Riesz representation theorem explained
The Riesz representation theorem, sometimes called the Riesz–Fréchet representation theorem after Frigyes Riesz and Maurice René Fréchet, establishes an important connection between a Hilbert space and its continuous dual space. If the underlying field is the real numbers, the two are isometrically isomorphic; if the underlying field is the complex numbers, the two are isometrically anti-isomorphic. The (anti-) isomorphism is a particular natural isomorphism.
Preliminaries and notation
Let
be a
Hilbert space over a field
where
is either the real numbers
or the complex numbers
If
(resp. if
) then
is called a (resp. a). Every real Hilbert space can be extended to be a
dense subset of a unique (up to
bijective isometry) complex Hilbert space, called its
complexification, which is why Hilbert spaces are often automatically assumed to be complex. Real and complex Hilbert spaces have in common many, but by no means all, properties and results/theorems.
This article is intended for both mathematicians and physicists and will describe the theorem for both. In both mathematics and physics, if a Hilbert space is assumed to be real (that is, if
) then this will usually be made clear. Often in mathematics, and especially in physics, unless indicated otherwise, "Hilbert space" is usually automatically assumed to mean "complex Hilbert space." Depending on the author, in mathematics, "Hilbert space" usually means either (1) a complex Hilbert space, or (2) a real complex Hilbert space.
Linear and antilinear maps
By definition, an (also called a)
is a map between
vector spaces that is :
and (also called or):
where
is the conjugate of the complex number
, given by
.
In contrast, a map
is
linear if it is additive and
:
Every constant
map is always both linear and antilinear. If
then the definitions of linear maps and antilinear maps are completely identical. A linear map from a Hilbert space into a
Banach space (or more generally, from any Banach space into any
topological vector space) is
continuous if and only if it is
bounded; the same is true of antilinear maps. The
inverse of any antilinear (resp. linear) bijection is again an antilinear (resp. linear) bijection. The composition of two linear maps is a map.
Continuous dual and anti-dual spaces
A on
is a function
whose
codomain is the underlying scalar field
Denote by
(resp. by
the set of all continuous linear (resp. continuous antilinear) functionals on
which is called the (resp. the) of
If
then linear functionals on
are the same as antilinear functionals and consequently, the same is true for such continuous maps: that is,
One-to-one correspondence between linear and antilinear functionals
Given any functional
the is the functional
This assignment is most useful when
because if
then
and the assignment
reduces down to the
identity map.
The assignment
defines an antilinear
bijective correspondence from the set of
all functionals (resp. all linear functionals, all continuous linear functionals
) on
onto the set of
all functionals (resp. all linear functionals, all continuous linear functionals
) on
Mathematics vs. physics notations and definitions of inner product
has an associated
inner product
valued in
's underlying scalar field
that is linear in one coordinate and antilinear in the other (as specified below).If
is a complex Hilbert space (
), then there is a crucial difference between the notations prevailing in mathematics versus physics, regarding which of the two variables is linear.However, for real Hilbert spaces (
), the inner product is a
symmetric map that is linear in each coordinate (
bilinear), so there can be no such confusion.
In mathematics, the inner product on a Hilbert space
is often denoted by
\left\langle ⋅ , ⋅ \right\rangle
or
\left\langle ⋅ , ⋅ \right\rangleH
while in
physics, the
bra–ket notation \left\langle ⋅ \mid ⋅ \right\rangle
or
\left\langle ⋅ \mid ⋅ \right\rangleH
is typically used. In this article, these two notations will be related by the equality:
These have the following properties:
- The map
\left\langle ⋅ , ⋅ \right\rangle
is linear in its first coordinate; equivalently, the map \left\langle ⋅ \mid ⋅ \right\rangle
is linear in its second coordinate. That is, for fixed
the map \left\langley\mid ⋅ \right\rangle=\left\langle ⋅ ,y\right\rangle:H\toF
with is a linear functional on
This linear functional is continuous, so \left\langley\mid ⋅ \right\rangle=\left\langle ⋅ ,y\right\rangle\inH*.
