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definition of canonical 'vectors spaces - linear morphisms' isomorphism between tensors space at point on \(C^\infty\) manifold with boundary and tensors space w.r.t. real numbers field and \(p\) cotangent vectors spaces and \(q\) tangent vectors spaces and field
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of canonical 'vectors spaces - linear morphisms' isomorphism between tensors space at point on \(C^\infty\) manifold with boundary and tensors space with respect to real numbers field and \(p\) cotangent vectors spaces and \(q\) tangent vectors spaces and field.
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:
\( M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\( m\): \(\in M\)
\( T_mM\): \(= \text{ the tangent vectors space at } m\)
\( B\): \(\in \{\text{ the bases for } T_mM\}\), \(= \{b_1, ..., b_d\}\)
\( T_mM^*\): \(= \text{ the cotangent vectors space at } m\)
\( B^*\): \(= \text{ the dual basis of } B \text{ for } T_mM^*\), \(= \{b^1, ..., b^d\}\)
\( {B^*}^*\): \(= \text{ the dual basis of } B^* \text{ for } {T_mM^*}^*\), \(= \{\widetilde{b}_1, ..., \widetilde{b}_d\}\)
\( p\): \(\in \mathbb{N}\)
\( q\): \(\in \mathbb{N}\)
\( T^p_q (T_mM)\): \(= T_mM \otimes ... \otimes T_mM \otimes T_mM^* \otimes ... \otimes T_mM^*\)
\( L (T_mM^*, ..., T_mM^*, T_mM, ..., T_mM: \mathbb{R})\):
\(*f\): \(: T^p_q (T_mM) \to L (T_mM^*, ..., T_mM^*, T_mM, ..., T_mM: \mathbb{R}), t^{j_1, ..., j_p}_{l_1, ..., l_q} [((b_{j_1}, ..., b_{j_p}, b^{l_1}, ..., b^{l_q}))] \mapsto t^{j_1, ..., j_p}_{l_1, ..., l_q} \widetilde{b}_{j_1} \otimes ... \otimes \widetilde{b}_{j_p} \otimes b^{l_1} \otimes ... \otimes b^{l_q}\)
//
Conditions:
//
\(f\) does not depend on the choice of \(B\) as is seen in Note, which is the reason why \(f\) is called "canonical".
2: Note
\(f\) is indeed a 'vectors spaces - linear morphisms' isomorphism, because \(\{[((b_{j_1}, ..., b_{j_p}, b^{k_1}, ..., b^{k_q}))]\}\) is a basis for \(T^p_q (T_mM)\), by the proposition that the tensor product of any \(k\) finite-dimensional vectors spaces has the basis that consists of the classes induced by any basis elements; \(\{\widetilde{b}_{j_1} \otimes ... \otimes \widetilde{b}_{j_p} \otimes b^{k_1} \otimes ... \otimes b^{k_q}\}\) is a basis for \(L (T_mM^*, ..., T_mM^*, T_mM, ..., T_mM: \mathbb{R})\), 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; and the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism applies.
Let us see that \(f\) does not depend on the choice of \(B\).
Let \(B' = \{b'_1, ..., b'_d\}\) be any other basis for \(T_mM\).
Let the map constructed by \(B'\) be \(f'\).
\(b'_j = b_l M^l_j\) for an invertible matrix, \(M\). So, \(b_j = b'_l {M^{-1}}^l_j\).
The dual basis of \(B'\), \(B'^* = \{b'^1, ..., b'^d\}\), is \(\{{M^{-1}}^j_l b^l\}\), 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. So, \(b^j = M^j_l b'^l\).
The dual basis of \(B'^*\), \({B'^*}^* = \{\widetilde{b'}_1, ..., \widetilde{b'}_d\}\), is \(\{\widetilde{b}_l M^l_j\}\), 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. So, \(\widetilde{b}_j = \widetilde{b'}_l {M^{-1}}^l_j\).
