description/proof of that for \((0, 2)\)-tensor of vectors space, linear map from vectors space into covectors space is induced, and when vectors space is finite-dimensional, induced map's representative matrix w.r.t. basis and dual basis is this
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of \((p, q)\)-tensors space of vectors space.
- The reader knows a definition of interior multiplication of tensor by vector.
- The reader knows a definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space.
- The reader knows a definition of representative matrix of linear map between finite-dimensional vectors spaces with respect to bases.
Target Context
- The reader will have a description and a proof of the proposition that for any \((0, 2)\)-tensor of any vectors space, a linear map from the vectors space into the covectors space is induced, and when the vectors space is finite-dimensional, the induced map's representative matrix with respect to any basis and the dual basis is this.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(T^0_2 (V)\):
\(t\): \(\in T^0_2 (V)\)
//
Statements:
\(\widehat{t}: V \to V^*, v \mapsto i_v (t) \in \{\text{ the linear maps }\}\)
\(\land\)
(
\(V \in \{\text{ the finite-dimensional vectors spaces }\}\)
\(\implies\)
\(\forall B = (b_1, ..., b_d) \in \{\text{ the bases for } V\} (M^j_l = t (b_l, b_j))\), where \(B^*\) is the dual basis of \(B\) for \(V^*\) and \(M\) is the representative matrix of \(\widehat{t}\) with respect to \(B\) and \(B^*\)
)
//
2: Proof
Whole Strategy: Step 1: see that \(\widehat{t}\) is valid; Step 2: see that \(\widehat{t}\) is \(F\)-linear; Step 3: suppose that \(V\) is finite-dimensional; Step 4: for each \(B\), take \(B^*\) and \(M\), and see that \(M^j_l = t (b_l, b_j)\).
Step 1:
Let us see that \(\widehat{t}\) is valid, which is about that \(\widehat{t} (v) \in V^*\).
For each \(w \in V\), \(\widehat{t} (v) (w) = i_v (t) (w) = t (v, w) \in F\).
For each \(w, w' \in V\) and each \(r, r' \in F\), \(\widehat{t} (v) (r w + r' w') = t (v, r w + r' w') = r t (v, w) + r' t (v, w') = r \widehat{t} (v) (w) + r' \widehat{t} (v) (w')\), which means that \(\widehat{t} (v)\) is a \(F\)-multilinear map.
So, \(\widehat{t} (v) \in V^*\).
Step 2:
Let us see that \(\widehat{t}\) is \(F\)-linear.
For each \(v, v' \in V\) and each \(r, r' \in F\), for each \(w \in V\), \(\widehat{t} (r v + r' v') (w) = t (r v + r' v', w) = r t (v, w) + r' t (v', w) = r \widehat{t} (v) (w) + r' \widehat{t} (v') (w) = (r \widehat{t} (v) + r' \widehat{t} (v')) (w)\).
That implies that \(\widehat{t} (r v + r' v') = r \widehat{t} (v) + r' \widehat{t} (v')\).
So, \(\widehat{t}\) is \(F\)-linear.
Step 3:
Let us suppose that \(V\) is finite-dimensional.
Step 4:
Let \(B = \{b_1, ..., b_d\}\) be any basis for \(V\).
Let \(B^* = \{b^1, ..., b^d\}\) be the dual basis of \(B\) for \(V^*\): refer to the definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space.
Let \(M\) be the representative matrix of \(\widehat{t}\) with respect to \(B\) and \(B^*\) with the canonical 'vectors spaces - linear morphisms' isomorphism from \(V\) onto the components vectors space with respect to \(B\), \(f_1: V \to F^d\), and the canonical 'vectors spaces - linear morphisms' isomorphism from \(V^*\) onto the components vectors space with respect to \(B^*\), \(f_2: V^* \to F^d\): refer to the definition of representative matrix of linear map between finite-dimensional vectors spaces with respect to bases.
\(\widehat{t} = {f_2}^{-1} \circ M \circ f_1\).
\(f_1 (b_l) = \begin{pmatrix} 0 \\ . \\ . \\ . \\ 0 \\ 1 \\ 0 \\ . \\ . \\ . \\ 0 \end{pmatrix}\) where \(1\) is the \(l\)-th component.
\(M \circ f_1 (b_l) = \begin{pmatrix} M^1_l \\ ... \\ M^d_l \end{pmatrix}\).
\({f_2}^{-1} \circ M \circ f_1 (b_l) = M^1_l b^1 + ... + M^d_l b^d\).
So, \(\widehat{t} (b_l) = M^1_l b^1 + ... + M^d_l b^d\).
So, \(\widehat{t} (b_l) (b_j) = (M^1_l b^1 + ... + M^d_l b^d) (b_j) = M^j_l\).
But the left hand side is \(t (b_l, b_j)\).
So, \(M^j_l = t (b_l, b_j)\).
