f_j A^j \right) v = \left( \sum f_j A^{k+j} \right) v = \left( \sum f_j A^j \right) A^k v = c_A(A) A^k v = 0.$$ Thus, $c(A)$ kills each of the basis elements $v$, $Av$, $A^2 v$, ..., $A^{n-1} v$, and we deduce that $c_A(A)=0$. $\square$

Lemma 2: Suppose that $A$ is a matrix with block form $\left[ \begin{smallmatrix} X&Y \\ 0&Z \end{smallmatrix} \right]$ and that $c_X(X)=0$ and $c_Z(Z)=0$. Then $c_A(A)=0$.

Proof: The determinant of a block matrix is the product of the determinants of the diagonal blocks, so $c_A(t) = c_X(t) c_Z(t)$, so we have $$c_A(A) = c_X(A) c_Z(A) = \begin{bmatrix} c_X(X) & \ast \\ 0 & c_X(Z) \end{bmatrix} \begin{bmatrix} c_Z(X) & \ast \\ 0 & c_Z(Z) \end{bmatrix}$$ $$=\begin{bmatrix} 0 & \ast \\ 0 & c_X(Z) \end{bmatrix} \begin{bmatrix} c_Z(X) & \ast \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \\ \end{bmatrix}. \qquad \square$$

Proof of the Cayley-Hamilton theorem: We induct on $\dim V$; if $\dim V=0$, the result is vacuously true.

Now, suppose $\dim V=n>0$ and choose a nonzero $v \in V$. Find the minimal $r$ such that there is a linear relation between $v$, $Av$, $A^2 v$, ..., $A^{r-1} v$, $A^r v$. Since $v \neq 0$, we have $r \geq 1$. If $r=n$, we are done by Lemma 1.

If not, complete $v$, $Av$, $A^2 v$, ..., $A^{r-1} v$ to a basis $v$, $Av$, $A^2 v$, ..., $A^{r-1} v$, $w_{r+1}$, $w_{r+2}$, $\dots$, $w_n$ of $k^n$. In this basis, $A$ has block form $\left[ \begin{smallmatrix} X&Y \\ 0&Z \end{smallmatrix} \right]$, with the $X$-block being $r \times r$ and the $Z$-block being $(n-r) \times (n-r)$. By induction, we have $c_X(X) = 0$ and $c_Z(Z)=0$, and we conclude by Lemma 2. $\square$

Answer 5

A few years ago I gave an "ugly" and long proof (more than 4 pages), which is purely computational.

Basically, I took an $n\times n$ matrix $A=(a_{i,j})_{i,j}$, where $a_{i,j}$ are variables. Then I wrote the characteristic polynomial as $P_A(X)=c_nX^n+\cdots +c_0$, where each $c_k$ is an explicit polynomial in the variables $a_{i,j}$. Then I wrote explicitly the $n^2$ entries of each $A^k$ with $0\leq k\leq n$ as polynomials in the variables $a_{i,j}$. Finally, I proved that each of the $n^2$ entries of $P_A(A)=c_nA^n+\cdots +c_0I_n$ is $0$. It's just playing with sums of monomials and proving that all of them reduce in the end.

You can find the proof in the 3-4/2014 issue of Gazeta Matematica, Seria A, pages 32-36. Available online at this address: https://ssmr.ro/gazeta/gma/2014/gma3-4-2014-continut.pdf

Answer 6

A more general version of the Cayley-Hamilton Theorem is given in Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry (1st Edition, Theorem 4.3, p.120). I give a slightly expanded version of the proof here.

Theorem. Suppose $A$ is a commutative ring and $M$ is a finitely generated $A$-module. If $\varphi:M\to M$ is an endomorphism, then there is a monic polynomial $p\in A[t]$ such that $p(\varphi)$ is the zero endomorphism of $M$.

Proof. We first endow $M$ with an $A[t]$-module structure by extending the scalar multiplication $A\times M\to M$ to a map $A[t]\times M\to M$. We let $t$ "act as $\varphi$", so that $$ (a_0+a_1t+\dots +a_nt^n)m=a_0m+a_1\varphi(m)+a_2\varphi^2(m)+\dots+a_n\varphi^n(m) \, , $$ where $\varphi^i$ denotes the $i$-fold composition of $\varphi$. (We can think of the scalar multiplication $A[t]\times M\to M$ as being induced by the ring homomorphism $A[t]\to\operatorname{End}_{\mathsf{Ab}}(M), p\mapsto p(\varphi)$.)

