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 $$ \begin{bmatrix} 1&0&0&\cdots&0&-f_n \\ 0&1&0&\cdots&0&-f_{n-1} \\ 0&0&1&\cdots&0&-f_{n-2} \\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots \\ 0&0&0&\cdots&1&-f_1 \\ \end{bmatrix}. $$ 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. $\square$