description/proof of that for 2 vectors spaces that share operations on intersection, intersection is vectors space
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note
- 4: Proof
Starting Context
- The reader knows a definition of %field name% vectors space.
Target Context
- The reader will have a description and a proof of the proposition that for any 2 vectors spaces that share the operations on the intersection, the intersection is a vectors space with the restriction of the shared operations.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with the operations, \(+_1, ._1\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with the operations, \(+_2, ._2\)
//
Statements:
(
\(V_1 \cap V_2 \neq \emptyset\)
\(\land\)
\(\forall r_1, r_2 \in F, \forall v_1, v_2 \in V_1 \cap V_2 (r_1 \text{ } ._{1} \text{ } v_1 \text{ } +_{1} \text{ } r_2 \text{ } ._{1} \text{ } v_2 = r_1 \text{ }._{2} \text{ } v_1 \text{ }+_{2} \text{ } r_2 \text{ }._{2} \text{ } v_2)\)
)
\(\implies\)
\(V_1 \cap V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with the operations, \(+, .\), which equals both the restriction of \(+_1, ._1\) and the restriction of \(+_2, ._2\).
//
2: Natural Language Description
For any field, \(F\), any \(F\) vectors space, \(V_1\), with the operations, \(+_1, ._1\), and any \(F\) vectors space, \(V_2\), with the operations, \(+_2, ._2\), if \(V_1 \cap V_2 \neq \emptyset\) and for each \(r_1, r_2 \in F\) and for each \(v_1, v_2 \in V_1 \cap V_2\), \(r_1 \text{ } ._{1} \text{ } v_1 \text{ } +_{1} \text{ } r_2 \text{ } ._{1} \text{ } v_2 = r_1 \text{ }._{2} \text{ } v_1 \text{ }+_{2} \text{ } r_2 \text{ }._{2} \text{ } v_2\), \(V_1 \cap V_2\) is a vectors space, with the operations, \(+, .\), which equals both the restriction of \(+_1, ._1\) and the restriction of \(+_2, ._2\).
3: Note
The condition, \(r_1 \text{ } ._{1} \text{ } v_1 \text{ } +_{1} \text{ } r_2 \text{ } ._{1} \text{ } v_2 = r_1 \text{ }._{2} \text{ } v_1 \text{ }+_{2} \text{ } r_2 \text{ }._{2} \text{ } v_2\), is crucial for this proposition.
As a counterexample, let us think of some disjoint \(V_1\) and \(V_2\) and replace some 2 points of \(V_2\) with some 2 points of \(V_1\), and change \(+_2, ._2\) to accommodate the replacements to preserve the vectors space structure (the new points are treated as the old points were). Then, \(V_1 \cap V_2\) consists of only those 2 points, which is not any vectors space.
As another counterexample, let us think of 2 \(\mathbb{R}^2\) Euclidean vectors spaces intersecting with a line without sharing the origins. Then, \(V_1 \cap V_2\), which is a line, is not any vectors space with the restriction of \(+_1, ._1\) or with the restriction of \(+_2, ._2\): for example, \(0 \text{ }._1\text{ } v = 0_1\) or \(0 \text{ }._2\text{ } v = 0_2\) is not contained in \(V_1 \cap V_2\). Certainly, an operations pair can be invented to make \(V_1 \cap V_2\) a vectors space, but this proposition is concerned with having the operations as the restriction of \(+_1, ._1\) or \(+_2, ._2\).
4: Proof
In fact, \(r_1 \text{ }._{1}\text{ } v_1 \text{ }+_{1}\text{ } r_2 \text{ }._{1}\text{ } v_2 = r_1 \text{ }._{2}\text{ } v_1 \text{ }+_{2}\text{ } r_2 \text{ }._{2}\text{ } v_2\) has already presupposed that \(r_1 \text{ }._{1}\text{ } v_1 \text{ }+_{1}\text{ } r_2 \text{ }._{1}\text{ } v_2, r_1 \text{ }._{2}\text{ } v_1 \text{ }+_{2}\text{ } r_2 \text{ }._{2}\text{ } v_2 \in V_1 \cap V_2\), because otherwise, \(r_1 \text{ }._{1}\text{ } v_1 \text{ }+_{1}\text{ } r_2 \text{ }._{1}\text{ } v_2\) and \(r_1 \text{ }._{2}\text{ } v_1 \text{ }+_{2}\text{ } r_2 \text{ }._{2}\text{ } v_2\) could not be even compared.