- The map
\left\langle ⋅ , ⋅ \right\rangle
is antilinear in its coordinate; equivalently, the map \left\langle ⋅ \mid ⋅ \right\rangle
is antilinear in its coordinate. That is, for fixed
the map\left\langle ⋅ \midy\right\rangle=\left\langley, ⋅ \right\rangle:H\toF
with is an antilinear functional on
This antilinear functional is continuous, so \left\langle ⋅ \midy\right\rangle=\left\langley, ⋅ \right\rangle\in\overline{H}*.
In computations, one must consistently use either the mathematics notation
\left\langle ⋅ , ⋅ \right\rangle
, which is (linear, antilinear); or the physics notation
\left\langle ⋅ \mid ⋅ \right\rangle
, whch is (antilinear | linear).
Canonical norm and inner product on the dual space and anti-dual space
If
then
\langlex\midx\rangle=\langlex,x\rangle
is a non-negative real number and the map
defines a canonical norm on
that makes
into a
normed space. As with all normed spaces, the (continuous) dual space
carries a canonical norm, called the, that is defined by
denoted by
is defined by using this same equation:
This canonical norm on
satisfies the
parallelogram law, which means that the
polarization identity can be used to define a which this article will denote by the notations
where this inner product turns
into a Hilbert space. There are now two ways of defining a norm on
the norm induced by this inner product (that is, the norm defined by
f\mapsto\sqrt{\left\langlef,f
}) and the usual
dual norm (defined as the supremum over the closed unit ball). These norms are the same; explicitly, this means that the following holds for every
As will be described later, the Riesz representation theorem can be used to give an equivalent definition of the canonical norm and the canonical inner product on
The same equations that were used above can also be used to define a norm and inner product on
's
anti-dual space
Canonical isometry between the dual and antidual
of a functional
which was defined above, satisfies
for every
and every
This says exactly that the canonical antilinear
bijection defined by
as well as its inverse
\operatorname{Cong}-1~:~\overline{H}*\toH*
are antilinear
isometries and consequently also
homeomorphisms. The inner products on the dual space
and the anti-dual space
denoted respectively by
and
are related by
and
If
then
and this canonical map
\operatorname{Cong}:H*\to\overline{H}*
reduces down to the identity map.
Riesz representation theorem
Two vectors
and
are if
which happens if and only if
for all scalars
The
orthogonal complement of a subset
is
which is always a
closed vector subspace of
The
Hilbert projection theorem guarantees that for any
nonempty closed
convex subset
of a
Hilbert space there exists a unique vector
such that
that is,
is the (unique) global minimum point of the function
defined by
Statement
Historically, the theorem is often attributed simultaneously to Riesz and Fréchet in 1907 (see references).
Let
denote the underlying scalar field of
Fix
Define
by
which is a linear functional on
since
is in the linear argument. By the
Cauchy–Schwarz inequality,
which shows that
is bounded (equivalently, continuous) and that
It remains to show that
By using
in place of
it follows that
(the equality
holds because
is real and non-negative). Thus that
The proof above did not use the fact that
is
complete, which shows that the formula for the norm
holds more generally for all
inner product spaces.
Suppose
are such that
\varphi(z)=\langlef|z\rangle
and
\varphi(z)=\langleg|z\rangle
for all
Then
which shows that
is the constant
linear functional. Consequently
0=\|\langlef-g| ⋅ \rangle\|=\|f-g\|,
which implies that
Let
K:=\ker\varphi:=\{m\inH:\varphi(m)=0\}.
If
(or equivalently, if
) then taking
completes the proof so assume that
and
The continuity of
implies that
is a closed subspace of
(because
and
is a closed subset of
). Let
denote the
orthogonal complement of
in
Because
is closed and
is a Hilbert space,
[1]
can be written as the direct sum
[2] (a proof of this is given in the article on the
Hilbert projection theorem). Because
there exists some non-zero
For any
which shows that
(\varphih)p-(\varphip)h~\in~\ker\varphi=K,
where now
implies
Solving for
shows that
which proves that the vector
} p satisfies
\varphih=\langlef\varphi|h\rangleforeveryh\inH.
