1098: For Vectors Space, Nonempty Subset Is Vectors Subspace iff Subset Is Closed Under Linear Combination

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description/proof of that for vectors space, nonempty subset is vectors subspace iff subset is closed under linear combination

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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: 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 vectors space, any nonempty subset of the vectors space is a vectors subspace if and only if the subset is closed under linear combination.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$S$: $\subseteq V$, $\ne \mathrm{\varnothing }$
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Statements:
$S\in \{\text{ the vectors subspaces of }V\}$
$⟺$
$\mathrm{\forall }{s}_1,s_2\in S,\mathrm{\forall }{r}_1,r_2\in F(r_1{s}_1+r_2{s}_2\in S)$
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2: Proof

Whole Strategy: Step 1: suppose that $r_1{s}_1+r_2{s}_2\in S$; Step 2: see that $S$ is a vectors subspace of $V$; Step 3: suppose that $S$ is a vectors subspace of $V$; Step 4: see that $r_1{s}_1+r_2{s}_2\in S$.

Step 1:

Let us suppose that $\mathrm{\forall }{s}_1,s_2\in S,\mathrm{\forall }{r}_1,r_2\in F(r_1{s}_1+r_2{s}_2\in S)$.

Let us see that $S$ satisfies the conditions for being a vectors space.

1) for any elements, $s_1,s_2\in S$, $s_1+s_2\in S$ (closed-ness under addition): $s_1+s_2=1s_1+1s_2\in S$.

2) for any elements, $s_1,s_2\in S$, $s_1+s_2=s_2+s_1$ (commutativity of addition): it holds because it holds on the ambient $V$.

3) for any elements, $s_1,s_2,s_3\in S$, $(s_1+s_2)+s_3=s_1+(s_2+s_3)$ (associativity of additions): it holds because it holds on the ambient $V$.

4) there is a 0 element, $0\in S$, such that for any $s\in S$, $s+0=s$ (existence of 0 vector): there is an $s\in S$, because $S\ne \mathrm{\varnothing }$, and $0=0s+0s\in S$, and $s+0=s$, because it holds on the ambient $V$.

5) for any element, $s\in S$, there is an inverse element, $s^{\prime }\in S$, such that $s^{\prime }+s=0$ (existence of inverse vector): $-1s+0s=-1s\in S$, and $-1s+s=0$, because it holds on the ambient $V$.

6) for any element, $s\in S$, and any scalar, $r\in F$, $r.s\in S$ (closed-ness under scalar multiplication): $r.s=r.s+0.s\in S$.

7) for any element, $s\in S$, and any scalars, $r_1,r_2\in F$, $(r_1+r_2).s=r_1.s+r_2.s$ (scalar multiplication distributability for scalars addition): it holds because it holds on the ambient $V$.

8) for any elements, $s_1,s_2\in S$, and any scalar, $r\in F$, $r.(s_1+s_2)=r.s_1+r.s_2$ (scalar multiplication distributability for vectors addition): it holds because it holds on the ambient $V$.

9) for any element, $s\in S$, and any scalars, $r_1,r_2\in F$, $(r_1{r}_2).s=r_1.(r_2.s)$ (associativity of scalar multiplications): it holds because it holds on the ambient $V$.

10) for any element, $s\in S$, $1.s=s$ (identity of 1 multiplication): it holds because it holds on the ambient $V$.

Step 3:

Let us suppose that $S$ is a vectors subspace of $V$.

Step 4:

$\mathrm{\forall }{s}_1,s_2\in S,\mathrm{\forall }{r}_1,r_2\in F(r_1{s}_1+r_2{s}_2\in S)$ holds, because the conditions for being a vectors space requires that.

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References

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