Then, \([((b_{j_1}, ..., b_{j_p}, b^{l_1}, ..., b^{l_q}))] = [((b'_{m_1} {M^{-1}}^{m_1}_{j_1}, ..., b'_{m_p} {M^{-1}}^{m_p}_{j_p}, M^{l_1}_{n_1} b'^{n_1}, ..., M^{l_q}_{n_q} b'^{n_q}))] = {M^{-1}}^{m_1}_{j_1} ... {M^{-1}}^{m_p}_{j_p} M^{l_1}_{n_1} ... M^{l_q}_{n_q} [((b'_{m_1}, ..., b'_{m_p}, b'^{n_1}, ..., b'^{n_q}))]\), which is mapped by \(f'\) to \({M^{-1}}^{m_1}_{j_1} ... {M^{-1}}^{m_p}_{j_p} M^{l_1}_{n_1} ... M^{l_q}_{n_q} \widetilde{b'}_{m_1} \otimes ... \otimes \widetilde{b'}_{m_p} \otimes b'^{n_1} \otimes ... \otimes b'^{n_q} = \widetilde{b'}_{m_1} {M^{-1}}^{m_1}_{j_1} \otimes ... \otimes \widetilde{b'}_{m_p} {M^{-1}}^{m_p}_{j_p} \otimes M^{l_1}_{n_1} b'^{n_1} \otimes ... \otimes M^{l_q}_{n_q} b'^{n_q} = \widetilde{b}_{j_1} \otimes ... \otimes \widetilde{b}_{j_p} \otimes b^{l_1} \otimes ... \otimes b^{l_q}\).
So, \(f' = f\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of canonical 'vectors spaces - linear morphisms' isomorphism between finite-dimensional vectors space and its double dual space
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of canonical 'vectors spaces - linear morphisms' isomorphism between finite-dimensional vectors space and its double 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:
\( F\): \(\in \{\text{ the fields }\}\)
\( V\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\( V^*\): \(= L (V: F)\)
\( {V^*}^*\): \(= L (L (V: F): F)\)
\( J\): \(\in \{\text{ the finite index sets }\}\)
\( B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_j \vert j \in J\}\)
\( B^*\): \(= \text{ the dual basis of } B\), \(= \{b^j \vert j \in J\}\)
\( {B^*}^*\): \(= \text{ the dual basis of } B^*\), \(= \{\widetilde{b}_j \vert j \in J\}\)
\(*f\): \(: V \to {V^*}^*, v^j b_j \mapsto \sum_{j \in J} v^j \widetilde{b}_j\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
//
Conditions:
//
\(f\) does not depend on the choice of \(B\): compare to the definition of canonical 'vectors spaces - linear morphisms' isomorphism between finite-dimensional vectors space and its covectors space with respect to original space basis with "with respect to original space basis".
2: Note
\(f\) is indeed a 'vectors spaces - linear morphisms' isomorphism, does not depend on the choice of \(B\), and has the property that for each \(v \in V\) and each \(w \in V^*\), \(w (v) = f (v) (w)\), by the proposition that between any finite-dimensional vectors space and its double dual, there is the canonical 'vectors spaces - linear morphisms' isomorphism.
Although it is sometimes sloppily expressed like "The double dual of a finite-dimensional vectors space is the original vectors space.", the double dual is not the same entity with the original vectors space, as the 2 have different meanings. They are just 'vectors spaces - linear morphisms' isomorphic, and having such a relation does not make 2 entities the same.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
definition of canonical 'vectors spaces - linear morphisms' isomorphism between finite-dimensional vectors space and its covectors space w.r.t. original space basis
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of canonical 'vectors spaces - linear morphisms' isomorphism between finite-dimensional vectors space and its covectors space with respect to original space basis.
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:
\( F\): \(\in \{\text{ the fields }\}\)
\( V\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\( V^*\): \(= L (V: F)\)
\( J\): \(\in \{\text{ the finite index sets }\}\)
\( B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_j \vert j \in J\}\)
\( B^*\): \(= \text{ the dual basis of } B\), \(= \{b^j \vert j \in J\}\)
\(*f\): \(: V \to V^*, v^j b_j \mapsto \sum_{j \in J} v^j b^j\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
//
Conditions:
//
\(f\) depends on the choice of \(B\) (as will be seen in Note), which is the reason why the concept is called with "with respect to original space basis".
2: Note
\(f\) is indeed a 'vectors spaces - linear morphisms' isomorphism, by the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism.
Let us see that \(f\) depends on the choice of \(B\).
Let \(B' = \{b'_j \vert j \in J\}\) be another basis for \(V\).
\(b'_j = b_l M^l_j\) for an invertible matrix, \(M\). So, \(b_j = b'_l {M^{-1}}^l_j\).
The dual basis of \(B'\), \(B'^* = \{b'^j \vert j \in J\}\), is \(\{{M^{-1}}^j_l b^l\}\), 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.
