Linear Algebra - Dual of a vector space

Dual Definition

The set of vectors u such that u · v = 0 for every vector v in V is called the dual of V. Dual is written as $V^*$ .

Definition: For a subspace V of $\mathbb{F}^n$ , the dual space of V, written $V^*$ , is: $$V^* = \{ u \in \mathbb{F}^n : u.v = 0 \text{ for every vector v} \in V\}\$$

The dual of Span {a1, . . . , am} is the solution set for a1 · x = 0, . . . , am · x = 0

Let $a_1, \dots , a_m$ be a basis for a vector space V. Let $b1, . . . , bk$ be a basis for the dual V* of the vector space V.

• Then $b_1.v = 0$ for every vector v in V
• Then $b_1.a_1 = 0, \dots, b_1.a_m = 0$
• Then $b_i.a_1 = 0, \dots, b_i.a_m = 0$ for i = 1,…, k

Computation

generators for a vector space V → Algorithm → generators for dual space $V^*$

Theorem

• (V*)* = V. The dual of the dual is the original space.

Matrix

Let $a_1, \dots , a_m$ be generators for V.

Let $A = \begin{bmatrix} \begin{array}{r} a_1 \\ \hline \\ \vdots \\ \hline \\ a_m \end{array} \end{bmatrix}$ then $V^* = \href{matrix#null space}{Null} A$

Example

over R

Let V = Span {[1, 0, 1], [0, 1, 0]}. Then V* = Span {[1, 0,−1]}:

• [1, 0,−1] · [1, 0, 1] = 0 and [1, 0,−1] · [0, 1, 0] = 0 therefore [1, 0,−1] · v = 0 for every vector v in Span {[1, 0, 1], [0, 1, 0]}.
• For any scalar $\beta$ : $\beta [1, 0, -1].v = \beta ([1, 0, -1].v) = 0$
• Which vectors u satisfy u · v = 0 for every vector v in Span {[1, 0, 1], [0, 1, 0]}? Only scalar multiples of [1, 0,−1].

over Gf2

Let V = Span {[1, 0, 1], [0, 1, 0]}. Then V* = Span {[1, 0, 1]}:

• [1, 0, 1] · [1, 0, 1] = 0 and [1, 0, 1] · [0, 1, 0] = 0.
• Therefore [1, 0, 1] · v = 0 for every vector v in Span {[1, 0, 1], [0, 1, 0]}.
• Of course [0, 0, 0] · v = 0 for every vector v in Span {[1, 0, 1], [0, 1, 0]}.
• [1, 0, 1] and [0, 0, 0] are the only such vectors.