Now suppose $x_1,\dots,x_n$ generate $M$ as an $A$-module. The action of $A[t]$ on $M$ gives rise to an action of $\mathcal M_n(A[t])$ on $M^n$; explicitly, we define $$ \begin{pmatrix} p_{11} & \dots & p_{1n} \\ \vdots & & \vdots \\ p_{1n} & \dots & p_{nn} \end{pmatrix} \begin{pmatrix} m_1 \\ \vdots \\ m_n \end{pmatrix} = \begin{pmatrix} p_{11}m_1+\dots+p_{1n}m_n \\ \vdots \\ p_{n1}m_1+\dots+p_{nn}m_n \end{pmatrix} \, . $$ We can write $\varphi(x_i)$ as a linear combination of the $x_j$, say $\varphi(x_i)=\sum_jb_{ij}x_j$ with $b_{ij}\in A$. In terms of the action of $\mathcal M_n(A[t])$ on $M^n$, these equations tell us that $$ (tI-B)x=0 \, , $$ where $B=(b_{ij})$ and $x=(x_1,\dots,x_n)$. We claim that $p=\det(tI-B)\in A[t]$ is the desired polynomial. By inspecting the Leibniz formula for the determinant, we see that $p$ is monic; it remains to be shown that $p(\varphi)$ is the zero endomorphism.

Recall that if $C$ is a matrix with entries in a commutative ring, then the adjugate $\DeclareMathOperator{\adj}{adj}\adj(C)$ satisfies $\adj(C)C=\det(C)I$. If we multiply the equation $(tI-B)x=0$ on the left by the adjugate of $tI-B$, then the equation becomes $[\det(tI-B)I]x=0$, hence $\det(tI-B)x_i=0$ for each $i$. It follows that $p(\varphi)$ is identically zero, completing the proof.

Remark. In the case where $M=A^n$ for some $n\in\mathbb N$, and the $x_i$ are the canonical basis vectors $e_i$, the matrix $B$ constructed in the above proof is simply the matrix representing $\varphi$ with respect to the canonical basis of $A^n$, and $p$ is the characteristic polynomial of $B$. Since $p(B)$ represents $p(\varphi)$, and $p(\varphi)$ is the zero operator, it follows that $p(B)$ is the zero matrix, from which we recover the "classical" Cayley-Hamilton Theorem.

Answer 7

The Cayley-Hamilton Theorem says where the dark box represents an antisymmetrizer of size $n+1$.

The proof of the theorem is a trivial application of the pigeonhole principle. What takes a bit more time and effort is to see that this is indeed the familiar Cayley-Hamilton Theorem. For more details, see Section 6.5 of the book by P. Cvitanović, Group theory. Birdtracks, Lie's, and exceptional groups. Princeton University Press, Princeton, NJ, 2008. In Lectures 4 and 5 of my course https://mabdesselam.github.io/MATH8450Spring2023.html you can see the $n=2$ case worked out.

This point of view was used by V. Lafforgue in his recent work https://arxiv.org/abs/1209.5352

Answer 8

For the sake of completeness the $1$-line proof should be mentioned here. The proof uses the adjugate identity

$$\det(tI - A) = (tI - A) \text{adj}(tI - A)$$

and then "just plugs in $t = A$"! See the link for a few different ways to justify this; one has to say some things about polynomials with matrix coefficients.

I really think this proof deserves to be better known; it is both more elementary than other proofs (you don't have to know what an eigenvalue or eigenvector is, you don't have to pass to an algebraic extension of the ground field, you don't have to use a density argument to reduce to the diagonalizable case) and proves a stronger result (this works over any commutative ring). It really clarifies that Cayley-Hamilton is "pure algebra" in some sense.

Answer 9

I think, I can add a peculiar proof for complex valued matrices based on purely analytic considerations.