As \(+_1, ._1\) and \(+_2, ._2\) are the same on \(V_1 \cap V_2\), let the same operations on \(V_1 \cap V_2\) be denoted as \(+, .\).
Let us prove that \(V_1 \cap V_2\) with the operations, \(+, .\), satisfies the conditions for being an \(F\) vectors space.
1) for any elements, \(v_1, v_2 \in V_1 \cap V_2\), \(v_1 + v_2 \in V_1 \cap V_2\), because that is what has been shown in the 1st paragraph of Proof.
2) for any elements, \(v_1, v_2 \in V_1 \cap V_2\), \(v_1 + v_2 = v_2 + v_1\), because \(v_1 + v_2 = v_1 +_1 v_2 = v_2 +_1 v_1 = v_2 + v_1\).
3) for any elements, \(v_1, v_2, v_3 \in V_1 \cap V_2\), \((v_1 + v_2) + v_3 = v_1 + (v_2 + v_3)\), because \((v_1 + v_2) + v_3 = (v_1 +_1 v_2) +_1 v_3 = v_1 +_1 (v_2 +_1 v_3) = v_1 + (v_2 + v_3)\).
4) there is a 0 element, \(0 \in V_1 \cap V_2\), such that for any \(v \in V_1 \cap V_2\), \(v + 0 = v\), because while there is a \(v' \in V_1 \cap V_2\), \(0 \text{ }._1\text{ } v' = 0_1 \in V_1 \cap V_2\), as has been shown in the 1st paragraph of Proof, and \(v + 0_1 = v \text{ }+_1\text{ } 0_1 = v\), and we will denote \(0_1\) as \(0\) hereafter.
5) for any element, \(v \in V_1 \cap V_2\), there is an inverse element, \(v' \in V_1 \cap V_2\), such that \(v' + v = 0\), because \(v' := -1 \text{ }._1\text{ } v \in V_1 \cap V_2\), by the 1st paragraph of Proof, and \(v' + v = v' \text{ }+_1\text{ } v = 0_1 = 0\).
6) for any element, \(v \in V_1 \cap V_2\), and any scalar, \(r \in F\), \(r . v \in V_1 \cap V_2\), because that is what has been shown in the 1st paragraph of Proof.
7) for any element, \(v \in V_1 \cap V_2\), and any scalars, \(r_1, r_2 \in F\), \((r_1 + r_2) . v = r_1 . v + r_2 . v\), because \((r_1 + r_2) . v = (r_1 + r_2) \text{ }._1\text{ } v = r_1 \text{ }._1\text{ } v +_1 r_2 \text{ }._1\text{ } v = r_1 . v + r_2 . v\).
8) for any elements, \(v_1, v_2 \in V_1 \cap V_2\), and any scalar, \(r \in F\), \(r . (v_1 + v_2) = r . v_1 + r . v_2\), because \(r . (v_1 + v_2) = r \text{ }._1\text{ } (v_1 +_1 v_2) = r \text{ }._1\text{ } v_1 +_1 r \text{ }._1\text{ } v_2 = r . v_1 + r . v_2\).
9) for any element, \(v \in V_1 \cap V_2\), and any scalars, \(r_1, r_2 \in F\), \((r_1 r_2) . v = r_1 . (r_2 . v)\), because \((r_1 r_2) . v = (r_1 r_2) \text{ }._1\text{ } v = r_1 \text{ }._1\text{ } (r_2 \text{ }._1\text{ } v) = r_1 . (r_2 . v)\).
10) for any element, \(v \in V_1 \cap V_2\), \(1 . v = v\), because \(1 . v = 1 \text{ }._1\text{ } v = v\).
