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The annihilator of the coordinate plane $z = 0$ in $R^{3}$ is the line spanned by the third coordinate functional

Example

For $U = {(x, y, 0) : x, y \in R} \leq R^{3}$ ,

$U^{\circ} = span {z^{*}}, z^{*} (x, y, z) = z .$

Facts & Assumptions

Given: The coordinate plane $U$ and coordinate functionals $x^{*}, y^{*}, z^{*}$ .

[L1]

The annihilator $U^{\circ}$ consists of the functionals that vanish on every vector of $U$ (The annihilator $U^{\circ} \leq V^{*}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{*}$ ).

[L2]

In finite dimension, $\dim U^{\circ} = \dim V - \dim U$ (Assuming choice, $^{\circ} (U^{\circ}) = U$ ; in finite dimension, $\dim U^{\circ} = \dim V - \dim U$ ).

[L3]

The three standard coordinate vectors form a basis of $R^{3}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

Verification

technique · direct coordinate computation

1.1

By [L3], a general functional is $f = a x^{*} + b y^{*} + c z^{*}$ . If $f \in U^{\circ}$ , evaluation at $(1, 0, 0)$ and $(0, 1, 0)$ gives $a = b = 0$ .

L1L3algebra

1.2

Conversely every $c z^{*}$ vanishes on $(x, y, 0)$ , so [L1] gives $U^{\circ} = span {z^{*}}$ .

L1algebra

2.1

As a check, $\dim U = 2$ and [L2] gives $\dim U^{\circ} = 3 - 2 = 1$ , agreeing with step 1.2.

step 1.2L2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 73 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

H. Pinkham, Linear Algebra, Chapter 6 (standard reference, not scraped)

The annihilator of the coordinate plane z=0z=0 in R3R3 is the line spanned by the third coordinate functional