Applying the norm formula that was proved above with
shows that
=\left\|\left\langlef\varphi| ⋅
=\left\|f\varphi\right\|H.
Also, the vector
has norm
and satisfies
f\varphi:=\overline{\varphi(u)}u.
It can now be deduced that
is
-dimensional when
Let
be any non-zero vector. Replacing
with
in the proof above shows that the vector
} q satisfies
\varphi(h)=\langleg|h\rangle
for every
The uniqueness of the (non-zero) vector
representing
implies that
which in turn implies that
and
} f_. Thus every vector in
is a scalar multiple of
The formulas for the inner products follow from the polarization identity.
Observations
If
then
So in particular,
\varphi\left(f\varphi\right)\geq0
is always real and furthermore,
\varphi\left(f\varphi\right)=0
if and only if
if and only if
Linear functionals as affine hyperplanes
A non-trivial continuous linear functional
is often interpreted geometrically by identifying it with the affine hyperplane
(the kernel
is also often visualized alongside
although knowing
is enough to reconstruct
because if
then
and otherwise
). In particular, the norm of
should somehow be interpretable as the "norm of the hyperplane
". When
then the Riesz representation theorem provides such an interpretation of
in terms of the affine hyperplane
as follows: using the notation from the theorem's statement, from
it follows that
C:=\varphi-1\left(\|\varphi\|2\right)=\|\varphi\|2\varphi-1(1)=\|\varphi\|2A
and so
\|\varphi\|=\left\|f\varphi\right\|=infc\|c\|
implies
\|\varphi\|=infa\|\varphi\|2\|a\|
and thus
This can also be seen by applying the
Hilbert projection theorem to
and concluding that the global minimum point of the map
defined by
is
The formulas
provide the promised interpretation of the linear functional's norm
entirely in terms of its associated affine hyperplane
(because with this formula, knowing only the
is enough to describe the norm of its associated linear). Defining
the
infimum formula
will also hold when
When the supremum is taken in
(as is typically assumed), then the supremum of the empty set is
but if the supremum is taken in the non-negative reals
(which is the
image/range of the norm
when
) then this supremum is instead
in which case the supremum formula
will also hold when
(although the atypical equality
is usually unexpected and so risks causing confusion).
Constructions of the representing vector
Using the notation from the theorem above, several ways of constructing
from
are now described. If
then
; in other words,
This special case of
is henceforth assumed to be known, which is why some of the constructions given below start by assuming
Orthogonal complement of kernel
If
then for any
0 ≠ u\in(\ker\varphi)\bot,
If
is a
unit vector (meaning
) then
(this is true even if
because in this case
f\varphi=\overline{\varphi(u)}u=\overline{0}u=0
). If
is a unit vector satisfying the above condition then the same is true of
which is also a unit vector in
However,
\overline{\varphi(-u)}(-u)=\overline{\varphi(u)}u=f\varphi
so both these vectors result in the same
Orthogonal projection onto kernel
If
is such that
and if
is the orthogonal projection of
onto
then
Orthonormal basis
of
and a continuous linear functional
the vector
can be constructed uniquely by
where all but at most countably many
will be equal to
and where the value of
does not actually depend on choice of orthonormal basis (that is, using any other orthonormal basis for
will result in the same vector). If
is written as
then
and
If the orthonormal basis
\left\{ei\right\}i=\left\{ei\right\}
is a sequence then this becomes
and if
is written as
then
Example in finite dimensions using matrix transformations
Consider the special case of
(where
is an integer) with the standard inner product
where
are represented as
column matrices \vec{w}:=\begin{bmatrix}w1\ \vdots\ wn\end{bmatrix}
and
\vec{z}:=\begin{bmatrix}z1\ \vdots\ zn\end{bmatrix}
