Page 131 - Linear Algebra Done Right

Chapter 6. Inner-Product Spaces
118
Now we prove that only one vector v ∈ V has the desired behavior.
Suppose v 1 ,v 2 ∈ V are such that
ϕ(u) = u, v 1 = u, v 2
for every u ∈ V. Then
0 = u, v 1 −Pu, v 2 = u, v 1 − v 2
for every u ∈ V. Taking u = v 1 − v 2 shows that v 1 − v 2 = 0. In other
words, v 1 = v 2 , completing the proof of the uniqueness part of the
theorem.

In addition to V, we need another finite-dimensional inner-product
space.

Let’s agree that for the rest of this chapter
W is a finite-dimensional, nonzero, inner-product space over F.

The word adjoint has Let T ∈L(V, W). The adjoint of T, denoted T , is the function from
∗
another meaning in W to V defined as follows. Fix w ∈ W. Consider the linear functional
∗
linear algebra. We will on V that maps v ∈ V to Tv, w . Let T w be the unique vector in V
not need the second such that this linear functional is given by taking inner products with
meaning, related to T w (6.45 guarantees the existence and uniqueness of a vector in V
∗
inverses, in this book. with this property). In other words, T w is the unique vector in V
∗
Just in case you such that
encountered the Tv, w = v, T w
∗
second meaning for
for all v ∈ V.
adjoint elsewhere, be
Let’s work out an example of how the adjoint is computed. Define
warned that the two
2
3
T : R → R by
meanings for adjoint
T(x 1 ,x 2 ,x 3 ) = (x 2 + 3x 3 , 2x 1 ).
are unrelated to one
3
2
another. Thus T ∗ will be a function from R to R . To compute T , fix a point
∗
2
(y 1 ,y 2 ) ∈ R . Then
∗
(x 1 ,x 2 ,x 3 ), T (y 1 ,y 2 ) = T(x 1 ,x 2 ,x 3 ), (y 1 ,y 2 )
= (x 2 + 3x 3 , 2x 1 ), (y 1 ,y 2 )
= x 2 y 1 + 3x 3 y 1 + 2x 1 y 2
= (x 1 ,x 2 ,x 3 ), (2y 2 ,y 1 , 3y 1 )
3
for all (x 1 ,x 2 ,x 3 ) ∈ R . This shows that