Linear Algebra - Dual of a vector space

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 <math>V^*</math> .

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

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

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

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

generators for a vector space V → Algorithm → generators for dual space <math>V^*</math>

• Linear Algebra - Dimension of a vector space
• (V*)* = V. The dual of the dual is the original space.

Let <math>a_1, \dots , a_m</math> be generators for V.

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

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 <math>\beta</math> : <math>\beta [1, 0, -1].v = \beta ([1, 0, -1].v) = 0</math>
• 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].

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.