with respect to the standard orthonormal basis
on
(here,
is
at its
th coordinate and
everywhere else; as usual,
will now be associated with the
dual basis) and where
\overline{\vec{z}}\operatorname{T
} := \left[\overline{z_1}, \ldots, \overline{z_n}\right] denotes the
conjugate transpose of
Let
be any linear functional and let
\varphi1,\ldots,\varphin\in\Complex
be the unique scalars such that
where it can be shown that
\varphii=\varphi\left(ei\right)
for all
Then the Riesz representation of
is the vector
To see why, identify every vector
w=\left(w1,\ldots,wn\right)
in
with the column matrix
\vec{w}:=\begin{bmatrix}w1\ \vdots\ wn\end{bmatrix}
so that
is identified with
} := \begin\overline \\ \vdots \\ \overline\end = \begin\overline \\ \vdots \\ \overline\end. As usual, also identify the linear functional
with its
transformation matrix, which is the
row matrix \vec{\varphi}:=\left[\varphi1,\ldots,\varphin\right]
so that
} := \overline^ and the function
is the assignment
\vec{w}\mapsto\vec{\varphi}\vec{w},
where the right hand side is
matrix multiplication. Then for all
w=\left(w1,\ldots,wn\right)\inH,
which shows that
satisfies the defining condition of the Riesz representation of
The bijective antilinear isometry
defined in the corollary to the Riesz representation theorem is the assignment that sends
z=\left(z1,\ldots,zn\right)\inH
to the linear functional
on
defined by
where under the identification of vectors in
with column matrices and vector in
with row matrices,
is just the assignment
As described in the corollary,
's inverse
is the antilinear isometry
which was just shown above to be:
where in terms of matrices,
is the assignment
Thus in terms of matrices, each of
and
is just the operation of
conjugate transposition \vec{v}\mapsto\overline{\vec{v}}\operatorname{T
} (although between different spaces of matrices: if
is identified with the space of all column (respectively, row) matrices then
is identified with the space of all row (respectively, column matrices).
This example used the standard inner product, which is the map
\langlez\midw\rangle:=\overline{\vec{z}}\operatorname{T
} \vec, but if a different inner product is used, such as
\langlez\midw\rangleM:=\overline{\vec{z}}\operatorname{T
} \, M \, \vec \, where
is any
Hermitian positive-definite matrix, or if a different orthonormal basis is used then the transformation matrices, and thus also the above formulas, will be different.
Relationship with the associated real Hilbert space
See also: Complexification.
Assume that
is a complex Hilbert space with inner product
\langle ⋅ \mid ⋅ \rangle.
When the Hilbert space
is reinterpreted as a real Hilbert space then it will be denoted by
where the (real) inner-product on
is the real part of
's inner product; that is:
The norm on
induced by
is equal to the original norm on
and the continuous dual space of
is the set of all -valued bounded
-linear functionals on
(see the article about the
polarization identity for additional details about this relationship). Let
\psi\R:=\operatorname{re}\psi
and
\psii:=\operatorname{im}\psi
denote the real and imaginary parts of a linear functional
so that
\psi=\operatorname{re}\psi+i\operatorname{im}\psi=\psi\R+i\psii.
The formula expressing a linear functional in terms of its real part is
where
for all
It follows that
and that
if and only if
It can also be shown that
\|\psi\|=\left\|\psi\R\right\|=\left\|\psii\right\|
where
\left\|\psi\R\right\|:=\sup\|h\|\left|\psi\R(h)\right|
and
\left\|\psii\right\|:=\sup\|h\|\left|\psii(h)\right|
are the usual
operator norms. In particular, a linear functional
is bounded if and only if its real part
is bounded.
Representing a functional and its real part
The Riesz representation of a continuous linear function
on a complex Hilbert space is equal to the Riesz representation of its real part
on its associated real Hilbert space.
