Page 234 - Linear Algebra Done Right

The right side of the equation above is, by definition, the characteristic
polynomial of T, completing the proof when V is a complex vector
space. Determinant of a Matrix 225
Now suppose V is a real vector space. Let λ 1 ,...,λ m denote the
eigenvalues of T and let (α 1 ,β 1 ),...,(α M ,β M ) denote the eigenpairs
of T, each repeated according to multiplicity. Thus for x ∈ R, the
eigenvalues of xI−T are x−λ 1 ,...,x−λ m and, by 10.16, the eigenpairs
of xI − T are
2
2
(−2x − α 1 ,x + α 1 x + β 1 ),...,(−2x − α M ,x + α M x + β M ),
each repeated according to multiplicity. Hence
2
2
det(xI −T) = (x −λ 1 )...(x −λ m )(x +α 1 x +β 1 )...(x +α M x +β M ).
The right side of the equation above is, by definition, the characteristic
polynomial of T, completing the proof when V is a real vector space.

Determinant of a Matrix

Most of this section is devoted to discovering how to calculate det T
from the matrix of T (with respect to an arbitrary basis). Let’s start with
the easiest situation. Suppose V is a complex vector space, T ∈L(V),
and we choose a basis of V with respect to which T has an upper-
triangular matrix. Then, as we noted in the last section, det T equals
the product of the diagonal entries of this matrix. Could such a simple
formula be true in general?
Unfortunately the determinant is more complicated than the trace.
In particular, det T need not equal the product of the diagonal entries
of M(T) with respect to an arbitrary basis. For example, the operator
3
on F whose matrix equals 10.8 has determinant 13, as we saw in the
last section. However, the product of the diagonal entries of that matrix
equals 0.
For each square matrix A, we want to define the determinant of A,
denoted det A, in such a way that det T = det M(T) regardless of which
basis is used to compute M(T). We begin our search for the correct def-
inition of the determinant of a matrix by calculating the determinants
of some special operators.
Let c 1 ,...,c n ∈ F be nonzero scalars and let (v 1 ,...,v n ) be a basis

of V. Consider the operator T ∈L(V) such that M T, (v 1 ,...,v n )
equals