721: For Vectors Space, Intersection of Finite-Dimensional Subspaces Is Subspace with Dimension Equal to or Smaller than Minimum Dimension of Subspaces

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description/proof of that for vectors space, intersection of finite-dimensional subspaces is subspace with dimension equal to or smaller than minimum dimension of subspaces

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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 dimension of vectors space.
• The reader admits 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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Target Context

• The reader will have a description and a proof of the proposition that for any vectors space, the intersection of any possibly uncountable number of finite-dimensional subspaces is a subspace with a dimension equal to or smaller than the minimum dimension of the subspaces.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$\{V_{\alpha }|\alpha \in A\}$: $V_{\alpha }\in \{\text{ the finite-dimensional vectors subspaces of }V\}$, $A\in \{\text{ the possibly uncountable index sets }\}$
$V^{\prime }$: $={\cap }_{\alpha \in A}{V}_{\alpha }$, $\subseteq V$
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Statements:
$V^{\prime }\in \{\text{ the subspaces of }V\}$
$\wedge$
$dim{V}^{\prime }\le min\{dim{V}_{\alpha }|\alpha \in A\}$
//

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2: Natural Language Description

For any field, $F$, any $F$ vectors space, $V$, and any set of finite-dimensional subspaces, $\{V_{\alpha }|\alpha \in A\}$, where $A$ is any possibly uncountable index set, the intersection, $V^{\prime }:={\cap }_{\alpha \in A}{V}_{\alpha }\subseteq V$, is a subspace of $V$ with a dimension equal to or less than the minimum dimension of the subspaces.

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3: Proof

Whole Strategy: Step 1: see that any linear combination of any 2 elements of $V^{\prime }$ belongs to $V^{\prime }$; Step 2: take any basis for each $V_{\alpha }$ and see that a subset of the basis is a basis of $V^{\prime }$; Step 3: see that the dimension of $V^{\prime }$ is equal to or less than the minimum of the dimensions of $V_{\alpha }$ s.

Step 1:

Let $v_1,v_2\in V^{\prime }$ be any and let $r^1,r^2\in F$ be any.

$v_1\in V_{\alpha }$ and $v_2\in V_{\alpha }$ for each $\alpha$. So, $\sum _{j\in \{1,2\}}{r}^j{v}_j\in V_{\alpha }$ for each $\alpha$. So, $\sum _{j\in \{1,2\}}{r}^j{v}_j\in V^{\prime }$. So, $V^{\prime }$ is a subspace of $V$.

Step 2:

Let a basis of $V_{\alpha }$ be $(e_{\alpha ,1},...,e_{\alpha ,d_{\alpha }})$. Any vector, $v\in V^{\prime }$, can be expressed as the linear combination of the basis, because $v$ is a vector on $V_{\alpha }$. That means that the basis spans $V^{\prime }$.

The basis can be reduced to be a basis of $V^{\prime }$, by the proposition that for any finite-dimensional vectors space, any subset that spans the space can be reduced to be a basis.

Step 3:

So, $dim{V}^{\prime }\le dim{V}_{\alpha }$ for each $\alpha \in A$.

There is the minimum of $\{dim{V}_{\alpha }|\alpha \in A\}$, because the set of the dimensions is a set of some cardinal numbers, which is a well-ordered set.

So, $min\{dim{V}_{\alpha }|\alpha \in A\}$ makes sense and $dim{V}^{\prime }\le min\{dim{V}_{\alpha }|\alpha \in A\}$.

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4: Note

The dimension does not need to equal the minimum. For example, $V={\mathbb{R}}^3$, $V_1,V_2$ are some distinct lines that go through the origin; $V_1\cap V_2=\{0\}$, which is $0$-dimensional while the minimum is $1$.

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References

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