Explicitly, let
and as above, let
be the Riesz representation of
obtained in
(H,\langle, ⋅ , ⋅ \rangle),
so it is the unique vector that satisfies
\varphi(x)=\left\langlef\varphi\midx\right\rangle
for all
The real part of
is a continuous real linear functional on
and so the Riesz representation theorem may be applied to
\varphi\R:=\operatorname{re}\varphi
and the associated real Hilbert space
\left(H\R,\langle, ⋅ , ⋅ \rangle\R\right)
to produce its Riesz representation, which will be denoted by
That is,
is the unique vector in
that satisfies
\varphi\R(x)=\left\langle
\midx\right\rangle\R
for all
The conclusion is
This follows from the main theorem because
\ker\varphi\R=\varphi-1(i\R)
and if
then
and consequently, if
then
\left\langlef\varphi\midm\right\rangle\R=0,
which shows that
f\varphi\in(\ker\varphi\R
.
Moreover,
\varphi(f\varphi)=\|\varphi\|2
being a real number implies that
\varphi\R(f\varphi)=\operatorname{re}\varphi(f\varphi)=\|\varphi\|2.
In other words, in the theorem and constructions above, if
is replaced with its real Hilbert space counterpart
and if
is replaced with
then
f\varphi=f\operatorname{re\varphi}.
This means that vector
obtained by using
\left(H\R,\langle, ⋅ , ⋅ \rangle\R\right)
and the real linear functional
is the equal to the vector obtained by using the origin complex Hilbert space
\left(H,\left\langle, ⋅ , ⋅ \right\rangle\right)
and original complex linear functional
(with identical norm values as well).
Furthermore, if
then
is perpendicular to
with respect to
where the kernel of
is be a
proper subspace of the kernel of its real part
Assume now that
Then
f\varphi\not\in\ker\varphi\R
because
\varphi\R\left(f\varphi\right)=\varphi\left(f\varphi\right)=\|\varphi\|2 ≠ 0
and
is a proper subset of
The vector subspace
has real codimension
in
while
has codimension
in
and
\left\langlef\varphi,\ker\varphi\R\right\rangle\R=0.
That is,
is perpendicular to
with respect to
Canonical injections into the dual and anti-dual
Induced linear map into anti-dual
The map defined by placing
into the coordinate of the inner product and letting the variable
vary over the coordinate results in an
functional:
This map is an element of
which is the continuous
anti-dual space of
The
is the
operator which is also an
injective isometry. The
Fundamental theorem of Hilbert spaces, which is related to Riesz representation theorem, states that this map is surjective (and thus
bijective). Consequently, every antilinear functional on
can be written (uniquely) in this form.
If
\operatorname{Cong}:H*\to\overline{H}*
is the canonical
linear bijective isometry
that was defined above, then the following equality holds:
Extending the bra–ket notation to bras and kets
See main article: Bra–ket notation.
Let
\left(H,\langle ⋅ , ⋅ \rangleH\right)
be a Hilbert space and as before, let
\langley|x\rangleH:=\langlex,y\rangleH.
Let
which is a bijective antilinear isometry that satisfies
Bras
Given a vector
let
denote the continuous linear functional
; that is,
so that this functional
is defined by
g\mapsto\left\langleh\midg\right\rangleH.
This map was denoted by
\left\langleh\mid ⋅ \right\rangle
earlier in this article.
The assignment
is just the isometric antilinear isomorphism
which is why
~\langlecg+h|~=~\overline{c}\langleg\mid~+~\langleh|~
holds for all
and all scalars
The result of plugging some given
into the functional
is the scalar
\langleh|g\rangleH=\langleg,h\rangleH,
which may be denoted by
[3] Bra of a linear functional
Given a continuous linear functional
let
denote the vector
; that is,
The assignment
\psi\mapsto\langle\psi\mid
is just the isometric antilinear isomorphism
which is why
~\langlec\psi+\phi\mid~=~\overline{c}\langle\psi\mid~+~\langle\phi\mid~
holds for all
and all scalars
The defining condition of the vector
is the technically correct but unsightly equality
which is why the notation
\left\langle\psi\midg\right\rangle
is used in place of
\left\langle\langle\psi\mid\midg\right\rangleH=\left\langleg,\langle\psi\mid\right\rangleH.