The canonical 'vectors spaces - linear morphisms' isomorphism with respect to \(B'\), \(f': V \to V^*\), maps \(b_j = {M^{-1}}^l_j b'_l\) to \(\sum_{l \in J} {M^{-1}}^l_j b'^l = \sum_{l \in J} {M^{-1}}^l_j {M^{-1}}^l_m b^m\), which does not equal \(b^j\) in general. In fact, when \(M\) is orthogonal, \(M^m_l = {M^{-1}}^l_m\), and \(= \sum_{m \in J} {M^{-1}}^l_j M^m_l b^m = \sum_{m \in J} \delta^m_j b^m = b^j\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that between vectors spaces, map that maps basis onto basis bijectively and expands mapping linearly is 'vectors spaces - linear morphisms' isomorphism
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(B_1\): \(= \{{b_1}_j \vert j \in J\}\), \(\in \{\text{ the bases for } V_1\}\)
\(B_2\): \(\in \{\text{ the bases for } V_2\}\)
\(f\): \(: B_1 \to B_2\), \(\in \{\text{ the bijections }\}\)
\(g\): \(: V_1 \to V_2, \sum_{j \in S} v^j {b_1}_j \mapsto \sum_{j \in S} v^j f ({b_1}_j)\), where \(S \in \{\text{ the finite subsets of } J\}\)
//
Statements:
\(g \in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
//
2: Note
\(V_1\) or \(V_2\) does not need to be finite-dimensional.
3: Proof
Whole Strategy: Step 1: see that \(g\) is well-defined; Step 2: see that \(g\) is a linear map; Step 3: see that \(g\) is injective; Step 4: see that \(g\) is surjective; Step 5: conclude the proposition.
Step 1:
Let us see that \(g\) is well-defined.
For each element, \(v \in V_1\), \(v = \sum_{j \in S} v^j {b_1}_j\) where \(S\) is a finite subset of \(J\), by the definition of basis of module.
The decomposition is unique, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
So, \(g\) is determined for each element of \(V_1\) unambiguously.
Step 2:
\(g\) is a linear map, by the proposition that from any module with any basis into any module, a linear map can be defined by mapping the basis and linearly expanding the mapping: \(f\) is mapping the basis, \(B_1\), and \(g\) is linearly expanding \(f\).
Step 3:
Let us see that \(g\) is injective.
Let \(v, v' \in V_1\) be any such that \(v \neq v'\).
\(v = \sum_{j \in S} v^j {b_1}_j\) and \(v' = \sum_{j \in S'} v'^j {b_1}_j\).
\(g (v) = \sum_{j \in S} v^j f ({b_1}_j)\) and \(g (v') = \sum_{j \in S'} v'^j f ({b_1}_j)\).
Let us suppose that \(g (v) = g (v')\).
\(\{f ({b_1}_j) \vert j \in S \cup S'\} \subseteq B_2\) was distinct, because \(f\) was bijective.
By the proposition that for any module with any basis, the components set of any element with respect to the basis is unique, \(S = S'\) and \(v^j = v'^j\), a contradiction against \(v \neq v'\).
So, \(g (v) \neq g (v')\).
Step 4:
Let us see that \(g\) is surjective.
For each \(v_2 \in V_2\), \(v_2 = \sum_{j \in \{1, ..., n\}} v^j {b_2}_j\), where \({b_2}_j \in B_2\).
As \(f\) is bijective, \(\{f^{-1} ({b_2}_1), ..., f^{-1} ({b_2}_n)\} = \{{b_1}_{j_1}, ..., {b_1}_{j_n}\} \subseteq B_1\) is valid and distinct.
\(v^1 {b_1}_{j_1} + ... + v^n {b_1}_{j_n} \in V_1\) and \(g (v^1 {b_1}_{j_1} + ... + v^n {b_1}_{j_n}) = v^1 f ({b_1}_{j_1}) + ... + v^n f ({b_1}_{j_n}) = v^1 {b_2}_1 + ... + v^n {b_2}_n = v_2\).
