Page 261 - Matrix Analysis & Applied Linear Algebra

256 Chapter 4 Vector Spaces

Equations (4.8.5) and (4.8.6) say that any two coordinate matrices of a
given linear operator must be similar. But must any two similar matrices be
coordinate matrices of the same linear operator? Yes, and here’s why. Suppose
C = Q −1 BQ, and let A(v)= Bv be the linear operator defined by matrix–
vector multiplication. If S is the standard basis, then it’s straightforward to see

that [A] S = B (Exercise 4.7.9). If B = {Q ∗1 , Q ∗2 ,..., Q ∗n } is the basis con-
sisting of the columns of Q, then (4.8.6) insures that [A] B =[I] −1 [A] S [I] B S ,
B S
where

[I] B S = [Q ∗1 ] S [Q ∗2 ] S ··· [Q ∗n ] S = Q.

Therefore, B =[A] S and C = Q −1 BQ = Q −1 [A] S Q =[A] B , so B and
C are both coordinate matrix representations of A. In other words, similar
matrices represent the same linear operator.
As stated at the beginning of this section, the goal is to isolate and study
coordinate-independent properties of linear operators. They are the ones de-
termined by sorting out those properties of coordinate matrices that are ba-
sis independent. But, as (4.8.5) and (4.8.6) show, all coordinate matrices for a
given linear operator must be similar, so the coordinate-independent properties
are exactly the ones that are similarity invariant (invariant under similarity
transformations). Naturally, determining and studying similarity invariants is an
important part of linear algebra and matrix theory.
Example 4.8.4
Problem: The trace of a square matrix C was defined in Example 3.3.1 to be
the sum of the diagonal entries

trace (C)= c ii .
i
Show that trace is a similarity invariant, and explain why it makes sense to talk
about the trace of a linear operator without regard to any particular basis. Then
2
determine the trace of the linear operator on that is defined by
A(x, y)=(y, −2x +3y). (4.8.7)
Solution: As demonstrated in Example 3.6.5, trace (BC)= trace (CB), when-
ever the products are defined, so
−1 −1
trace Q CQ = trace CQQ = trace (C),
and thus trace is a similarity invariant. This allows us to talk about the trace of
a linear operator A without regard to any particular basis because trace ([A] B )
is the same number regardless of the choice of B. For example, two coordinate
matrices of the operator A in (4.8.7) were computed in Example 4.8.2 to be

01 10
[A] S = and [A] S = ,
−23 02
and it’s clear that trace ([A] S )= trace ([A] S )=3. Since trace ([A] B )=3 for
all B, we can legitimately define trace (A)=3.