11.3 Riesz Representation Theorem
The next theorem is one of the most important results in the theory of inner product spaces. It is
called the Riesz representation theorem.
Theorem 11.3.1 Let f ∈ℒ
where X is an inner product space of dimension n. Then
there exists a unique z ∈ X such that for all x ∈ X,
Proof: First I will verify uniqueness. Suppose zj works for j = 1,2. Then for all
x ∈ X,
and so z1 = z2.
It remains to verify existence. By Lemma 11.2.1, there exists an orthonormal basis,
If there is such a z,
then you would need f
and so you would need
Also you must have z
. Therefore, define
Then using Lemma 11.2.3,
Corollary 11.3.2 Let A ∈ℒ
where X and Y are two inner product spaces of finite
dimension. Then there exists a unique A∗∈ℒ
for all x ∈ X and y ∈ Y. The following formula holds
Proof: Let fy ∈ℒ
be defined as
Then by the Riesz representation theorem, there exists a unique element of X, A∗
It only remains to verify that A∗ is linear. Let a and b be scalars. Then for all x ∈ X,
Since this holds for every x, it follows
which shows A∗ is linear as claimed.
Consider the last assertion that ∗ is conjugate linear.
and since this is true for all y,
Definition 11.3.3 The linear map, A∗ is called the adjoint of A. In the case when
A : X → X and A = A∗, A is called a self adjoint map. Such a map is also called Hermitian.
Theorem 11.3.4 Let M be an m×n matrix. Then M∗ =
T in words, the transpose
of the conjugate of M is equal to the adjoint.
Proof: Using the definition of the inner product in ℂn,
Since x,y are arbitrary vectors, it follows that Mji =
ij and so, taking conjugates of both
The next theorem is interesting. You have a p dimensional subspace of Fn where F = ℝ or ℂ.
Of course this might be “slanted”. However, there is a linear transformation Q which preserves
distances which maps this subspace to Fp.
Theorem 11.3.5 Suppose V is a subspace of Fn having dimension p ≤ n. Then there exists a
for all x. Also
Proof: By Lemma 11.2.1 there exists an orthonormal basis for V,
By using the
Gram Schmidt process this may be extended to an orthonormal basis of the whole space
Now define Q ∈ℒ
and extend linearly. If ∑
is an arbitrary
element of Fn,
It remains to verify that Q∗Q = QQ∗ = I. To do so, let x,y ∈ Fn. Then let ω be a complex
number such that
and since Q preserves norms, it follows that for all x,y ∈ Fn,
for all x,y. Letting x = Q∗Qy − y, it follows Q∗Qy = y. Similarly QQ∗ = I. ■
Note that is is actually shown that QV = span
and that in case
that a linear transformation which maps an orthonormal basis to an orthonormal basis is