description/proof of that for vectors space, nonempty subset is vectors subspace iff subset is closed under linear combination
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space, any nonempty subset of the vectors space is a vectors subspace if and only if the subset is closed under linear combination.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(S\): \(\subseteq V\), \(\neq \emptyset\)
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Statements:
\(S \in \{\text{ the vectors subspaces of } V\}\)
\(\iff\)
\(\forall s_1, s_2 \in S, \forall r_1, r_2 \in F (r_1 s_1 + r_2 s_2 \in S)\)
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2: Proof
Whole Strategy: Step 1: suppose that \(r_1 s_1 + r_2 s_2 \in S\); Step 2: see that \(S\) is a vectors subspace of \(V\); Step 3: suppose that \(S\) is a vectors subspace of \(V\); Step 4: see that \(r_1 s_1 + r_2 s_2 \in S\).
Step 1:
Let us suppose that \(\forall s_1, s_2 \in S, \forall r_1, r_2 \in F (r_1 s_1 + r_2 s_2 \in S)\).
Let us see that \(S\) satisfies the conditions for being a vectors space.
1) for any elements, \(s_1, s_2 \in S\), \(s_1 + s_2 \in S\) (closed-ness under addition): \(s_1 + s_2 = 1 s_1 + 1 s_2 \in S\).
2) for any elements, \(s_1, s_2 \in S\), \(s_1 + s_2 = s_2 + s_1\) (commutativity of addition): it holds because it holds on the ambient \(V\).
3) for any elements, \(s_1, s_2, s_3 \in S\), \((s_1 + s_2) + s_3 = s_1 + (s_2 + s_3)\) (associativity of additions): it holds because it holds on the ambient \(V\).
4) there is a 0 element, \(0 \in S\), such that for any \(s \in S\), \(s + 0 = s\) (existence of 0 vector): there is an \(s \in S\), because \(S \neq \emptyset\), and \(0 = 0 s + 0 s \in S\), and \(s + 0 = s\), because it holds on the ambient \(V\).
5) for any element, \(s \in S\), there is an inverse element, \(s' \in S\), such that \(s' + s = 0\) (existence of inverse vector): \(-1 s + 0 s = -1 s \in S\), and \(-1 s + s = 0\), because it holds on the ambient \(V\).
6) for any element, \(s \in S\), and any scalar, \(r \in F\), \(r . s \in S\) (closed-ness under scalar multiplication): \(r . s = r . s + 0 . s \in S\).
7) for any element, \(s \in S\), and any scalars, \(r_1, r_2 \in F\), \((r_1 + r_2) . s = r_1 . s + r_2 . s\) (scalar multiplication distributability for scalars addition): it holds because it holds on the ambient \(V\).
8) for any elements, \(s_1, s_2 \in S\), and any scalar, \(r \in F\), \(r . (s_1 + s_2) = r . s_1 + r . s_2\) (scalar multiplication distributability for vectors addition): it holds because it holds on the ambient \(V\).
9) for any element, \(s \in S\), and any scalars, \(r_1, r_2 \in F\), \((r_1 r_2) . s = r_1 . (r_2 . s)\) (associativity of scalar multiplications): it holds because it holds on the ambient \(V\).
10) for any element, \(s \in S\), \(1 . s = s\) (identity of 1 multiplication): it holds because it holds on the ambient \(V\).
Step 3:
Let us suppose that \(S\) is a vectors subspace of \(V\).
Step 4:
\(\forall s_1, s_2 \in S, \forall r_1, r_2 \in F (r_1 s_1 + r_2 s_2 \in S)\) holds, because the conditions for being a vectors space requires that.