With this notation, the defining condition becomes
Kets
For any given vector
the notation
is used to denote
; that is,
The assignment
is just the identity map
\operatorname{Id}H:H\toH,
which is why
~\midcg+h\rangle~=~c\midg\rangle~+~\midh\rangle~
holds for all
and all scalars
The notation
and
is used in place of
\left\langleh\mid\midg\rangle\right\rangleH~=~\left\langle\midg\rangle,h\right\rangleH
and
\left\langle\psi\mid\midg\rangle\right\rangleH~=~\left\langleg,\langle\psi\mid\right\rangleH,
respectively. As expected,
~\langle\psi\midg\rangle=\psig~
and
really is just the scalar
~\langleh\midg\rangleH~=~\langleg,h\rangleH.
Adjoints and transposes
Let
be a
continuous linear operator between
Hilbert spaces \left(H,\langle ⋅ , ⋅ \rangleH\right)
and
\left(Z,\langle ⋅ , ⋅ \rangleZ\right).
As before, let
\langley\midx\rangleH:=\langlex,y\rangleH
and
\langley\midx\rangleZ:=\langlex,y\rangleZ.
Denote bythe usual bijective antilinear isometries that satisfy:
Definition of the adjoint
See main article: Hermitian adjoint and Conjugate transpose.
For every
the scalar-valued map
\langlez\midA( ⋅ )\rangleZ
on
defined by
is a continuous linear functional on
and so by the Riesz representation theorem, there exists a unique vector in
denoted by
such that
\langlez\midA( ⋅ )\rangleZ=\left\langleA*z\mid ⋅ \right\rangleH,
or equivalently, such that
The assignment
thus induces a function
called the of
whose defining condition is
The adjoint
is necessarily a
continuous (equivalently, a
bounded)
linear operator.
If
is finite dimensional with the standard inner product and if
is the
transformation matrix of
with respect to the standard orthonormal basis then
's
conjugate transpose \overline{M\operatorname{T
}} is the transformation matrix of the adjoint
Adjoints are transposes
See main article: Transpose of a linear map.
See also: Transpose.
It is also possible to define the or of
which is the map
defined by sending a continuous linear functionals
to
where the
composition
is always a continuous linear functional on
and it satisfies
\|A\|=\left\|{}tA\right\|
(this is true more generally, when
and
are merely
normed spaces). So for example, if
then
sends the continuous linear functional
\langlez\mid ⋅ \rangleZ\inZ*
(defined on
by
g\mapsto\langlez\midg\rangleZ
) to the continuous linear functional
\langlez\midA( ⋅ )\rangleZ\inH*
(defined on
by
h\mapsto\langlez\midA(h)\rangleZ
); using bra-ket notation, this can be written as
{}tA\langlez\mid~=~\langlez\midA
where the juxtaposition of
with
on the right hand side denotes function composition:
H\xrightarrow{A}Z\xrightarrow{\langlez\mid}\Complex.
The adjoint
is actually just to the transpose
when the Riesz representation theorem is used to identify
with
and
with
Explicitly, the relationship between the adjoint and transpose is:
which can be rewritten as:
Alternatively, the value of the left and right hand sides of at any given
can be rewritten in terms of the inner products as:
so that
{}tA~\circ~\PhiZ~=~\PhiH~\circ~A*
holds if and only if
\langlez\midA( ⋅ )\rangleZ=\langleA*z\mid ⋅ \rangleH
holds; but the equality on the right holds by definition of
The defining condition of
can also be written
if bra-ket notation is used.
Descriptions of self-adjoint, normal, and unitary operators
Assume
and let
Let
be a continuous (that is, bounded) linear operator.