Step 5:
So, \(g\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for \(C^\infty\) manifold with boundary and tangent vectors space at point, transition of standard bases w.r.t. charts is this
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any \(C^\infty\) manifold with boundary and the tangent vectors space at any point, the transition of the standard bases with respect to any charts 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:
\(M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\(m\): \(\in M\)
\(T_mM\): \(= \text{ the tangent vectors space at } m\)
\((U_m \subseteq M, \phi_m)\): \(\in \{\text{ the charts for } M \text{ around } m\}\)
\((U'_m \subseteq M, \phi'_m)\): \(\in \{\text{ the charts for } M \text{ around } m\}\)
\(B\): \(= \text{ the standard basis for } T_mM\) with respect to \((U_m \subseteq M, \phi_m)\), \(= \{\partial / \partial x^1, ..., \partial / \partial x^d\}\)
\(B'\): \(= \text{ the standard basis for } T_mM\) with respect to \((U'_m \subseteq M, \phi'_m)\), \(= \{\partial / \partial x'^1, ..., \partial / \partial x'^d\}\)
//
Statements:
\(\partial / \partial x'^j = \partial x^k / \partial x'^j \partial / \partial x^k\)
//
\(x\) as a function of \(x'\) is \(\phi_m \circ {\phi'_m}^{-1} \vert_{\phi'_m (U_m \cap U'_m)}: \phi'_m (U_m \cap U'_m) \to \phi_m (U_m \cap U'_m)\).
2: Proof
Whole Strategy: Step 1: let \(\partial / \partial x'^j = M^k_j \partial / \partial x^k\); Step 2: make the both sides of it operate on \(x^l: U_m \cap U'_m \to \mathbb{R}, p \mapsto {\phi_m (p)}^l\), and see that \(M^k_j = \partial x^k / \partial x'^j\).
Step 1:
Let \(\partial / \partial x'^j = M^k_j \partial / \partial x^k\), which is possible because \(B\) is a basis for \(T_mM\) and \(\partial / \partial x'^j \in T_mM\).
Step 2:
\(x^l: U_m \cap U'_m \to \mathbb{R}, p \mapsto {\phi_m (p)}^l\) is a \(C^\infty\) function.
\(\partial / \partial x'^j (x^l) = M^k_j \partial / \partial x^k (x^l) = M^k_j \partial_k (x^l \circ {\phi_m}^{-1}) = M^k_j \delta^l_k = M^l_j\). So, \(M^k_j = \partial x^k / \partial x'^j\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for finite-dimensional vectors space and vectors space endomorphism, transition of endomorphism matrix w.r.t. change of bases is this
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space and any vectors space endomorphism, the transition of the endomorphism matrix with respect to any change of bases 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(B\): \(\in \{\text{ the bases for } V\} = \{b_s \vert 1 \le s \le dim V\}\)
\(B'\): \(\in \{\text{ the bases for } V\} = \{b'_s = b_j M_s^j \vert 1 \le s \le dim V\}\)
\(f\): \(: V \to V\)
\(N\): \(= \text{ the matrix of } f \text{ with respect to } B\)
\(N'\): \(= \text{ the matrix of } f \text{ with respect to } B'\)
//
Statements:
\(N = M N' M^{-1}\)
//
2: Note
A motivation for this proposition is to get \(N\) from a simple \(N'\).
For example, for \(V = \mathbb{R}^3\) with the Euclidean inner product, in order to get \(N\) for the \(\theta\) rotation around the axis, \((n^1, n^2, n^3)\), which is with respect to \(B\), we can get any orthonormal \(B'\) with \((n^1, n^2, n^3)\) as \(b'_3\), then, \(N'\) is simple because it is the rotation around the \(b'_3\) axis, and we can get \(N\) from \(N'\).
3: Proof
Whole Strategy: Step 1: for each \(v \in V\), let the components column vectors with respect to \(B\) and \(B'\) be \(\overline{v}\) and \(\overline{v}'\); Step 2: see that \(\overline{f (v)}' = N' \overline{v}'\); Step 3: see that \(\overline{v}' = M^{-1} \overline{v}\) and \(\overline{f (v)}' = M^{-1} \overline{f (v)}\); Step 4: see that \(M^{-1} \overline{f (v)} = N' M^{-1} \overline{v}\), and conclude the proposition.
Step 1:
For each \(v \in V\), let the components column vectors with respect to \(B\) and \(B'\) be \(\overline{v}\) and \(\overline{v}'\).
Step 2:
\(\overline{f (v)}' = N' \overline{v}'\).
Step 3:
\(\overline{v}' = M^{-1} \overline{v}\) and \(\overline{f (v)}' = M^{-1} \overline{f (v)}\), by the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
Step 4:
So, \(M^{-1} \overline{f (v)} = N' M^{-1} \overline{v}\).
