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:
We first begin with a fascinating reduction argument, which I first saw in a lecture of Paul Balmer at the ethz:
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
for .
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:
By the first lemma, it suffices to show the theorem over . But this is accomplished by the previous theorem.
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