Whether or not
is
self-adjoint,
normal, or
unitary depends entirely on whether or not
satisfies certain defining conditions related to its adjoint, which was shown by to essentially be just the transpose
Because the transpose of
is a map between continuous linear functionals, these defining conditions can consequently be re-expressed entirely in terms of linear functionals, as the remainder of subsection will now describe in detail. The linear functionals that are involved are the simplest possible continuous linear functionals on
that can be defined entirely in terms of
the inner product
on
and some given vector
Specifically, these are
\left\langleAh\mid ⋅ \right\rangle
and
\langleh\midA( ⋅ )\rangle
where
Self-adjoint operators
See also: Self-adjoint operator, Hermitian matrix and Symmetric matrix.
A continuous linear operator
is called
self-adjoint if it is equal to its own adjoint; that is, if
Using, this happens if and only if:
where this equality can be rewritten in the following two equivalent forms:
Unraveling notation and definitions produces the following characterization of self-adjoint operators in terms of the aforementioned continuous linear functionals:
is self-adjoint if and only if for all
the linear functional
\langlez\midA( ⋅ )\rangle
is equal to the linear functional
; that is, if and only if
where if bra-ket notation is used, this is
Normal operators
See also: Normal operator and Normal matrix.
A continuous linear operator
is called
normal if
which happens if and only if for all
Using and unraveling notation and definitions produces the following characterization of normal operators in terms of inner products of continuous linear functionals:
is a normal operator if and only if
where the left hand side is also equal to
\overline{\langleAh\midAz\rangle}H=\langleAz\midAh\rangleH.
The left hand side of this characterization involves
only linear functionals of the form
while the right hand side involves
only linear functions of the form
\langleh\midA( ⋅ )\rangle
(defined as above). So in plain English, characterization says that an operator is
normal when the inner product of any two linear functions of the first form is equal to the inner product of their second form (using the same vectors
for both forms).In other words, if it happens to be the case (and when
is injective or self-adjoint, it is) that the assignment of linear functionals
\langleAh\mid ⋅ \rangle~\mapsto~\langleh|A( ⋅ )\rangle
is well-defined (or alternatively, if
\langleh|A( ⋅ )\rangle~\mapsto~\langleAh\mid ⋅ \rangle
is well-defined) where
ranges over
then
is a normal operator if and only if this assignment preserves the inner product on
The fact that every self-adjoint bounded linear operator is normal follows readily by direct substitution of
into either side of
This same fact also follows immediately from the direct substitution of the equalities into either side of .
Alternatively, for a complex Hilbert space, the continuous linear operator
is a normal operator if and only if
\|Az\|=\left\|A*z\right\|
for every
which happens if and only if
Unitary operators
See also: Unitary transformation and Unitary matrix.
An invertible bounded linear operator
is said to be
unitary if its inverse is its adjoint:
By using, this is seen to be equivalent to
\Phi\circA-1={}tA\circ\Phi.
Unraveling notation and definitions, it follows that
is unitary if and only if
The fact that a bounded invertible linear operator
is unitary if and only if
(or equivalently,
) produces another (well-known) characterization: an invertible bounded linear map
is unitary if and only if
Because
is invertible (and so in particular a bijection), this is also true of the transpose
This fact also allows the vector
in the above characterizations to be replaced with
or
thereby producing many more equalities. Similarly,
can be replaced with
or
Notes
Proofs
Bibliography
Notes and References
- Showing that there is a non-zero vector
in
relies on the continuity of
and the Cauchy completeness of
This is the only place in the proof in which these properties are used.
- Technically,
means that the addition map
defined by
is a surjective linear isomorphism and homeomorphism. See the article on complemented subspaces for more details.
- The usual notation for plugging an element
into a linear map
is
and sometimes
Replacing
with
produces
or
which is unsightly (despite being consistent with the usual notation used with functions). Consequently, the symbol
is appended to the end, so that the notation
is used instead to denote this value