Introduction
Constructions with linear subspace
Basic constructions
There are two basic construction operations with linear subspaces, intersecting and adding. Let \(\mathcal{V}\) be a vector space over a field \(\mathbb{F}\), let \(J\) be an index set and \(\{ \mathcal{U}_j\}_{j \in J}\) be a family of linear subspaces of \(\mathcal{V}\).
I) Then the intersection of these linear spaces is defined by:
II) The sum of linear subspaces is constructed by:
Since the axiom of choice assures us the existence of a basis for a vector space we can think of the sum by taking a basis, \(\mathcal{B}_1\), of \(\mathcal{U}_1\) and then extend it by basis elements of the basis of all the other linear subspaces which are linearly independent of all the basis elements in $ \mathcal{B}_1 $ with the help of the basis extension theorem:
Let \(\{ \mathcal{B}_j\}_{j \in J}\) be the corresponding family bases for each linear subspace. Then we obtain the basis, \(\mathcal{B}\), of \(\sum_{j \in J} \mathcal{U}_{j}\) by setting:
\[\mathcal{B} = \mathcal{B}_1\] Now, let \(\mathcal{\hat{B}}_2 \subseteq \mathcal{B}_2\) be the subset with basis elements that a linearly independent of \(\mathcal{B}\).
Define the new basis as the union of the previous two:
We continue repeating that method for each basis such that we obtain:
\[\mathcal{B} = \Biggl( \bigcup_{j \in J \\ j \neq 1} \mathcal{\hat{B}}_j \Biggl) \ \cup \ \mathcal{B}_1\]Then a vector \(v \in \sum_{j \in J} \mathcal{U}_{j}\) is simply a linear combination of \(\mathcal{B}\). This perspective makes it also evident that the vector space created by the intersection \(\bigcap_{j \in J} \mathcal{U}_{j}\) is a subspace of the summed space since:
\[\bigcap_{j \in J} \mathcal{B}_{j} \subseteq \mathcal{B}_{1} \subseteq \mathcal{B}_1 \ \cup \ \Biggl( \bigcup_{j \in J \\ j \neq 1} \mathcal{\hat{B}}_j \Biggl) = \mathcal{B}\]Relation between Sum and Product
Let \(\mathcal{V}(\mathbb{F})\) be a vector space with basis, \(\mathcal{B} = \{ b_1, ..., b_n \}\), then for any choice of basis elements the generating system of it is a linear subspace of \(\mathcal{V}\):
\[\textbf{span} ( \{ b_{\sigma(1)}, ..., b_{\sigma(m)} \} ) = \left\{ v \in \mathcal{V} \ | \ v = \sum_{l=1}^{m} \lambda_l b_{\sigma(l)} \ , \ \lambda_l \in \mathbb{F} \ , \ l = 1, ..., m \right\} \ , \quad m \leq n\]In that way we can decomposes \(\mathcal{V}\) into a sum of linear subspaces created by dividing the basis into j smaller bases and constructing generating systems out of them. Let us denote these newly created subspaces as \(\mathcal{U}_k\) and their corresponding basis \(\mathcal{B}_k\) with \(k = 1, ..., j\). Then the intersection of these spans contains only the zero vector:
\[\bigcap_{k = 1}^{j} \mathcal{B}_{k} = \emptyset \implies \bigcap_{k = 1}^{j} \mathcal{U}_{k} = \{ 0_{\mathcal{V}} \}\]Because of this property we name this sum a direct sum and write it in the following way:
\[\mathcal{V} = \bigoplus_{k = 1}^{j} \mathcal{U}_{k}\]Let \(v = (v_1, ..., v_n) \in \mathcal{V}\). An now \(v = ((v_1, ..., v_k), (v_{k+1}, ..., v_n)) = (v_{w_1}, v_{w_2})\)
Recall that the product of a family of linear subspaces, \(\{ \mathcal{U}_j \}_{j \in J}\), of a vector space, \(\mathcal{V}\), over a field, \(\mathbb{F}\), is defined by:
\[\prod_{j \in J} \mathcal{U}_j := \left\{ v = (u_1, ..., u_{|J|}) \in \mathcal{V} \ | \ \forall \ j \in J : u_j \in \mathcal{U}_{j} \right\}\]