description/proof of that for infinite-dimensional vectors space, "dual" of basis is never basis for covectors (dual) space
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of covectors (dual) space of vectors space.
- The reader knows a definition of basis of module.
- The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
Target Context
- The reader will have a description and a proof of the proposition that for any infinite-dimensional vectors space, the "dual" of any basis is never any basis for the covectors (dual) space.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V^*\): \(= L (V: F)\)
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_j \vert j \in J\}\)
\(B^*\): \(= \{b^j \vert j \in J\}\) such that \(\forall j \in J (\forall l \in J (b^j (b_l) = \delta^j_l))\)
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Statements:
\(B^* \notin \{\text{ the bases for } V^*\}\)
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2: Note
This is saying that \(B^*\) is never any basis, not "not necessarily a basis".
Compare with the definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space
This article uses the expression, "dual", with double-quotations because \(B^*\) is not usually called "dual" because it is not any dual basis.
3: Proof
Whole Strategy: Step 1: see that \(B^*\) is well-defined; Step 2: take \(w \in V^*\) as for each \(v = v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n}\), \(w (v) = v^{j_1} + ... + v^{j_n}\), and see that \(w\) is not spanned by \(B^*\).
Step 1:
Let us see that \(B^*\) is well-defined.
Let \(v = v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n} \in V\) be any, which is the unique decomposition with nonzero components, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(b^j (v) = b^j (v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n}) = v^{j_1} b^j (b_{j_1}) + ... + v^{j_n} b^j (b_{j_n}) = v^{j_1} \delta^j_{j_1} + ... + v^{j_n} \delta^j_{j_n}\), determined uniquely.
\(b^j\) is linear, because for each \(v = v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n}, v' = v'^{l_1} b_{l_1} + ... + v'^{l_m} b_{l_m} \in V\) and each \(r, r' \in F\), when \(b_j\) is not contained in \(v\) nor \(v'\), \(b^j (r v + r' v') = 0 = 0 + 0 = r b^j (v) + r' b^j (v')\); when \(b_j\) is contained only in \(v\) as \(b_{j_p}\), \(b^j (r v + r' v') = r v^{j_p} = r v^{j_p} + 0 = r b^j (v) + r' b^j (v')\); when \(b_j\) is contained only in \(v'\) as \(b_{l_q}\), \(b^j (r v + r' v') = r' v'^{l_q} = 0 + r' v'^{l_q} = r b^j (v) + r' b^j (v')\); when \(b_j\) is contained both in \(v\) as \(b_{j_p}\) and in \(v'\) as \(b_{l_q}\), \(b^j (r v + r' v') = r v^{j_p} + r' v^{l_q} = r b^j (v) + r' b^j (v')\).
So, \(b^j \in V^*\).
Step 2:
Let us take \(w \in V^*\) as for each \(v = v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n}\), \(w (v) = v^{j_1} + ... + v^{j_n}\).
\(w\) is well-defined, because the decomposition is unique with nonzero components, as before.
Let us see that \(w\) is indeed linear.
Let \(v = v^{k_1} b_{k_1} + ... + v^{k_m} b_{k_m}, v' = v'^{l_1} b_{l_1} + ... + v'^{l_n} b_{l_n} \in V\) and \(r, r' \in F\) be any.
Let \(K := \{k_1, ... k_m\}\) and \(L := \{l_1, ... l_n\}\).
\(K \cap L\) may be empty or nonempty, but anyway, \(v = \sum_{k \in K \setminus L} v^k b_k + \sum_{k \in K \cap L} v^k b_k\) and \(v' = \sum_{l \in L \setminus K} v^l b_l + \sum_{k \in K \cap L} v^k b_k\).
\(r v + r' v' = r (\sum_{k \in K \setminus L} v^k b_k + \sum_{k \in K \cap L} v^k b_k) + r' (\sum_{l \in L \setminus K} v'^l b_l + \sum_{k \in K \cap L} v'^k b_k) = r \sum_{k \in K \setminus L} v^k b_k + r' \sum_{l \in L \setminus K} v'^l b_l + \sum_{k \in K \cap L} (r v^k + r' v'^k) b_k\).
So, \(w (r v + r' v') = r \sum_{k \in K \setminus L} v^k + r' \sum_{l \in L \setminus K} v'^l + \sum_{k \in K \cap L} (r v^k + r' v'^k) = r \sum_{k \in K \setminus L} v^k + \sum_{k \in K \cap L} r v^k + r' \sum_{l \in L \setminus K} v'^l + \sum_{k \in K \cap L} r' v'^k = r (\sum_{k \in K \setminus L} v^k + \sum_{k \in K \cap L} v^k) + r' (\sum_{l \in L \setminus K} v'^l + \sum_{k \in K \cap L} v'^k) = r \sum_{k \in K} v^k + r' \sum_{l \in L} v'^l = r w (v) + r' w (v')\).
So, indeed, \(w \in V^*\).
Let us see that \(w\) is not spanned by \(B^*\).
Let us suppose that \(w = w^1 b^{j_1} + ... + w^m b^{j_m}\).
There is a \(b^l \in B^* \setminus \{b^{j_1}, ..., b^{j_m}\}\).
Then, \(w (b_l) = 1\) but \((w^1 b^{j_1} + ... + w^m b^{j_m}) (b_l) = 0 + ... + 0 = 0\), a contradiction.
So, \(w\) is not spanned by \(B^*\).
So, \(B^*\) is not any basis.
