The Brouwer fixed point theorem is one of the most significant theorems in mathematics.
There exist relatively easy proofs of this important theorem. The proof I am giving here is
the one given in Evans [11]. I think it is one of the shortest and easiest proofs of this
important theorem. It is based on the following lemma which is an interesting result about
cofactors of a matrix.
Recall that for A an p×p matrix, cof
(A)
_{ij} is the determinant of the matrix which results
from deleting the i^{th} row and the j^{th} column and multiplying by
(− 1)
^{i+j}. In the proof and in
what follows, I am using Dg to equal the matrix of the linear transformation Dg taken with
respect to the usual basis on ℝ^{p}. Thus
(Dg )
_{ij} = ∂g_{i}∕∂x_{j} where g = ∑_{i}g_{i}e_{i} for the e_{i} the
standard basis vectors.
Lemma 9.11.1Let g : U → ℝ^{p}be C^{2}where U is an open subset of ℝ^{p}. Then
p
∑ cof (Dg ) = 0,
j=1 ij,j
where here
(Dg )
_{ij}≡ g_{i,j}≡
∂gi
∂xj
. Also,cof
(Dg )
_{ij} =
∂det(Dg)
∂gi,j
.
Proof: From the cofactor expansion theorem,
∑p
det(Dg ) = gi,jcof (Dg )ij
i=1
and so
∂det(Dg )
---------= cof (Dg )ij (9.22)
∂gi,j
(9.22)
which shows the last claim of the lemma. Also
δkjdet(Dg ) = ∑ gi,k(cof (Dg)) (9.23)
i ij
(9.23)
because if k≠j this is just the cofactor expansion of the determinant of a matrix in which the
k^{th} and j^{th} columns are equal. Differentiate 9.23 with respect to x_{j} and sum on j. This
yields
∑ ∑ ∑
δkj∂(detDg-)gr,sj = gi,kj(cof (Dg ))ij + gi,kcof (Dg )ij,j.
r,s,j ∂gr,s ij ij