117: Linear Range of Finite-Dimensional Vectors Space Is Vectors Space

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description/proof of that linear range of finite-dimensional vectors space 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: 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 finite-dimensional vectors space under any linear map is a vectors space.

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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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f$: $V_1\to V_2$, $\in \{\text{ the linear maps }\}$
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Statements:
$f(V_1)\in \{\text{ the }F\text{ vectors spaces }\}$
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2: Natural Language Description

For any field, $F$, any finite-dimensional $F$ vectors space, $V_1$, any $F$ vectors space, $V_2$, and any linear map, $f:V_1\to V_2$, the range of $f$, $f(V_1)$, is a $F$ vectors space.

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3: Proof

Whole Strategy: Step 1: take any basis for $V_1$, $\{b_1,b_2,...,b_d\}$; Step 2: take any maximal linearly independent subset of $\{f(b_1),f(b_2),...,f(b_d)\}$, $\{f(b_1),f(b_2),...,f(b_r)\}$, with the basis for $V_1$ re-indexed; Step 3: see that the space spanned by $\{f(b_1),f(b_2),...,f(b_r)\}$ is $f(V_1)$.

Step 1:

Let us take any basis for $V_1$, $\{b_1,b_2,...,b_d\}$.

Step 2:

Let us take any maximal linearly independent subset of $\{f(b_1),f(b_2),...,f(b_d)\}$.

What that means is this: if $\{f(b_1),f(b_2),...,f(b_d)\}$ is linearly independent, it is the thing; if not, an $f(b_j)$ is a linear combination of the rest, remove $f(b_j)$, and if the leftover is linearly independent, it is the thing; if not, an $f(b_k)$ is a linear combination of the rest, remove $f(b_k)$, and if the leftover is linearly independent, it is the thing; and so on. If nothing is left, $f(V_1)$ is the 0 vectors space, which is a kind of vectors space, and we are done. Let us suppose otherwise hereafter.

We have a linearly independent subset, but by re-indexing $\{b_1,b_2,...,b_d\}$, we can say that $\{f(b_1),f(b_2),...,f(b_r)\}$ is the linearly independent subset.

Step 3:

Let us see that the space spanned by $\{f(b_1),f(b_2),...,f(b_r)\}$, $V$, is $f(V_1)$.

Let $f(\sum _{j\in \{1,...,d\}}{c}^j{b}_j)\in f(V_1)$ be any. $f(\sum _{j\in \{1,...,d\}}{c}^j{b}_j)=\sum _{j\in \{1,...,d\}}{c}^{j}f(b_j)$, but as $f(b_j)$ for each $r<j$ is a linear combination of $\{f(b_1),f(b_2),...,f(b_r)\}$, it is $\sum _{j\in \{1,...,r\}}{c}^{\prime j}f(b_j)\in V$. So, $f(V_1)\subseteq V$.

Let $\sum _{j\in \{1,...,r\}}{c}^{\prime j}f(b_j)\in V$ be any. Let us take $\sum _{j\in \{1,...,r\}}{c}^{\prime j}{b}_j\in V_1$. Then, $f(\sum _{j\in \{1,...,r\}}{c}^{\prime j}{b}_j)=\sum _{j\in \{1,...,r\}}{c}^{\prime j}f(b_j)$. So, $V\subseteq f(V_1)$.

So, $f(V_1)=V$.

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References

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