680: For Finite-Dimensional Vectors Space, Proper Subspace Has Lower Dimension

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description/proof of that for finite-dimensional vectors space, proper subspace has lower dimension

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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 dimension of vectors space.
• The reader admits the proposition that for any finite-dimensional vectors space, there is no linearly independent subset that has more than the dimension number of elements.

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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 proper subspace has a lower dimension.

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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:
$F$: $\in \{\text{ the fields }\}$
$V^{\prime }$: $\in \{\text{ the }{d}^{\prime }\text{ -dimensional }F\text{ vectors spaces }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional proper vectors subspaces of }{V}^{\prime }\}$
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Statements:
$d<d^{\prime }$
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2: Natural Language Description

For any field, $F$, any $d^{\prime }$-dimensional $F$ vectors space, $V^{\prime }$, and any $d$-dimensional proper vectors subspace, $V\subseteq V^{\prime }$, $d<d^{\prime }$.

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3: Proof

Whole Strategy: Step 1: suppose that $d^{\prime }\le d$; Step 2: find a contradiction: find a set of more than $d^{\prime }$ linearly independent elements on $V^{\prime }$, which would contradict the proposition that for any finite-dimensional vectors space, there is no linearly independent subset that has more than the dimension number of elements.

Step 1:

Let us suppose that $d^{\prime }\le d$.

Step 2:

Step 2 Strategy: Step 2-1: take a $d$-cardinality basis of $V$; Step 2-2: take any element of $V^{\prime }\setminus V$; Step 2-3: show that the basis with the element added would be linearly independent on $V^{\prime }$.

Step 2-1:

$V$ would have a basis, $\{e_1,...,e_d\}$.

Step 2-2:

There would be a $v\in V^{\prime }\setminus V$.

Step 2-3:

Let us prove that $\{e_1,...,e_d,v\}$ would be linearly independent on $V^{\prime }$.

Let us suppose otherwise.

$c^1{e}_1+...+c^d{e}_d+cv=0$ would have a nonzero solution for $(c^1,...,c^d,c)$. In fact, $c\ne 0$, because otherwise, $c^1{e}_1+...+c^d{e}_d=0$, which would imply $c^j=0$, because the basis was linearly independent. So, $v=-c^{-1}(c^1{e}_1+...+c^d{e}_d)$, a contradiction against $v\notin V$.

So, $\{e_1,...,e_d,v\}$ would be linearly independent on $V^{\prime }$

So, $V^{\prime }$ would have a set of $d+1$ linearly independent vectors, but $d^{\prime }<d+1$, a contradiction against the proposition that for any finite-dimensional vectors space, there is no linearly independent subset that has more than the dimension number of elements.

So, $d<d^{\prime }$.

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References

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