So, \(\overline{f (v)} = M M^{-1} \overline{f (v)} = M N' M^{-1} \overline{v}\), which means that \(N = M N' M^{-1}\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for tensor product of \(k\) finite-dimensional vectors spaces over field, transition of components of element w.r.t. standard bases w.r.t. bases for vectors spaces is this
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for the tensor product of any \(k\) finite-dimensional vectors spaces over any field, the transition of the components of any element w.r.t. the standard bases w.r.t. any bases for the vectors 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(\{V_1, ..., V_k\}\): \(\subseteq \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(V_1 \otimes ... \otimes V_k\): \(= \text{ the tensor product }\)
\(\{B_1, ..., B_k\}\): \(B_j \in \{\text{ the bases for } V_j\} = \{{b_j}_l \vert 1 \le l \le dim V_j\}\)
\(\{B'_1, ..., B'_k\}\): \(B'_j \in \{\text{ the bases for } V_j\} = \{{b_j}_l \vert 1 \le l \le dim V_j\}\)
\(B\): \(= \{[(({b_1}_{l_1}, ..., {b_k}_{l_k}))] \vert {b_j}^{l_j} \in B_j\}\), \(\in \{\text{ the bases for } V_1 \otimes ... \otimes V_k\}\)
\(B'\): \(= \{[(({b'_1}_{l_1}, ..., {b'_k}_{l_k}))] \vert {b'_j}^{l_j} \in B'_j\}\), \(\in \{\text{ the bases for } V_1 \otimes ... \otimes V_k\}\)
//
Statements:
\({b'_j}_l = {b_j}_m {M_j}^m_l\)
\(\implies\)
\(\forall f = f^{l_1, ..., l_k} [(({b_1}_{l_1}, ..., {b_k}_{l_k}))] = f'^{m_1, ..., m_k} [(({b'_1}_{m_1}, ..., {b'_k}_{m_k}))] \in V_1 \otimes ... \otimes V_k (f'^{l_1, ..., l_k} = {{M_1}^{-1}}^{l_1}_{m_1} ... {{M_k}^{-1}}^{l_k}_{m_k} f^{m_1, ..., m_k})\)
//
2: Proof
Whole Strategy: just apply the proposition that for the tensor product of any \(k\) finite-dimensional vectors spaces over any field, the transition of the standard bases with respect to any bases for the vectors spaces is this, the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this, and the proposition that for the tensors space with respect to any field and any finite number of finite-dimensional the field vectors spaces and the field or the tensor product of any finite-dimensional vectors spaces over any field, the transition of any standard bases or the components is a square matrix, and the inverse matrix is the product of the inverses; Step 1: see that \([(({b'_1}_{l_1}, ..., {b'_k}_{l_k}))] = [(({b_1}_{m_1}, ..., {b_k}_{m_k}))] {M_1}^{m_1}_{l_1} ... {M_k}^{m_k}_{l_k}\); Step 2: conclude the proposition.
Step 1:
\([(({b'_1}_{l_1}, ..., {b'_k}_{l_k}))] = [(({b_1}_{m_1}, ..., {b_k}_{m_k}))] {M_1}^{m_1}_{l_1} ... {M_k}^{m_k}_{l_k}\), by the proposition that for the tensor product of any \(k\) finite-dimensional vectors spaces over any field, the transition of the standard bases with respect to any bases for the vectors spaces is this.