Let $A \in \mathbb C^{n\times n}$ and let $R(z) = (zI - A)^{-1}$ for $z \in \mathbb C$ such that $z$ is not an eigenvalue of $A$. Clearly, for large $z$ we have an expansion $R(z) = \sum_{n = 1}^\infty z^{-n} A^{n-1}$. Thus, for a contour $\gamma$ that encloses all eigenvalues of $A$ we obtain $\int_\gamma z^kR(z)dz = 2\pi iA^k$ for $k = 0,1,\dots$ As a consequence we get $\int_\gamma p(z)R(z)dz = 2\pi i p(A)$ for all polynomials $p \in \mathbb C[x]$.

Now the proof of Cayley-Hamilton theorem finishes as follows. Let $f(z) = \det(zI - A)$ be a characteristic polynomial of $A$ and observe that $R(z)f(z)$ is an entire function (moreover, this matrix function is polynomial in $z$ and $A$). Thus, $2\pi i f(A) = \int_\gamma f(z)R(z)dz = 0$ by Cauchy theorem.

Answer 10

[This is just a version of Sh.R’s answer above]

Consider ${ A \in \mathbb{C} ^{n \times n } }.$ By upper triangularization, there exists a basis ${ [P _1, \ldots, P _n] }$ such that $${ A [P _1, \ldots, P _n] = [P _1, \ldots, P _n] U }$$ and $${ U = \begin{pmatrix} \lambda _1 &* &* &\ldots &* \\ &\lambda _2 &* &\ldots &* \\ & &\ddots & &\vdots \\ & & &\lambda _{n-1} &* \\ & & & & \lambda _n \end{pmatrix} }.$$

Writing ${ P = [P _1, \ldots, P _n ] },$ since $${ \det(tI - A) = \det(P ^{-1} (tI - A) P) = \det(tI - P ^{-1} A P) }$$ we have polynomial $${ f(t) = \det(tI - A) = \det(tI - U) = (t - \lambda _1) \ldots (t - \lambda _n) }.$$

We are interested in the matrix $${ f(A) = P f(P ^{-1} A P) P ^{-1} = P f(U) P ^{-1} . }$$

We can show matrix ${ f(U) = 0 }$ (and hence ${ f(A) = 0 }$).

Thm: Let ${ U }$ be an upper triangular matrix $${ U = \begin{pmatrix} \lambda _1 &* &* &\ldots &* \\ &\lambda _2 &* &\ldots &* \\ & &\ddots & &\vdots \\ & & &\lambda _{n-1} &* \\ & & & & \lambda _n \end{pmatrix} }.$$ Then ${ f(U) = (U - \lambda _1 I) \ldots (U - \lambda _n I) = 0 }.$

Pf: We have ${ U = \begin{pmatrix} \lambda _1 &* &\ldots &* \\ 0& &\tilde{U} \end{pmatrix} }$ with submatrix ${ \tilde{U} = \begin{pmatrix} \lambda _2 &* &* &\ldots &* \\ &\ddots & & &\vdots \\ & & &\lambda _{n-1} &* \\ & & & & \lambda _n \end{pmatrix} }.$
By induction hypothesis, $${ (\tilde{U} - \lambda _2 I) \ldots (\tilde{U} - \lambda _n I) = 0 }.$$ Hence $${ (U - \lambda _2 I) \ldots (U - \lambda _n I) }$$ $${ \begin{align*} &= \begin{pmatrix} * &* &\ldots &* \\ 0& &\tilde{U} - \lambda _2 I \end{pmatrix} \ldots \begin{pmatrix} * &* &\ldots &* \\ 0& &\tilde{U} - \lambda _n I \end{pmatrix} \\ &= \begin{pmatrix} * &* &\ldots &* \\ 0 & &(\tilde{U} - \lambda _2 I) \ldots (\tilde{U} - \lambda _n I) \end{pmatrix} = \begin{pmatrix} * &* &\ldots &* \\ 0 & &0 & \end{pmatrix}. \end{align*}}$$ Hence $${ (U - \lambda _1 I) (U - \lambda _2 I) \ldots (U - \lambda _n I) }$$ $${ \begin{align*} &= \begin{pmatrix} 0 &* &\ldots &* \\ 0 & &\tilde{U}-\lambda _1 I & \end{pmatrix} \begin{pmatrix} * &* &\ldots &* \\ 0 & &0 & \end{pmatrix} = 0 , \end{align*} }$$ as needed.