934: Range of Linear Map Between Vectors Spaces Is Sub-'Vectors Space' of Codomain

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description/proof of that range of linear map between vectors spaces is sub-'vectors space' of codomain

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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.
• The reader knows a definition of linear map.

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Target Context

• The reader will have a description and a proof of the proposition that the range of any linear map between any vectors spaces is a sub-'vectors space' of the codomain.

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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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the linear maps }\}$


Statements:
$f(V_1)\in \{\text{ the sub-'vectors space's of }{V}_2\}$
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2: Proof

Whole Strategy: Step 1: see that $f(V_1)$ satisfies the requirements to be a vectors space.

Step 1:

1) for any elements, $f(v),f(v^{\prime })\in f(V_1)$, $f(v)+f(v^{\prime })\in f(V_1)$ (closed-ness under addition): $f(v)+f(v^{\prime })=f(v+v^{\prime })\in f(V_1)$.

2) for any elements, $f(v),f(v^{\prime })\in f(V_1)$, $f(v)+f(v^{\prime })=f(v^{\prime })+f(v)$ (commutativity of addition): it holds on ambient $V_2$.

3) for any elements, $f(v),f(v^{\prime }),f(v^{″})\in f(V_1)$, $(f(v)+f(v^{\prime }))+f(v^{″})=f(v)+(f(v^{\prime })+f(v^{″}))$ (associativity of additions): it holds on ambient $V_2$.

4) there is a 0 element, $0\in f(V_1)$, such that for any $f(v)\in f(V_1)$, $f(v)+0=f(v)$ (existence of 0 vector): $f(0)=0\in f(V_1)$ is the one.

5) for any element, $f(v)\in f(V_1)$, there is an inverse element, $f(v^{\prime })\in f(V_1)$, such that $f(v^{\prime })+f(v)=0$ (existence of inverse vector): $f(-v)=-f(v)\in f(V_1)$ is the one.

6) for any element, $f(v)\in f(V_1)$, and any scalar, $r\in F$, $r.f(v)\in f(V_1)$ (closed-ness under scalar multiplication): $f(rv)=r.f(v)\in f(V_1)$.

7) for any element, $f(v)\in f(V_1)$, and any scalars, $r_1,r_2\in F$, $(r_1+r_2).f(v)=r_1.f(v)+r_2.f(v)$ (scalar multiplication distributability for scalars addition): it holds on ambient $V_2$.

8) for any elements, $f(v),f(v^{\prime })\in f(V_1)$, and any scalar, $r\in F$, $r.(f(v)+f(v^{\prime }))=r.f(v)+r.f(v^{\prime })$ (scalar multiplication distributability for vectors addition): it holds on ambient $V_2$.

9) for any element, $f(v)\in f(V_1)$, and any scalars, $r_1,r_2\in F$, $(r_1{r}_2).f(v)=r_1.(r_2.f(v))$ (associativity of scalar multiplications): it holds on ambient $V_2$.

10) for any element, $f(v)\in f(V_1)$, $1.f(v)=f(v)$ (identity of 1 multiplication): it holds on ambient $V_2$.

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References

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