Step 2:
Let us apply the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
By the proposition that for the tensors space with respect to any field and any finite number of finite-dimensional the field vectors spaces and the field or the tensor product of any finite-dimensional vectors spaces over any field, the transition of any standard bases or the components is a square matrix, and the inverse matrix is the product of the inverses, \(f'^{l_1, ..., l_k} = {{M_1}^{-1}}^{l_1}_{m_1} ... {{M_k}^{-1}}^{l_k}_{m_k} f^{m_1, ..., m_k}\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for tensors space w.r.t. field and \(k\) finite-dimensional vectors spaces over field and field, transition of components of tensor w.r.t. standard bases w.r.t. bases for vectors spaces is this
Topics
About:
vectors space
The table of contents of this article
Starting Context
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 components of any tensor with respect to the standard bases w.r.t. any bases for the vectors 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(\{V_1, ..., V_k\}\): \(\subseteq \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(L (V_1, ..., V_k: F)\): \(= \text{ the tensors space }\)
\(\{B_1, ..., B_k\}\): \(B_j \in \{\text{ the bases for } V_j\} = \{{b_j}_l \vert 1 \le l \le dim V_j\}\)
\(\{B'_1, ..., B'_k\}\): \(B'_j \in \{\text{ the bases for } V_j\} = \{{b_j}_l \vert 1 \le l \le dim V_j\}\)
\(\{B^*_1, ..., B^*_k\}\): \(B^*_j = \text{ the dual basis of } B_j = \{{b_j}^l \vert 1 \le l \le dim V_j\}\)
\(\{B'^*_1, ..., B'^*_k\}\): \(B'^*_j = \text{ the dual basis of } B_j = \{{b_j}^l \vert 1 \le l \le dim V_j\}\)
\(B^*\): \(= \{{b_1}^{j_1} \otimes ... \otimes {b_k}^{j_k} \vert {b_l}^{j_l} \in B^*_l\}\), \(\in \{\text{ the bases for } L (V_1, ..., V_k: F)\}\)
\(B'^*\): \(= \{{b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} \vert {b'_l}^{j_l} \in B'^*_l\}\), \(\in \{\text{ the bases for } L (V_1, ..., V_k: F)\}\)
//
Statements:
\({b'_j}_l = {b_j}_m {M_j}^m_l\)
\(\implies\)
\(\forall f = f_{j_1, ..., j_k} {b_1}^{j_1} \otimes ... \otimes {b_k}^{j_k} = f'_{l_1, ..., l_k} {b'_1}^{l_1} \otimes ... \otimes {b'_k}^{j_l} \in L (V_1, ..., V_k: F) (f'_{l_1, ..., l_k} = f_{j_1, ..., j_k} {M_1}^{j_1}_{l_1} ... {M_k}^{j_k}_{l_k})\)
//
2: Proof
Whole Strategy: just apply 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 vectors spaces is this, the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this, and the proposition that for the tensors space with respect to any field and any finite number of finite-dimensional the field vectors spaces and the field or the tensor product of any finite-dimensional vectors spaces over any field, the transition of any standard bases or the components is a square matrix, and the inverse matrix is the product of the inverses; Step 1: see that \({b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} = {{M_1}^{-1}}^{j_1}_{l_1} ... {{M_k}^{-1}}^{j_k}_{l_k} {b_1}^{l_1} \otimes ... \otimes {b_k}^{l_k}\); Step 2: conclude the proposition.
Step 1:
\({b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} = {{M_1}^{-1}}^{j_1}_{l_1} ... {{M_k}^{-1}}^{j_k}_{l_k} {b_1}^{l_1} \otimes ... \otimes {b_k}^{l_k}\), by 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 vectors spaces is this.
Step 2:
Let us apply the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
By the proposition that for the tensors space with respect to any field and any finite number of finite-dimensional the field vectors spaces and the field or the tensor product of any finite-dimensional vectors spaces over any field, the transition of any standard bases or the components is a square matrix, and the inverse matrix is the product of the inverses, \(f'_{l_1, ..., l_k} = f_{j_1, ..., j_k} {M_1}^{j_1}_{l_1} ... {M_k}^{j_k}_{l_k}\).
