720: For Finite-Dimensional Vectors Space, Subset That Spans Space Can Be Reduced to Be Basis

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description/proof of that for finite-dimensional vectors space, subset that spans space can be reduced to be basis

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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 basis of module.
• The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space, any subset that spans the space can be reduced to be a basis.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$S$: $\subseteq V$
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Statements:
$\mathrm{\forall }v\in V(\mathrm{\exists }{S}^{\prime }\in \{\text{ the finite subsets of }S\},\mathrm{\exists }{r}^j\in F(v=\sum _{v_j\in S^{\prime }}{r}^j{v}_j)))$
$⟹$
$\mathrm{\exists }B\in \{\text{ the }d\text{ -ordered subsets of }S\}(B\in \{\text{ the bases of }V\})$
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2: Natural Language Description

For any field, $F$, any $d$-dimensional $F$ vectors space, $V$, and any subset, $S\subseteq V$, such that $\mathrm{\forall }v\in V(\mathrm{\exists }{S}^{\prime }\in \{\text{ the finite subsets of }S\},\mathrm{\exists }{r}^j\in F(v=\sum _{v_j\in S^{\prime }}{r}^j{v}_j)))$, there is a $d$-ordered subset, $B\subseteq S$, that is a basis of $V$.

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3: Proof

Whole Strategy: Step 1: deal with the case, $V=\{0\}$, and suppose otherwise thereafter; Step 2: choose any nonzero element, $e_1\in S$; Step 3: choose any another element, $e_2\in S\setminus \{e_1\}$, such that $\{e_1,e_2\}$ is linearly independent; Step 4: and so on to get a linearly independent $\{e_1,...,e_d\}$; Step 5: see that $\{e_1,...,e_d\}$ is a basis.

Step 1:

Let us suppose that $V=\{0\}$.

Inevitably, $S=\{0\}$.

$B=\mathrm{\varnothing }\subseteq S$ will do.

Let us suppose otherwise hereafter.

Step 2:

Let us choose any nonzero element, $e_1\in S$. That is possible, because if $S=\{0\}$, $S$ could not span $V$.

Step 3:

If $2\le d$, let us choose any another element, $e_2\in S\setminus \{e_1\}$, such that $\{e_1,e_2\}$ is linearly independent. That is possible, because otherwise, for each $v\in S\setminus \{e_1\}$, $\{e_1,v\}$ would be linearly dependent, which would mean that $c^1{e}_1+c^{2}v=0$ would have not-all-zero $(c^1,c^2)$, but $c^2\ne 0$, because otherwise, $c^1{e}_1=0$, which would mean that $e_1=0$, a contradiction, so, $v=-{c^2}^{-1}{c}^1{e}_1$, which would mean that $\{e_1\}$ spanned $V$ and $\{e_1\}$ was a basis, a contradiction against $V$'s being $d$-dimensional.

Step 4:

And so on, and we get a linearly independent $\{e_1,...,e_d\}$. That is possible, because supposing that we have already chosen a linearly independent $\{e_1,...,e_j\}$ for any $j<d$, we can choose an element, $e_{j+1}\in S\setminus \{e_1,...,e_j\}$, such that $\{e_1,...,e_{j+1}\}$ is linearly independent, because otherwise, for each $v\in S\setminus \{e_1,...,e_j\}$, $\{e_1,...,e_j,v\}$ would be linearly dependent, which would mean that $c^1{e}_1+...+c^j{e}_j+c^{j+1}v=0$ would have not-all-zero $(c^1,...,c^{j+1})$, but $c^{j+1}\ne 0$, because otherwise, $c^1{e}_1+...+c^j{e}_j=0$, which would mean that $(c^1,...,c^j)=(0,...,0)$, because $\{e_1,...,e_j\}$ was linearly independent, a contradiction, so, $v=-{c^{j+1}}^{-1}(c^1{e}_1+...+c^j{e}_j)$, which would mean that $\{e_1,...,e_j\}$ spanned $V$ and $\{e_1,...,e_j\}$ was a basis, a contradiction against $V$'s being $d$-dimensional.

Step 5:

$\{e_1,...,e_d\}$ is a basis, by the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.

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References

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