Worksheet: dual basis

Worksheet 3.4 Worksheet: dual basis
A4 US

Let

V

be a vector space over

R .

(That is, scalars are real numbers, rather than, say, complex.) A linear transformation

ϕ : V \to R

is called a linear functional.

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Here are some examples of linear functionals:

The map $ϕ : R^{3} \to R$ given by $ϕ (x, y, z) = 3 x - 2 y + 5 z .$
The evaluation map $e v_{a} : P_{n} (R) \to R$ given by $e v_{a} (p) = p (a) .$ (For example, $e v_{2} (3 - 4 x + 5 x^{2}) = 2 - 4 (2) + 5 (2^{2}) = 14 .$ )
The map $ϕ : C [a, b] \to R$ given by $ϕ (f) = \int_{a}^{b} f (x) d x,$ where $C [a, b]$ denotes the space of all continuous functions on $[a, b] .$

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Note that for any vector spaces

V, W,

the set

L (V, W)

of linear transformations from

V

W

is itself a vector space, if we define

(S + T) (v) = S (v) + T (v), and (k T) (v) = k (T (v)) .

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In particular, given a vector space

V,

we denote the set of all linear functionals on

V

V^{*} = L (V, R),

and call this the dual space of

V .

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We make the following observations:

If $\dim V = n$ and $\dim W = m,$ then $L (V, W)$ is isomorphic to the space $M_{m n}$ of $m \times n$ matrices, so it has dimension $m n .$
Since $\dim R = 1,$ if $V$ is finite-dimensional, then $V^{*} = L (V, R)$ has dimension $1 n = n .$
Since $\dim V^{*} = \dim V,$ $V$ and $V^{*}$ are isomorphic.

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Here is a basic example that is intended as a guide to your intuition regarding dual spaces. Take

V = R^{3} .

Given any

v \in V,

define a map

ϕ_{v} : V \to R

ϕ_{v} (w) = v \cdot w

(the usual dot product).

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One way to think about this: if we write

v \in V

as a column vector

[\begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix}],

then we can identify

ϕ_{v}

with

v^{T},

where the action is via multiplication:

ϕ_{v} (w) = [\begin{matrix} v_{1} & v_{2} & v_{3} \end{matrix}] [\begin{matrix} w_{1} \\ w_{2} \\ w_{3} \end{matrix}] = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} .

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It turns out that this example can be generalized, but the definition of

ϕ_{v}

involves the dot product, which is particular to

R^{n} .

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There is a generalization of the dot product, known as an inner product. (See Chapter 10 of Nicholson, for example.) On any inner product space, we can associate each vector

v \in V

to a linear functional

ϕ_{v}

using the procedure above.

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Another way to work concretely with dual vectors (without the need for inner products) is to define things in terms of a basis.

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Given a basis

{v_{1}, v_{2}, \dots, v_{n}}

V,

we define the corresponding dual basis

{ϕ_{1}, ϕ_{2}, \dots, ϕ_{n}}

V^{*}

ϕ_{i} (v_{j}) = {\begin{cases} 1, & if i = j \\ 0, & if i \neq j \end{cases} .

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Note that each

ϕ_{j}

is well-defined, since any linear transformation can be defined by giving its values on a basis.

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For the standard basis on

R^{n},

note that the corresponding dual basis functionals are given by

ϕ_{j} (x_{1}, x_{2}, \dots, x_{n}) = x_{j} .

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That is, these are the coordinate functions on

R^{n} .

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1.

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Show that the dual basis is indeed a basis for

V^{*} .

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Next, let

V

and

W

be vector spaces, and let

T : V \to W

be a linear transformation. For any such

T,

we can define the dual map

T^{*} : W^{*} \to V^{*}

T^{*} (ϕ) = ϕ \circ T

for each

ϕ \in W^{*} .

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2.

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Confirm that (a)

T^{*} (ϕ)

does indeed define an element of

V^{*};

that is, a linear map from

V

R,

and (b) that

T^{*}

is linear.

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3.

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Let

V = P (R)

be the space of all polynomials, and let

D : V \to V

be the derivative transformation

D (p (x)) = p^{'} (x) .

Let

ϕ : V \to R

be the linear functional defined by

ϕ (p (x)) = \int_{0}^{1} p (x) d x .

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What is the linear functional

D^{*} (ϕ) ?

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4.

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Show that dual maps satisfy the following properties: for any

S, T \in L (V, W)

and

k \in R,

$(S + T)^{*} = S^{*} + T^{*}$
$(k S)^{*} = k S^{*}$
$(S T)^{*} = T^{*} S^{*}$

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In item Item 3.4.4.c, assume

S \in L (V, W)

and

T \in L (U, V) .

(Reminder: the notation

S T

is sometimes referred to as the “product” of

S

and

T,

in analogy with matrices, but actually represents the composition

S \circ T .

)

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We have one topic remaining in relation to dual spaces: determining the kernel and image of a dual map

T^{*}

(in terms of the kernel and image of

T

). Let

V

be a vector space, and let

U

be a subspace of

V .

Any such subspace determines an important subspace of

V^{*} :

the annihilator of

U,

denoted by

U^{0}

and defined by

U^{0} = {ϕ \in V^{*} | ϕ (u) = 0 for all u \in U} .

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5.

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Determine a basis (in terms of the standard dual basis for

(R^{4})^{*}

) for the annihilator

U^{0}

of the subspace

U \subseteq R^{4}

given by

U = {(2 a + b, 3 b, a, a - 2 b) | a, b \in R} .

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Here is a fun theorem about annihilators that I won’t ask you to prove.

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Theorem 3.4.1.

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Let

V

be a finite dimensional vector space. For any subspace

U

V,

\dim U + \dim U^{0} = \dim V .

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Here’s an outline of the proof. For any subspace

U \subseteq V,

we can define the inclusion map

i : U \to V,

given by

i (u) = u .

(This is not the identity on

V

since it’s only defined on

U .

In particular, it is not onto unless

U = V,

although it is clearly one-to-one.)

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Then

i^{*}

is a map from

V^{*}

U^{*} .

Moreover, note that for any

ϕ \in V^{*},

i^{*} (ϕ) \in U^{*}

satisfies, for any

u \in U,

i^{*} (ϕ) (u) = ϕ (i (u)) = ϕ (u) .

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Thus,

ϕ \in \ker i^{*}

if and only if

i^{*} (ϕ) = 0,

which is if and only if

ϕ (u) = 0

for all

u \in U,

which is if and only if

ϕ \in U^{0} .

Therefore,

\ker i^{*} = U^{0} .

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By the dimension theorem, we have:

\dim V^{*} = \dim \ker i^{*} + \dim im i^{*} .

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With a bit of work, one can show that

im i^{*} = U^{*},

and we get the result from the fact that

\dim V^{*} = \dim V

and

\dim U^{*} = \dim U .

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There are a number of interesting results of this flavour. For example, one can show that a map

T

is injective if and only if

T^{*}

is surjective, and vice-versa.

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One final, optional task: return to the example of

R^{n},

viewed as column vectors, and consider a matrix transformation

T_{A} : R^{n} \to R^{m}

given by

T_{A} (\vec{x}) = A \vec{x}

as usual. Viewing

(R^{n})^{*}

as row vectors, convince yourself that

(T_{A})^{*} = T_{A^{T}};

that is, that what we’ve really been talking about all along is just the transpose of a matrix!

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Worksheet 3.4 Worksheet: dual basis A4US

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Theorem 3.4.1.

Worksheet 3.4 Worksheet: dual basis
A4 US