Page 229 - Linear Algebra Done Right

Chapter 10. Trace and Determinant
220
Proof: Suppose (u 1 ,...,u n ) and (v 1 ,...,v n ) are bases of V. Let

A =M I, (u 1 ,...,u n ), (v 1 ,...,v n ) . Then

−1
The third equality here trace M T, (u 1 ,...,u n ) = trace A M T, (v 1 ,...,v n ) A
depends on the −1
= trace M T, (v 1 ,...,v n ) A A
associative property of

matrix multiplication. = trace M T, (v 1 ,...,v n ) ,
where the first equality follows from 10.3 and the second equality fol-
lows from 10.9. The third equality completes the proof.
The theorem below states that the trace of an operator equals the
sum of the diagonal entries of the matrix of the operator. This theorem
does not specify a basis because, by the corollary above, the sum of
the diagonal entries of the matrix of an operator is the same for every
choice of basis.

10.11 Theorem: If T ∈L(V), then trace T = trace M(T).

Proof: Let T ∈L(V). As noted above, trace M(T) is independent
of which basis of V we choose (by 10.10). Thus to show that

trace T = trace M(T)

for every basis of V, we need only show that the equation above holds
for some basis of V. We already did this (on page 218), choosing a basis
of V with respect to which M(T) is an upper-triangular matrix (if V is a
complex vector space) or an appropriate block upper-triangular matrix
(if V is a real vector space).

If we know the matrix of an operator on a complex vector space, the
theorem above allows us to find the sum of all the eigenvalues without
finding any of the eigenvalues. For example, consider the operator
5
on C whose matrix is
 
0 0 0 0 −3


 1 0 0 0 6 
 
 0 1 0 0 0  .
 


 0 0 1 0 0 
0 0 0 1 0