description/proof of that for tensors space w.r.t. field and k finite-dimensional vectors spaces over field and field, transition of standard bases w.r.t. bases for vector spaces is this
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of tensors space with respect to field and k vectors spaces and vectors space over field.
- The reader knows a definition of tensor product of tensors.
- The reader admits the proposition that for any field and any k finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces.
- The reader admits the proposition that for any finite-dimensional vectors space, the transition of the dual bases for the covectors space with respect to any bases for the vectors space is this.
Target Context
- The reader will have a description and a proof of the proposition that for the tensors space with respect to any field and any k finite-dimensional vectors spaces over the field and the field, the transition of the standard bases with respect to any bases for the vector spaces 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(\{V_1, ..., V_k\}\): \(\subseteq \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(L (V_1, ..., V_k: F)\): \(= \text{ the tensors space }\)
\(\{B_1, ..., B_k\}\): \(B_j \in \{\text{ the bases for } V_j\} = \{{b_j}_l \vert 1 \le l \le dim V_j\}\)
\(\{B'_1, ..., B'_k\}\): \(B'_j \in \{\text{ the bases for } V_j\} = \{{b'_j}_l \vert 1 \le l \le dim V_j\}\)
\(\{B^*_1, ..., B^*_k\}\): \(B^*_j = \text{ the dual basis of } B_j = \{{b_j}^l \vert 1 \le l \le dim V_j\}\)
\(\{B'^*_1, ..., B'^*_k\}\): \(B'^*_j = \text{ the dual basis of } B_j = \{{b'_j}^l \vert 1 \le l \le dim V_j\}\)
\(B^*\): \(= \{{b_1}^{j_1} \otimes ... \otimes {b_k}^{j_k} \vert \forall l \in \{1, ..., k\} (1 \le j_l \le dim V_l)\}\), \(\in \{\text{ the bases for } L (V_1, ..., V_k: F)\}\)
\(B'^*\): \(= \{{b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} \vert \forall l \in \{1, ..., k\} (1 \le j_l \le dim V_l)\}\), \(\in \{\text{ the bases for } L (V_1, ..., V_k: F)\}\)
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Statements:
\({b'_j}_l = {b_j}_m {M_j}^m_l\)
\(\implies\)
\({b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} = {{M_1}^{-1}}^{j_1}_{l_1} ... {{M_k}^{-1}}^{j_k}_{l_k} {b_1}^{l_1} \otimes ... \otimes {b_k}^{l_k}\)
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2: Proof
Whole Strategy: Step 1: see that \(B^*\) and \(B'^*\) are some bases for \(L (V_1, ..., V_k: F)\); Step 2: see that \({b'_j}^{l_j} = {{M_j}^{-1}}^{l_j}_{m_j} {b_j}^{m_j}\); Step 3: conclude the proposition.
Step 1:
\(B^*\) and \(B'^*\) are indeed some bases for \(L (V_1, ..., V_k: F)\), by the proposition that for any field and any k finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces.
Step 2:
\({b'_j}^{l_j} = {{M_j}^{-1}}^{l_j}_{m_j} {b_j}^{m_j}\), by the proposition that for any finite-dimensional vectors space, the transition of the dual bases for the covectors space with respect to any bases for the vectors space is this.
Step 3:
\({b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} = ({{M_1}^{-1}}^{j_1}_{m_1} {b_1}^{m_1}) \otimes ... \otimes ({{M_k}^{-1}}^{j_k}_{m_k} {b_k}^{m_k})\).
We note the fact that in general, for each \(f_j, f'_j \in L (V_{j, 1}, ..., V_{j, k_j}: F)\) and each \(r, r' \in F\), \(f_1 \otimes ... \otimes (r f_j + r' f'_j) \otimes ... \otimes f_l = r f_1 \otimes ... \otimes f_j \otimes ... \otimes f_l + r' f_1 \otimes ... \otimes f'_j \otimes ... \otimes f_l\): refer to Note for the definition of tensor product of tensors.
So, \(({{M_1}^{-1}}^{j_1}_{m_1} {b_1}^{m_1}) \otimes ... \otimes ({{M_k}^{-1}}^{j_k}_{m_k} {b_k}^{m_k}) = {{M_1}^{-1}}^{j_1}_{m_1} ... {{M_k}^{-1}}^{j_k}_{m_k} {b_1}^{m_1} \otimes ... \otimes {b_k}^{m_k}\).
