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 13.2.1Let f ∈ℒ
(H, F)
where H is a Hilbert space and f is continuous. Recall thatin finite dimensions, this is automatic. Then there exists a unique z ∈ H such that for allx ∈ H,
f (x ) = (x,z) .
Proof:First I will verify uniqueness. Suppose z_{j} works for j = 1,2. Then for all x ∈ H,
0 = f (x)− f (x) = (x,z1 − z2)
and so z_{1} = z_{2}.
If f
(H )
= 0, let z = 0 and this works. Otherwise, let u
∕∈
f^{−1}
(0)
which is a closed subspace of H. Let
w = u − Pu≠0. Then
f (f (w )x− f (x)w ) = f (w )f (x)− f (x)f (w) = 0
Definition 13.2.3The linear map, A^{∗}is called the adjoint of A. In the case when A : X → Xand A = A^{∗}, A is called a self adjoint map. Such a map is also called Hermitian.
Theorem 13.2.4Let M be an m × n matrix. Then M^{∗} =
(M-)
^{T}in words, the transpose of theconjugate of M is equal to the adjoint.
Proof:Using the definition of the inner product in ℂ^{n},
∗ ∑ ∑---------- ∑ --------
(M x,y) = (x,M y) ≡ xi (M ∗)ijyj = (M ∗)ijyjxi.
i j i,j
Also
∑ ∑ --
(M x,y) = Mjiyjxi.
j i
Since x,y are arbitrary vectors, it follows that M_{ji} =
(M ∗)
_{ij} and so, taking conjugates of both
sides,
∗ ----
M ij = Mji ■
Some linear transformations preserve distance. Something special can be asserted about these which is
in the next lemma.
Lemma 13.2.5Suppose R ∈ ℒ
(X,Y )
where X,Y are inner product spaces and R preservesdistances. Then R^{∗}R = I.
= 0 for all v and so R^{∗}Ru = u for all u which implies
R^{∗}R = I. ■
The next theorem is interesting. You have a p dimensional subspace of F^{n} 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 F^{p}.
Theorem 13.2.6Suppose V is a subspace of F^{n}having dimension p ≤ n. Then there exists aQ ∈ℒ
(Fn, Fn)
such that
QV ⊆ span(e1,⋅⋅⋅,ep)
and
|Qx|
=
|x |
for allx.Also
Q∗Q = QQ ∗ = I.
Proof:By Lemma 13.1.6 there exists an orthonormal basis for V,
{vi}
_{i=1}^{p}. By using the Gram
Schmidt process this may be extended to an orthonormal basis of the whole space F^{n},
{v1,⋅⋅⋅,vp,vp+1,⋅⋅⋅,vn} .
Now define Q ∈ℒ
(Fn,Fn )
by Q
(v )
i
≡ e_{i} and extend linearly. If ∑_{i=1}^{n}x_{i}v_{i} is an arbitrary element of
F^{n},
Thus Q preserves lengths and so, by Lemma 13.2.5, it follows that Q^{∗}Q = I. Also, this shows that Q maps
V onto V and so a generic element of V is of the form Qx. Now
showing that Q^{∗} also preserves lengths. Hence it is also the case that QQ^{∗} = I because from the definition
of the adjoint,
(Q∗)
^{∗} = Q. ■
Definition 13.2.7If U ∈ ℒ
(X,X )
for X an inner product space, then U is calledunitary ifU^{∗}U = UU^{∗} = I.
Note that it is actually shown that QV = span
(e1,⋅⋅⋅,ep)
and that in case p = n one obtains that a
linear transformation which maps an orthonormal basis to an orthonormal basis is unitary. Unitary
matrices are also characterized by preserving length. More generally
Corollary 13.2.8Suppose U ∈ℒ
(X, X )
where X is an inner product space. Then U is unitary ifand only if
|Ux|
=
|x|
for all x so it preserves distance.
Proof:⇒ If U is unitary, then
|Ux |
^{2} =
(Ux,U x)
=
(U ∗Ux,x)
=
(x,x )
=
|x|
^{2}.
⇐ If
|U x|
=
|x|
for all x then by Lemma 13.2.5, U^{∗}U = I. Thus U is onto since it is one to one and so
a generic element of X is Ux. Note how this would fail if you had U ∈ℒ
(X, Y)
where the dimension of Y is
larger than the dimension of X. Then as above,
Thus also UU^{∗} = I because U^{∗} preserves distances and
(U ∗)
^{∗} = U from the definition. ■
Now here is an important result on factorization of an m × n matrix. It is called a QR
factorization.
Theorem 13.2.9Let A be an m × n complex matrix. Then there exists a unitary Q and R, allzero below the main diagonal (R_{ij} = 0if i > j) such that A = QR.
Proof: This is obvious if m = 1.
( ) ( )
a1 ⋅⋅⋅ an = (1) a1 ⋅⋅⋅ an
Suppose true for m − 1 and let
A = ( a ⋅⋅⋅ a ) , A is m × n
1 n
Using Theorem 13.2.6, there exists Q_{1} a unitary matrix such that Q_{1}
(a1∕|a1|)
= e_{1} in case a_{1}≠0. Thus
Q_{1}a_{1} =
|a1|
e_{1}. If a_{1} = 0, let Q_{1} = I. Thus
( T )
Q1A = a b
0 A1
where A_{1} is
(m − 1)
×
(n− 1)
. If n = 1, this obtains
( ) ( )
a ∗ a ∗
Q1A = 0 , A = Q1 0 , let Q = Q1.
That which is desired is obtained. So assume n > 1. By induction, there exists Q_{2}^{′} an
(m − 1)
×
(n − 1)
unitary matrix such that Q_{2}^{′}A_{1} = R^{′}, R_{ij}^{′} = 0 if i > j. Then
( 1 0 ) ( a bT )
′ Q1A = ′ = R
0 Q2 0 R
Since the product of unitary matrices is unitary, there exists Q unitary such that Q^{∗}A = R and so
A = QR. ■