In this post, I want to present a very elegant proof of the Cayley-Hamilton Theorem which works over all commutative unitary rings by reducing to the case over the complex numbers, where a topological argument is used to reduce to the case of diagonalizable matrices. First of all, let us state the definitions and the theorem itself.
Let be a commutative unitary ring and a -matrix over . The characteristic polynomial of is the polynomial .
Then the theorem says:
The Theorem of Cayley-Hamilton holds over any commutative unitary ring if, and only if, it holds over the complex numbers.
Clearly, if the theorem holds for all rings, so it does for the special case . So assume that it holds for .
Let be any commutative unitary ring and , . Set and consider the ring homomorphism , . Over , consider the matrix . Now induces -algebra homomorphisms and with . Clearly, they satisfy and . Therefore, it suffices to prove .
Now has infinite transcendence degree over (otherwise, it could be countable), whence there exists an embedding ; simply choose algebraically independent elements in and map the to them. Again, we get maps and which are injective and satisfy and . But by assumption, Cayley-Hamilton holds over , whence . Since is injective, , which implies as mentioned above.
Now we can concentrate on showing the Theorem of Cayley-Hamilton for the complex numbers. We begin with a special case, namely the diagonalizable matrices.
A matrix is said to be diagonalizable if there exists an invertible matrix such that
We then have:
The Theorem of Cayley-Hamilton holds for diagonalizable matrices.
We first assume that . Then one gets , and since
one gets .
Now assume that is diagonalizable, and let such that . Clearly, and, therefore,
Now write with . Then
whence . But now , whence we get and, hence, .
We now get to the main piece of proving Cayley-Hamilton over :
For this proof, we need two facts from linear algebra:
- Every matrix over is equivalent to a triagonal matrix; this can be done if, and only if, the characteristic polynomial of the matrix splits into linear factors. But, by the Fundamental Theorem of Algebra, this is always the case over .
- An -matrix with distinct eigenvalues is diagonalizable.
Let be an arbitrary matrix. Then there exists a matrix such that
with . As the transcendence degree of over is infinite, there exist elements such that for every , the set has exactly elements. Define , . Then for and has distinct eigenvalues for every , namely . But this implies that , whence we found a sequence in converging to .
Now, we are able to conclude:
Set . Clearly, and is dense in by the previous lemma. Hence, it suffices to show that is closed.
But note that the map , is defined by polynomials; hence, it is continuous. Now is the preimage of a closed set, whence is closed itself.
This completes the proof of the theorem: