15B. Isometries and Functionals
1
Isometries
If the important functions between vector spaces V
and W are those that preserve linearity (i.e., linear
transformations), then the important functions
between inner product spaces should be those that
preserve the inner product. To this end, we make
the following natural definition: a linear
transformation T:V → W between inner product
spaces is isometric if it satisfies the condition
€
T(X),T(Y) = X, Y
for every X and Y in V. (Note that the inner
product€on the left is the one in W while the one on
the right is the one in V. Also, the term “isometric”
refers to having “the same distance”: an isometric
map preserves lengths of vectors as it passes from
domain to codomain space. Of course, it also
preserves angles.)
Proposition An isometric linear transformation
has trivial kernel.
Proof Suppose T(X) = 0. Then |X|2= X, X =
T(X),T(X) =|T(X)|2 =|0|2 = 0, so X = 0. //
€
€
€
15B. Isometries and Functionals
€
2
This last proposition shows that all isometric maps
are one-to-one. (Smith calls such maps isometric
embeddings.) So if the linear transformation
T:V → W is isometric and V and W have the
same dimension, the map must be an isomorphism
as well; we call an isometric isomorphism simply
an isometry, and we say that the spaces V and W
are isometric.
Proposition Any two inner product spaces of the
same dimension are isometric.
€
Proof Suppose inner product spaces V and W are
both of dimension n. Then we can find orthonormal
bases for each, say {V1, V2,…, Vn } for V and
{W1, W2,…, Wn } for W. Let T:V → W be the linear
transformation that is the extension of the
assignments
€ T(Vi ) = Wi . Then T is isometric, for
any X and Y in V can
€ be written the form
€
X = a1V1 + a2 V2 + L+ an Vn
Y = b1V1 + b2 V2 +L + bn Vn
so that
€
15B. Isometries and Functionals
3
T(X),T(Y) = T( ∑ a i Vi ),T( ∑ b j V j )
1≤i ≤n
=
∑ aiT(Vi ), ∑ b jT(V j )
1≤i≤n
=
1≤ j ≤n
1≤ j ≤n
∑ aib j T(Vi ),T(V j )
1≤i, j ≤n
=
∑ aib j Vi , V j
1≤i, j ≤n
=
∑ ai Vi , ∑ b j V j
1≤i≤n
= X, Y
1≤ j ≤n
.
//
Corollary Every n-dimensional inner product
€
space is isomorphic to R n (endowed with the
standard dot product). //
€
15B. Isometries and Functionals
4
Functionals
Recall from Exercise 9.9(3) that any finitedimensional vector space V is isomorphic to its
dual space V* = L(V ,R). We will investigate next
how having a scalar product helps to make an
important connection between the inner product
space V€and its dual V*.
The vectors in V* are real-valued linear functions
on V. They are often called (linear) functionals
on V; we will denote them with letters like f, g,
etc., to emphasize their nature as functions on V.
One example of a functional makes use of the inner
product on V. Let A be a fixed vector in V and
define for arbitrary vectors X in V, f(X) = X, A ;
€
€
then f is a functional (why is it linear?). We will
see that every functional on V can be represented
this way. For instance, consider the
functional
€
f :R 3 → R given by f (x, y, z) = 3x − 2y + z; we
represent it in the above form by writing
f (x, y, z) = (x, y, z) ⋅ (3,−2,1), i.e., f corresponds to the
dot product with the fixed vector A = (3,–2,1).
€
15B. Isometries and Functionals
5
Observe that we have a geometric interpretation
for this as well: the kernel of the functional f
corresponds to the plane 3x − 2y + z = 0 in R 3 , and A
is a normal vector to this plane (every vector in the
kernel of f is orthogonal to A). We now propose to
work out the details of how this situation
€
€
generalizes in abstract
inner product spaces.
If S is any set of vectors in an inner product space
V, we define the orthogonal complement of S to
be the set S ⊥ of vectors in V orthogonal to all
vectors in S: S ⊥ = {X ∈ V | X, Y = 0 for all Y ∈ S }.
€
Proposition
If S is any set of vectors in an inner
⊥
product
space
V,
then
S
is a subspace of V.
€
Further, only 0 can lie in both S and S ⊥ .
Proof If X and€X′ lie in S ⊥ , then X + X′, Y =
X, Y + X′, Y = 0 and rX, Y €
= r X, Y = 0 for all
Y ∈S ⊥ . €So S ⊥ is a€subspace of V. Also, if X lies in
both S and S ⊥ , then X, X€= 0, so |X|= 0, whence
€
€
X = 0. //
€
€
€
€
15B. Isometries and Functionals
6
Corollary Let W be a subspace of the inner
product space V. Then V = W ⊕ W ⊥ .
Proof By the proposition, W ⊥ is also subspace of
V and their intersection
€
To prove the
€ is trivial.
result, we need only show that V = W + W ⊥ . So let
{W1, W2,…, Wk } be an€orthonormal basis for W and
suppose that V is any vector in V. Let
X = V, W1 W1 +L + V, Wk Wk and €
set Y = V – X.
Then necessarily V = X + Y with X in W. But since
any W vector in W has the form
€
€
W = a1W1 + L+ a kWk
for suitable scalars a1,…, ak , we can compute
€
Y, W = Y,a1W1 + L+ akWk
€
= a1 Y, W1 + L+ ak Y, Wk
[
]
= a1 V, W1 − X, W1 +L
[
+ a k V, Wk − X, Wk
and since
€
]
15B. Isometries and Functionals
X, Wi =
7
V, W1 W1 +L + V, Wk Wk, Wi
= V, W1 W1, Wi + L+ V, Wk Wk, Wi
= V, Wi Wi , Wi
= V, Wi
it follows that Y, W = 0 for every W in W. That
€
is, Y lies in W ⊥ . So V = W + W ⊥ , completing the
proof. //
€
€
€
Corollary Let W be a k-dimensional subspace of
the n-dimensional inner product space V. Then
W ⊥ has dimension n – k. //
€
The results we have just worked out provide the
technical details that allow us to prove the central
theorem of this discussion:
15B. Isometries and Functionals
8
Theorem [The Riesz Representation Theorem]
Let V be an inner product space. Then for every
functional f ∈ V* , there exists a corresponding
unique vector A ∈ V with the property that
€
f(X) = X, A .
€
That is, every functional can be represented as an
inner product against some vector in V.
€
Proof If f is the zero map, then we can take A = 0
and we’re done. So from here on, we may assume
that f is not always zero on V. Since Im(f) is not
trivial and is also a subspace of R , it follows that
Im(f) = R . Thus, dim(Im(f)) = 1, so by the
Dimension Theorem, dim(ker(f)) = n – 1 where
dimV = n.
€
€
Let {V1, V2,…, Vn−1 } be an orthonormal basis for
ker(f). Since the subspace ker(f) ⊥ must satisfy
V = ker(f) ⊕ ker(f) ⊥ , it follows that ker(f) ⊥ is one€ dimensional. If V0 is a unit vector in ker(f) ⊥ , then
{V0, V1, V2,…, Vn−1 } must€be an orthonormal basis
€
€
for€all of V.
€
€
15B. Isometries and Functionals
9
So if X is any vector in V, we can write
X = X, V0 V0 + X, V1 V1 +L + X, Vn−1 Vn−1
so that
€
f (X) = X, V0 f (V0 ) + X, V1 f (V1 ) +L + X, Vn−1 f (Vn−1 )
= X, V0 f (V0 )
= X, f (V0 )V0
It follows that the vector A = f (V0 )V0 satisfies the
condition desired: f(X) = X, A .
