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Linear functional. Matrix representation. Dual space, conjugate space, adjoint space. Basis for dual space. Annihilator. Transpose of a linear mapping.

Def. Functional. Let V be an abstract vector space over a field F. A functional T is a function T:V → F that assigns a number from field F to each vector x ε V.

Def. Linear functional. A functional T is linear if

T(av_{1} + bv_{2}) = aTv_{1} + bTv_{2}

for all vectors v_{1} and v_{2} and scalars a and b.

Examples.

1. Let V be the vector space of polynomials in t over R, the field of reals. Let T:V → R be the integral operator defined by

This integral effects a linear mapping from the space of polynomials to the field of reals and hence T is a linear functional.

2. Let V be the vector space of n-square matrices over F. Let T:V → R be the trace mapping

T(A) = a_{11} + a_{22} + .... + a_{nn}

where matrix A = (a_{ij}). That is, T assigns to a matrix A the sum of its diagonal elements. This
mapping can be shown to be linear and hence T is a linear functional.

3. Let π_{i}:R^{n }→^{ }R be the i-th *projection mapping* i.e. for any vector X = (a_{1}, a_{2}, ..... , a_{n}) ε R^{n}, π_{i} = a_{i},
the i-th coordinate of X. This mapping is linear and π_{i} is a linear functional on R^{n}.

The domain V of a linear functional T: V → F can be either infinite dimensional or finite dimensional. We will consider here only linear functionals in which the domain V is finite dimensional.

Matrix representation of a linear functional whose domain is finite dimensional. Any linear mapping from one finite dimensional abstract vector space to another is represented by a matrix. A linear mapping from an n-dimensional vector space over a field F to an m-dimensional vector space over F is represented by an mxn matrix.over F. A linear functional T: V → F whose domain V is finite dimensional is a linear mapping from an n-dimensional vector space to a 1-dimensional vector space and is represented by a 1xn matrix i.e. an n-element row vector. The matrix representation of the mapping is

T(v) = Av

where v is an n-element coordinate vector and A is a 1xn matrix representation of T. Thus the linear functional has the form

or

T(v) = a_{1}v_{1} + a_{2}v_{2} + .... + a_{n}v_{n}

Dual Space. If V is some abstract vector space over a field F, then the dual space of V is the vector space V* consisting of all linear functionals with domain V and range contained in F. The dual space V*, of a space V, is the vector space Hom (V,F). Linear functionals whose domain is finite dimensional and of dimension n are represented by 1xn matrices and dual space [ Hom (V,F) ] corresponds to the set of all 1xn matrices over F. If V is of dimension n then the dual space has dimension n.

Syn. conjugate space, adjoint space

Example. Let V be column space consisting of all n-element column vectors over R. Let

T: V → F be

T(v) = [a_{1}, a_{2}, .... , a_{n}] v

where a_{1}, a_{2}, .... , a_{n} are real numbers and v is any element in V. The row vector [a_{1}, a_{2}, .... ,
a_{n}] can be viewed as a linear operator operating on vectors in V. It is a linear functional which
maps elements of V into field R. The dual space V* of V is then the vector space of all n-element row vectors. Thus row space is the dual space of column space V.

Basis for Dual Space. Suppose V is some abstract vector space of dimension n over a field
F. Suppose {v_{1}, v_{2}, .... , v_{n}} is a basis for V. Then a basis for the dual space V* of V is the set of
n linear functionals f_{1}, f_{2}, .... , f_{n}
V* defined by

f_{i}(v_{j}) = 1 if i = j

f_{i}(v_{j}) = 0 if i ≠ j

where i = 1,n; j = 1,n. This is the Kronecker delta mapping

f_{i}(v_{j}) = δ(i,j)

where δ(i,j) is the Kronecker delta.

More explicitly, it is the following n mappings:

f_{1} mapping: v_{1}
1, v_{2 }
0, v_{3}
0, ..... , v_{n}
0

f_{2} mapping: v_{1}
0, v_{2 }
1, v_{3}
0, ..... , v_{n}
0

...........................................................................

f_{n} mapping: v_{1}
0, v_{2 }
0, v_{3}
0, ..... , v_{n}
1

The basis {f_{1}, f_{2}, .... , f_{n}} is called the basis dual to {v_{1}, v_{2}, ..... , v_{n}} or the dual basis. There are
infinitely many possible bases for V and each basis has a dual basis as defined above.

See

Hom(V,W). Vector space of all mxn matrices.

Theorem 1. Let {v_{1}, v_{2}, ..... , v_{n}} be a basis for V and let {f_{1}, f_{2}, .... , f_{n}} be the basis of V*
(i.e. dual basis). Then for any vector u
V,

u = f_{1}(u)v_{1} + f_{2}(u)v_{2 + .... +} f_{n}(u)v_{n }

and for any linear functional σ V*

σ = σ(v_{1})f_{1} + σ(v_{2})f_{2 + .... + }σ(v_{n})f_{n }

Thus we see that the coordinates of u are f_{1}(u), f_{2}(u), .... , f_{n}(u) and the coordinates of σ

are σ(v_{1}), σ(v_{2}), .... ,_{ }σ(v_{n}) .

Annihilator. Let W be a subset (not necessarily a subspace) of a vector space V. A linear
functional f
V* is called an annihilator of W if f(w) = 0 for every w
W. The set of all
such mappings, denoted by W^{0} and called the annihilator of W, is a subspace of V*.

Example. Pass a line through the origin of an x-y-z Cartesian coordinate system and label it L.
Let W be the set of all vectors in line L. Pass a plane through the origin of the coordinate system
perpendicular to line L and label it K. Let S represent the set of all vectors in plane K. Then S is
an annihilator of W. Why? If s is any vector in S and w is any vector in W then the dot product
s∙w = 0. The vector s can be viewed as a linear operator (linear functional) mapping the vectors
of W into the field of reals and it maps all the elements of W into zero. By the same logic W is
an annihilator of S. So the annihilator W^{0} of W is the set S and the annihilator S^{0} of the set S is
the set W.

Theorem 2. Suppose V has finite dimension and W is a subspace of V. Then

1) dimW + dim W^{0} = dim V

and

2) W^{00} = W

Transpose of a linear mapping. Let T:V → U be an arbitrary linear mapping from a vector space V into a vector space U. Now for any linear functional ω ε U*, the composition mapping ω T is a linear mapping from V into F. See Fig. 1. Thus ω T ε V*. We thus have a one-to-one correspondence between ω ε U* and ω T ε V*. The linear mapping

T ^{t} (ω) = ω
T

that maps ω ε U* into ω T ε V* is called the transpose of T.

Thus [T ^{t} (ω)]v = ω(Tv) for every v ε V.

In summary, if T is a linear mapping from V into U, then T ^{t} is a linear mapping from U* into
V*:

Theorem 3. Let T:V → U be linear, and let A be the matrix representation of T relative to
bases {v_{i}} of V and {u_{i}} of U. Then the transpose matrix A^{t} is the matrix representation of T
^{t}:U* → V* relative to the bases dual to {u_{i}} and {v_{i}}.

References

Lipschutz. Linear Algebra. p. 249-251

Taylor. Introduction to Functional Analysis. p. 33

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