Page 231 - Linear Algebra Done Right

Chapter 10. Trace and Determinant
222
The statement of this
corollary does not
ST − TS = I.
involve traces, though 10.13 Corollary: There do not exist operators S, T ∈L(V) such that
the short proof uses Proof: Suppose S, T ∈L(V). Then
traces. Whenever
something like this trace(ST − TS) = trace(ST) − trace(TS)
happens in
= 0,
mathematics, we can be
sure that a good where the second equality comes from 10.12. Clearly the trace of I
definition lurks in the equals dim V, which is not 0. Because ST − TS and I have different
background. traces, they cannot be equal.
Determinant of an Operator

Note that det T For T ∈L(V), we define the determinant of T, denoted det T,to
depends only on T and be (−1) dim V times the constant term in the characteristic polynomial
not on a basis of V of T. The motivation for the factor (−1) dim V in this definition comes
because the from 10.6.
characteristic If V is a complex vector space, then det T equals the product of
polynomial of T does the eigenvalues of T, counting multiplicity; this follows immediately
not depend on a choice from 10.6. Recall that if V is a complex vector space, then there is
of basis. a basis of V with respect to which T has an upper-triangular matrix
(see 5.13); thus det T equals the product of the diagonal entries of this
matrix (see 8.10).
If V is a real vector space, then det T equals the product of the
eigenvalues of T times the product of the second coordinates of the
eigenpairs of T, each repeated according to multiplicity—this follows
from 10.7 and the observation that m = dim V − 2M (in the notation
of 10.7), and hence (−1) m = (−1) dim V .
3
For example, suppose T ∈L(C ) is the operator whose matrix is
given by 10.8. As we noted in the last section, the eigenvalues of T are
1, 2 + 3i, and 2 − 3i, each with multiplicity 1. Computing the product
of the eigenvalues, we have det T = (1)(2 + 3i)(2 − 3i); in other words,
det T = 13.
3
As another example, suppose T ∈L(R ) is the operator whose ma-
trix is also given by 10.8 (note that in the previous paragraph we were
working on a complex vector space; now we are working on a real vec-
tor space). Then, as we noted earlier, 1 is the only eigenvalue of T (it