927: For 2 Vectors Spaces That Share Operations on Intersection, Intersection Is Vectors Space

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description/proof of that for 2 vectors spaces that share operations on intersection, intersection is vectors space

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Topics

About: vectors space

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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

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Starting Context

• The reader knows a definition of %field name% vectors space.

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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.

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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.

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Main Body

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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, ${+}_2,{.}_2$
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Statements:
(
$V_1\cap V_2\ne \mathrm{\varnothing }$
$\wedge$
$\mathrm{\forall }{r}_1,r_2\in F,\mathrm{\forall }{v}_1,v_2\in V_1\cap V_2(r_1{.}_1{v}_1{+}_1{r}_2{.}_1{v}_2=r_1{.}_2{v}_1{+}_2{r}_2{.}_2{v}_2)$
)
$⟹$
$V_1\cap V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$, with the operations, $+,.$, which equals both the restriction of ${+}_1,{.}_1$ and the restriction of ${+}_2,{.}_2$.
//

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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 $V_1\cap V_2\ne \mathrm{\varnothing }$ and for each $r_1,r_2\in F$ and for each $v_1,v_2\in V_1\cap V_2$, $r_1{.}_1{v}_1{+}_1{r}_2{.}_1{v}_2=r_1{.}_2{v}_1{+}_2{r}_2{.}_2{v}_2$, $V_1\cap V_2$ is a vectors space, with the operations, $+,.$, which equals both the restriction of ${+}_1,{.}_1$ and the restriction of ${+}_2,{.}_2$.

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3: Note

The condition, $r_1{.}_1{v}_1{+}_1{r}_2{.}_1{v}_2=r_1{.}_2{v}_1{+}_2{r}_2{.}_2{v}_2$, 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, $0{.}_{1}v=0_1$ or $0{.}_{2}v=0_2$ 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 ${+}_2,{.}_2$.

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4: Proof

In fact, $r_1{.}_1{v}_1{+}_1{r}_2{.}_1{v}_2=r_1{.}_2{v}_1{+}_2{r}_2{.}_2{v}_2$ has already presupposed that $r_1{.}_1{v}_1{+}_1{r}_2{.}_1{v}_2,r_1{.}_2{v}_1{+}_2{r}_2{.}_2{v}_2\in V_1\cap V_2$, because otherwise, $r_1{.}_1{v}_1{+}_1{r}_2{.}_1{v}_2$ and $r_1{.}_2{v}_1{+}_2{r}_2{.}_2{v}_2$ 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 $v^{\prime }\in V_1\cap V_2$, $0{.}_1{v}^{\prime }=0_1\in V_1\cap V_2$, as has been shown in the 1st paragraph of Proof, and $v+0_1=v{+}_1{0}_1=v$, 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, $v^{\prime }\in V_1\cap V_2$, such that $v^{\prime }+v=0$, because $v^{\prime }:=-1{.}_{1}v\in V_1\cap V_2$, by the 1st paragraph of Proof, and $v^{\prime }+v=v^{\prime }{+}_{1}v=0_1=0$.

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 $(r_1+r_2).v=(r_1+r_2){.}_{1}v=r_1{.}_{1}v{+}_1{r}_2{.}_{1}v=r_1.v+r_2.v$.

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 $r.(v_1+v_2)=r{.}_1(v_1{+}_1{v}_2)=r{.}_1{v}_1{+}_{1}r{.}_1{v}_2=r.v_1+r.v_2$.

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 $(r_1{r}_2).v=(r_1{r}_2){.}_{1}v=r_1{.}_1(r_2{.}_{1}v)=r_1.(r_2.v)$.

10) for any element, $v\in V_1\cap V_2$, $1.v=v$, because $1.v=1{.}_{1}v=v$.

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References

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