proof of Cayley-Hamilton theorem by formal substitutions
Proof (Popular fake proof):
In the expression
substitute ; then .
It is clear why the argument is faulty. But interestingly, there is a way to rescue it using clever formal substitution arguments. For the moment, we assume that the matrix is over the complex field.
Since the notation and can be confusing at first sight, as one expression takes scalar values and the other matrix values, we will change it. From now on, we will use the notation when applying the polymial to a matrix . In this sense is a function of and is function of matrices. Also, by definition,
Of course, we intend to prove that .
Proof (Proof of the complex case):
For each let be the classical adjoint to . We have
for some constant coefficient matrices .
Now, we would like to be defined for matrices (just like from the polynomial we have considered ), so we define for every matrix
Now consider the following function of matrices:
The above expression may look strange, but if we think that the matrix commutes with all and , then expression (2) is easily seen to be equal to
This means that
The reason for not defining by the expression in (3) is that we want to be some kind of ”polynomial” in (with matrix coefficients on the left of each ).
We now state some properties that can be easily checked by straightforward calculation:
Notice that matrices of the form with commute with every other matrix, so that
Now is also a matrix whose entries are polynomials in . Since we must have for all . This means that is the zero polynomial and, since this occurs for all , it follows that the matrix coefficients of occurring in (2) are all zero, i.e. is the zero matrix for all matrices .
Taking we can also see that
Hence , which finishes the proof.
Proof (Proof for an arbitrary commutative ring with identity):
Let , where the entries are in commutative ring with identity . First notice that, since , where , we have that the coefficients are polynomials in .
Hence is a matrix whose entries are also polynomials in . These polynomials vanish for every assignment of to numbers in , because the complex case of the theorem has already been proven (this would be the same as substituting the matrix by a matrix with complex entries). Therefore these polynomials are zero polynomials and we conclude that as we inteded to prove.
Comments on other proofs
Yet another proof of the Cayley-Hamilton Theorem (http://planetmath.org/ProofOfCayleyHamiltonTheorem) is to establish it for diagonalizable matrices, and then by a density argument (i.e. every matrix can be approximated by diagonalizable ones in an algebraically closed field), we conclude that is an identity for all matrices over a field. This kind of proof is also presented as an exercise in .
The two other proofs just mentioned can be extended to matrices over an arbitrary commutative ring simply by repeating the last argument in our proof.
|Title||proof of Cayley-Hamilton theorem by formal substitutions|
|Date of creation||2013-03-22 15:27:28|
|Last modified on||2013-03-22 15:27:28|
|Last modified by||asteroid (17536)|