Let be a matrix over a field . Prove that the constant term of the characteristic polynomial is and that the term is the negative of the sum of the diagonal entries of (called the trace of ). Deduce that is the product of the eigenvalues of , and that (the trace of ) is the sum of the eigenvalues of .
Recall that , where is 1 if and 0 otherwise. So .
Now the constant term of is merely .
Consider now the polynomial . Note in particular that the degree of is just the number of elements in which are fixed by . No can fix elements, so that except for , the polynomial does not contribute to the term of . That is, the term of is just the term of , which is just as desired.
Now recall that the eigenvalues of are (by definition) the roots of . It then follows that the constant term of is times the product of the eigenvalues and that the term is the negative of the sum of the eigenvalues.