€
The choice of A is€unique, for if B is another vector
with the property that f(X) = X, B , then X, A =
€
X, B for all X in V, so in particular,
A − B, A = A −€B, B ⇒ A −€B, A − B = 0
⇒|A − B|2= 0
⇒ A−B= 0
⇒A=B
€
//
€
€
15B. Isometries and Functionals
10
The Riesz Representation Theorem, applied to the
inner product space R 3 (under the standard dot
product), guarantees that for every linear map
(functional) f :R 3 → R there is an associated vector
A = (a,b,c) so
€ that f has the form
€
f (x, y, z) = (x, y, z) ⋅ (a,b, c) = ax + by + cz.
In fact, as the proof of the theorem indicates, A
3
determines
a
1-dimensional
subspace
of
R
€
orthogonal to the 2-dimensional kernel of f, namely
the plane with equation
€
(x, y, z) ⋅ (a,b, c) = ax + by + cz = 0;
that is, A is a vector normal to the plane.
€ none of this is new information, we can
While
produce some new results by looking at how the
theorem applies in other inner product spaces.
Consider the inner product space Pk(R) (with the
integral metric p(x), q(x) = ∫ 1−1 p(x)q(x) dx ). In this
setting, the Riesz Representation Theorem
guarantees that to any functional, like the
evaluation-at-a map fa ( p(x)) = p(a), there
€
corresponds a fixed polynomial qa (x) so that
€
€
15B. Isometries and Functionals
(*)
€
€
11
1
p(a) = f a ( p(x)) = p(x),q a (x) = ∫−1
p(x)qa (x) dx.
That is, the polynomial qa (x) has the remarkable
property that computation of the integral
1
∫−1 p(x)qa (x) dx is identical to evaluating the
polynomial p(x)€at x = a, and this is true
simultaneously for all polynomials p(x) of degree up
to k.
For instance, with k = 2 and a = 0, we want a
polynomial q0 (x) = a0 + a1x + a2 x 2 satisfying the
condition (*) above. Following the procedure
outlined in the proof of the theorem, we first find a
polynomial r(x) which is orthogonal to the kernel of
the€functional. The p(x) in this kernel satisfy
f0 ( p(x)) = p(0) = 0, so p(x) must have the form
b1x + b2 x 2. So r(x) satisfies
€
€
€
1
1
0 = b1x + b2 x 2, r(x) = b1 ∫−1
x ⋅ r(x) dx + b2 ∫−1
x 2 ⋅ r(x) dx
for every choice of b1,b2 . Thus, both integrals above
must vanish: if we put r(x) = c0 + c1x + c2x 2 , then
1
1
1
2
1
3
0 = ∫−1
x ⋅ r(x)
€ dx = c0 ∫−1 x dx + c1 ∫−1 x dx + c2 ∫−1 x dx
1 2
1 2
1
1
0 = ∫−1
x ⋅ r(x) dx
€ = c0 ∫−1
x dx + c1 ∫−1
x 3 dx + c2 ∫−1
x 4 dx
€
15B. Isometries and Functionals
12
becomes
0 = c0 ⋅0 + c1 ⋅ 23 + c 2 ⋅0
0 = c0 ⋅ 23 + c1 ⋅ 0 + c 2 ⋅ 25
whence c1 = 0 and c2 = − 53 c0 . So r(x) = c(1 − 53 x 2 ) for
any nonzero
choice of c. The procedure requires
€
that we normalize r(x), so we compute its length:
€
€
€
r(x), r(x) = c 2 ∫ 1−1 (1 − 53 x 2 )2dx
2
25 x 4 ) dx
= c 2 ∫ 1−1 (1 − 10
x
+
3
9
10 )
= c 2 (2 − 20
+
9
9
= 89 c 2
€
To make this length equal 1, we force c 2 = 98 .
Finally, according to the procedure, we put
2
q0 (x) = f 0(r(x)) ⋅ r(x) = r(0) ⋅ r(x) = c€⋅c(1 − 53 x 2 ) = 98 − 15
x
8
€
It follows that every polynomial p(x) in P2(R)
satisfies the formula
1
2
p(0) = ∫−1
p(x)( 89 − 15
x
) dx .
8
€