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
Topics
About: vectors space
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
Starting Context
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.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(\{V_\alpha \vert \alpha \in A\}\): \(V_\alpha \in \{\text{ the finite-dimensional vectors subspaces of } V\}\), \(A \in \{\text{ the possibly uncountable index sets }\}\)
\(V'\): \(= \cap_{\alpha \in A} V_\alpha\), \(\subseteq V\)
//
Statements:
\(V' \in \{\text{ the subspaces of } V\}\)
\(\land\)
\(dim V' \le min \{dim V_\alpha \vert \alpha \in A\}\)
//
2: Natural Language Description
For any field, \(F\), any \(F\) vectors space, \(V\), and any set of finite-dimensional subspaces, \(\{V_\alpha \vert \alpha \in A\}\), where \(A\) is any possibly uncountable index set, the intersection, \(V' := \cap_{\alpha \in A} V_\alpha \subseteq V\), is a subspace of \(V\) with a dimension equal to or less than the minimum dimension of the subspaces.
3: Proof
Whole Strategy: Step 1: see that any linear combination of any 2 elements of \(V'\) belongs to \(V'\); Step 2: take any basis for each \(V_\alpha\) and see that a subset of the basis is a basis of \(V'\); Step 3: see that the dimension of \(V'\) is equal to or less than the minimum of the dimensions of \(V_\alpha\) s.
Step 1:
Let \(v_1, v_2 \in V'\) 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, \(\sum_{j \in \{1, 2\}} r^j v_j \in V'\). So, \(V'\) is a subspace of \(V\).
Step 2:
Let a basis of \(V_\alpha\) be \((e_{\alpha, 1}, ..., e_{\alpha, d_\alpha})\). Any vector, \(v \in V'\), can be expressed as the linear combination of the basis, because \(v\) is a vector on \(V_\alpha\). That means that the basis spans \(V'\).
The basis can be reduced to be a basis of \(V'\), 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, \(dim V' \le dim V_\alpha\) for each \(\alpha \in A\).
There is the minimum of \(\{dim V_\alpha \vert \alpha \in A\}\), because the set of the dimensions is a set of some cardinal numbers, which is a well-ordered set.
So, \(min \{dim V_\alpha \vert \alpha \in A\}\) makes sense and \(dim V' \le min \{dim V_\alpha \vert \alpha \in A\}\).
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 \(1\).
