676: For Finite-Dimensional Vectors Space Basis, Replacing Element by Linear Combination of Elements with Nonzero Coefficient for Element Forms Basis

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description/proof of that for finite dimensional vectors space basis, replacing element by linear combination of elements with nonzero coefficient for element forms 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.

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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 basis, replacing any element by any linear combination of the elements with any nonzero coefficient for the element forms 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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$B$: $=\{e_1,...,e_d\}$, $\in \{\text{ the bases of }V\}$
$e_k^{\prime }$: $=c^j{e}_j$, with the Einstein convention, where $c^k\ne 0$
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Statements:
$B^{\prime }:=(B\setminus \{e_k\})\cup \{e_k^{\prime }\}\in \{\text{ the bases of }V\}$
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2: Natural Language Description

For any field, $F$, any $d$-dimensional $F$ vectors space, $V$, any basis of $V$, $B=\{e_1,...,e_d\}$, and any $e_k^{\prime }:=c^j{e}_j$, with the Einstein convention, where $c^k\ne 0$, $B^{\prime }:=(B\setminus \{e_k\})\cup \{e_k^{\prime }\}$ is a basis of $V$.

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3: Proof

Whole Strategy: Step 1: show that $B^{\prime }$ is linearly independent; Step 2: show that $B^{\prime }$ spans $V$.

Step 1:

Let us prove that $B^{\prime }$ is linearly independent.

Let us suppose that $d^1{e}_1+d^2{e}_2+...+d^{k-1}{e}_{k-1}+d^k{e}_k^{\prime }+d^{k+1}{e}_{k+1}+...+d^d{e}_d=0$. $d^1{e}_1+d^2{e}_2+...+d^{k-1}{e}_{k-1}+d^k{c}^j{e}_j+d^{k+1}{e}_{k+1}+...+d^d{e}_d=0$. As the coefficient of $e_k$ is $d^k{c}^k$ and $B$ is linearly independent, $d^k{c}^k=0$, and $d^k=0$ as $c^k\ne 0$. Then, $d^1{e}_1+d^2{e}_2+...+d^{k-1}{e}_{k-1}+0+d^{k+1}{e}_{k+1}+...+d^d{e}_d=0$, and each $d^j$ is $0$ as $B$ is linearly independent.

So, $B^{\prime }$ is linearly independent.

Step 2:

Let us prove that any vector in $V$ is a linear combination of the elements of $B^{\prime }$.

$e_k^{\prime }/c_k=c^j{e}_j/c_k=e_k+\sum _{j\ne k}(c^j/c_k)e_j$, so, $e_k=e_k^{\prime }/c_k-\sum _{j\ne k}(c^j/c_k)e_j$.

For any vector, $v=d^j{e}_j$, $e_k$ can be replaced by $e_k^{\prime }/c_k-\sum _{j\ne k}(c^j/c_k)e_j$, then, it is a linear combination of $B^{\prime }$.

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References

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