678: For Finite-Dimensional Vectors Space, There Is No Basis That Has More Than Dimension Elements

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description/proof of that for finite-dimensional vectors space, there is no basis that has more than dimension elements

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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
• 4: Note

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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 basis, replacing an element by any linear combination of the elements with any nonzero coefficient for the element forms 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, there is no basis that has more than the dimension number of elements.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
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Statements:
$\mathrm{\neg }\mathrm{\exists }B\in \{\text{ the bases of }V\}(d<|B|)$, where $|B|$ denotes the cardinality of $B$
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2: Natural Language Description

For any field, $F$, and any $d$-dimensional $F$ vectors space, $V$, there is no basis of $V$, $B$, whose cardinality is larger than $d$.

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3: Proof

Whole Strategy: Step 1: take a $d$-cardinality basis; Step 2: suppose any other basis with any cardinality larger than $d$; Step 3: find a contradiction: the another basis was not indeed any basis.

Step 1:

Let us remember that the dimension is the minimum cardinality of the bases of the space. So, there is a basis, $B=\{e_1,e_2,...,e_d\}$.

Step 2:

Let us suppose that a basis was a $B^{\prime }=\{e_{\alpha }^{\prime }|\alpha \in A\}$, where $A$ was a possibly uncountable index set whose cardinality was larger than $d$.

Step 3:

Step 3 Strategy: modify $B$ by replacing each element of $B$ by an element of $B^{\prime }$ to be still a basis, then the new basis would be a proper subset of $B^{\prime }$, which could not be any basis.

Each $e_{\alpha }^{\prime }$ would be a linear combination of $e_j$ s, and there would be an $e_{\alpha }^{\prime }$ whose coefficient for $e_1$ was not $0$, because otherwise, $V$ would be spanned by only $\{e_2,e_3,...,e_d\}$, a contradiction against $\{e_1,e_2,...,e_d\}$'s being a basis. So, $e_1$ could be replaced by the $e_{\alpha }^{\prime }$ to form a basis, by the proposition that for any finite dimensional vectors space basis, replacing an element by any linear combination of the elements with any nonzero coefficient for the element forms a basis. Let us denote the $e_{\alpha }^{\prime }$ as $e_1^{″}$.

Let us suppose that $e_1,...,e_k$ had been already replaced by $e_1^{″},...,e_k^{″}$ to form a basis. Each $e_{\alpha }^{\prime }$ would be a linear combination of the elements of the new basis, and there would be an $e_{\alpha }^{\prime }$ whose coefficient for $e_{k+1}$ was not $0$, because otherwise, $V$ would be spanned by only $(e_1^{″},...,e_k^{″},e_{k+2},...,e_d)$, a contradiction against $(e_1^{″},...,e_k^{″},e_{k+1},...,e_d)$'s being a basis. In fact, $e_{\alpha }^{\prime }$ was not any of $e_j^{″}$ s, because otherwise, $e_{\alpha }^{\prime }=e_j^{″}=c^1{e}_1^{″}+...+c^k{e}_k^{″}+c^{k+1}{e}_{k+1}+...+d^d{e}_d$, and $c^1{e}_1^{″}+...+(c^j-1)e_j^{″}+...+c^k{e}_k^{″}+c^{k+1}{e}_{k+1}+...+c^d{e}_d=0$, where $c^{k+1}\ne 0$, a contradiction against the elements of the new basis being linearly independent. So, $e_{k+1}$ could be replaced by the $e_{\alpha }^{\prime }$ as $e_{k+1}^{″}$ to form a basis, by the proposition that for any finite dimensional vectors space basis, replacing an element by any linear combination of the elements with any nonzero coefficient for the element forms a basis.

After all, $e_1,...,e_d$ had been replaced by $e_1^{″},...,e_d^{″}$ to form a basis.

There would be an $e_{\alpha }^{\prime }$ that was not chosen to be an $e_j^{″}$, and $e_{\alpha }^{\prime }$ would be a linear combination of $e_1^{″},...,e_d^{″}$, a contradiction against $B^{\prime }$'s being a basis.

So, the number of the elements of $B^{\prime }$ is equal to or smaller than $d$.

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4: Note

In fact, the number of the elements of $B^{\prime }$ cannot be smaller than $d$ by the definition of dimension of vectors space, so, the number of the elements of any basis of any finite-dimensional vectors space is guaranteed to be the dimension.

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References

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