description/proof of that for finite-dimensional vectors space and finite number of vectors subspaces, if intersection of each subspace and sum of rest is \(\{0\}\) and sum of dimensions of subspaces is dimension of space, space is space as direct sum of subspaces
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of module as direct sum of finite number of submodules.
- The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space and any finite number of vectors subspaces, if the intersection of each subspace and the sum of the rest is \(\{0\}\) and the sum of the dimensions of the subspaces is the dimension of the space, the space is a space as the direct sum 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(d\): \(\in \mathbb{N}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(\{V_1, ..., V_n\}\): \(V_j \in \{\text{ the vectors subspaces of } V\}\)
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Statements:
(
\(\forall V_j \in \{V_1, ..., V_n\} (V_j \cap (V_1 + ... + V_{j - 1} + \widehat{V_j} + V_{j + 1} + ... + V_n) = \{0\})\)
\(\land\)
\(Dim (V) = Dim (V_1) + ... + Dim (V_n)\)
)
\(\implies\)
\(V\) is a vectors space as the direct sum of \(\{V_1, ..., V_n\}\)
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2: Proof
Whole Strategy: Step 1: take any basis for each \(V_j\), \(B_j = \{b_{j, 1}, ..., b_{j, d_j}\}\); Step 2: see that \(B := B_1 \cup ... \cup B_n\) is a basis for \(V\); Step 3: conclude the proposition.
Step 1:
Let us take any basis for each \(V_j\), \(B_j = \{b_{j, 1}, ..., b_{j, d_j}\}\).
Step 2:
Let us take \(B := B_1 \cup ... \cup B_n\).
Let us see that \(B\) is a basis for \(V\).
\(\{B_1, ..., B_n\}\) has no duplication, because if \(b_{j, l} = b_{p, q}\) for some \(j \neq p\), \(b_{j, l} = b_{p, q} \in V_j \cap V_p\), a contradiction against \((V_j \cap (V_1 + ... + V_{j - 1} + \widehat{V_j} + V_{j + 1} + ... + V_n) = \{0\})\), because \(V_p \subseteq V_1 + ... + V_{j - 1} + \widehat{V_j} + V_{j + 1} + ... + V_n\) and \(b_{j, l} = b_{p, q} \neq 0\).
So, \(B\) has the \(d = Dim (V_1) + ... + Dim (V_n)\) elements.
Let \(r^{1, 1} b_{1, 1} + ... + r^{1, d_1} b_{1, d_1} + ... + r^{n, 1} b_{n, 1} + ... + r^{n, d_n} b_{n, d_n} = 0\).
For each \(j \in \{1, ..., n\}\), \(r^{j, 1} b_{j, 1} + ... + r^{j, d_j} b_{j, d_j} = - (r^{1, 1} b_{1, 1} + ... + r^{1, d_1} b_{1, d_1} + ... + \widehat{(r^{j, 1} b_{j, 1} + ... + r^{j, d_j} b_{j, d_j})} + ... + r^{n, 1} b_{n, 1} + ... + r^{n, d_n} b_{n, d_n})\).
But the left hand side is on \(V_j\) and the right hand side is on \(V_1 + ... + V_{j - 1} + \widehat{V_j} + V_{j + 1} + ... + V_n\), so, the left hand side is on \(V_j \cap (V_1 + ... + V_{j - 1} + \widehat{V_j} + V_{j + 1} + ... + V_n) = \{0\}\), so, \(r^{j, 1} b_{j, 1} + ... + r^{j, d_j} b_{j, d_j} = 0\).
As \(B_j\) is linearly independent, all the \(r^{j, l}\) s are \(0\).
As \(j\) is arbitrary, all the \(r^{j, l}\) s are \(0\).
So, \(B\) is linearly independent.
\(B\) is a basis for \(V\), by the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.
Step 3:
For each \(v \in V\), \(v = v^{1, 1} b_{1, 1} + ... + v^{1, d_1} b_{1, d_1} + ... + v^{n, 1} b_{n, 1} + ... + v^{n, d_n} b_{n, d_n}\), because \(B\) is a basis for \(V\).
That means that \(V = V_1 + ... + V_n\).
So, \(V\) is a vectors space as the direct sum of \(\{V_1, ..., V_n\}\).
