definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors 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.
Target Context
- The reader will have a definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors 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:
\( F\): \(\in \{\text{ the fields }\}\)
\( V\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\( V^*\): \(= L (V: F)\)
\( J\): \(\in \{\text{ the finite index sets }\}\)
\( B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_j \vert j \in J\}\)
\(*B^*\): \(\in \{\text{ the bases for } V^*\}\), \(= \{b^j \vert j \in J\}\)
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Conditions:
\(\forall j \in J (\forall k \in J (b^j (b_k) = \delta^j_k))\)
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2: Note
Let us see that \(B^*\) is indeed a basis for \(V^*\).
Let \(w \in V^*\) be any. For each \(j \in J\), \(w_j := w (b_j) \in F\). For each \(v \in V\), \(v = v^j b_j\), and \(w (v) = w (v^j b_j) = v^j w (b_j) = v^j w_j\). \(w_j b^j (v) = w_j b^j (v^k b_k) = w_j v^k b^j (b_k) = w_j v^k \delta^j_k = w_j v^j\). So, \(w = w_j b^j\), which means that \(B^*\) spans \(V^*\).
Let \(c_j b^j = 0\). \(c_j b^j (b_k) = 0 (b_k) = 0\), but \(c_j b^j (b_k) = c_j \delta^j_k = c_k\), so, \(c_k = 0\) for each \(k \in J\), which means that \(B^*\) is linearly independent.
\(V^*\) inevitably has the dimension of \(V\).
For each \(w \in V^*\), the components of \(w\) with respect to \(B^*\) are \(w (b_1), ..., w (b_d)\).
It is crucial that \(V\) is finite-dimensional: otherwise, \(w_j b^j\) may not make sense, because some infinitely many \(w_j\) s may be nonzero.