References
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description/proof of that for tensors space or tensor product of vectors spaces, transition of standard bases or components is square matrix, and inverse is product of inverses
Topics
About:
vectors space
The table of contents of this article
Starting Context
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 finite number of finite-dimensional the field vectors spaces and the field or the tensor product of any finite-dimensional vectors spaces over any field, the transition of any standard bases or the components is a square matrix, and the inverse matrix is the product of the inverses.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(\{V_1, ..., V_k\}\): \(\subseteq \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(L (V_1, ..., V_k: F)\): \(= \text{ the tensors space }\)
\(V_1 \otimes ... \otimes V_k\): \(= \text{ the tensor product }\)
\(\{B_1, ..., B_k\}\): \(B_j \in \{\text{ the bases of } V_j\} = \{{b_j}_l \vert 1 \le l \le dim V_j\}\)
\(\{B'_1, ..., B'_k\}\): \(B'_j \in \{\text{ the bases of } V_j\} = \{{b'_j}_l \vert 1 \le l \le dim V_j\}\)
\(\{B^*_1, ..., B^*_k\}\): \(B^*_j = \text{ the dual basis of } B_j = \{{b_j}^l \vert 1 \le l \le dim V_j\}\)
\(\{B'^*_1, ..., B'^*_k\}\): \(B'^*_j = \text{ the dual basis of } B'_j = \{{b'_j}^l \vert 1 \le l \le dim V_j\}\)
\(B^*\): \(= \{{b_1}^{j_1} \otimes ... \otimes {b_k}^{j_k} \vert {b_l}^{j_l} \in B^*_l\}\), \(\in \{\text{ the bases for } L (V_1, ..., V_k: F)\}\)
\(B'^*\): \(= \{{b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} \vert {b'_l}^{j_l} \in B'^*_l\}\), \(\in \{\text{ the bases for } L (V_1, ..., V_k: F)\}\)
\(B\): \(= \{[(({b_1}_{j_1}, ..., {b_k}_{j_k}))] \vert {b_l}_{j_l} \in B_l\}\), \(\in \{\text{ the bases for } V_1 \otimes ... \otimes V_k\}\)
\(B'\): \(= \{[(({b'_1}_{j_1}, ..., {b'_k}_{j_k}))] \vert {b'_l}_{j_l} \in B'_l\}\), \(\in \{\text{ the bases for } V_1 \otimes ... \otimes V_k\}\)
//
Statements:
\({b'_j}_l = {b_j}_m {M_j}^m_l\)
\(\implies\)
(
(
\({b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} = {{M_1}^{-1}}^{j_1}_{l_1} ... {{M_k}^{-1}}^{j_k}_{l_k} {b_1}^{l_1} \otimes ... \otimes {b_k}^{l_k}\)
\(\land\)
\(M^{j_1, ..., j_k}_{l_1, ..., l_k} := {{M_1}^{-1}}^{j_1}_{l_1} ... {{M_k}^{-1}}^{j_k}_{l_k}\) is a square matrix
\(\land\)
\({M^{-1}}^{j_1, ..., j_k}_{l_1, ..., l_k} = {M_1}^{j_1}_{l_1} ... {M_k}^{j_k}_{l_k}\)
)
\(\land\)
(
\([(({b'_1}_{j_1}, ..., {b'_k}_{j_k}))] = [(({b_1}_{l_1}, ..., {b_k}_{l_k}))] {M_1}^{l_1}_{j_1} ... {M_k}^{l_k}_{j_k}\)
\(\land\)
\(M^{l_1, ..., l_k}_{j_1, ..., j_k} := {M_1}^{l_1}_{j_1} ... {M_k}^{l_k}_{j_k}\) is a square matrix
\(\land\)
\({M^{-1}}^{l_1, ..., l_k}_{j_1, ..., j_k} = {{M_1}^{-1}}^{l_1}_{j_1} ... {{M_k}^{-1}}^{l_k}_{j_k}\)
)
)
//
2: Proof
Whole Strategy: Step 1: see that the transition for the bases for \(L (V_1, ..., V_k: F)\) holds; Step 2: see that \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) is a square matrix; Step 3: see that the inverse of \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) is as is claimed; Step 4: see that the transition for the bases for \(V_1 \otimes ... \otimes V_k\) holds; Step 5: see that \(M^{l_1, ..., l_k}_{j_1, ..., j_k}\) is a square matrix; Step 6: see that the inverse of \(M^{l_1, ..., l_k}_{j_1, ..., j_k}\) is as is claimed; Step 7: see that also the components transitions are some square matrices.
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.
\({b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} = {{M_1}^{-1}}^{j_1}_{l_1} ... {{M_k}^{-1}}^{j_k}_{l_k} {b_1}^{l_1} \otimes ... \otimes {b_k}^{l_k}\) holds, by 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.
Step 2:
\(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) may not look like any matrix unless \(k = 1\), because it is a multi-dimensional array.
But the set of the combinations, \(J := \{(j_1, ..., j_k) \vert 1 \le j_1 \le dim V_1, ..., 1 \le j_k \le dim V_k\}\), whose order is \(dim V_1 * ... * dim V_k\), can be regarded as a single index set. And \(J = \{(l_1, ..., l_k) \vert 1 \le l_1 \le dim V_1, ..., 1 \le l_k \le dim V_k\}\) can be regarded as a single index set.
So, \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) can be regarded to be a \((dim V_1 * ... * dim V_k) \times (dim V_1 * ... * dim V_k)\) square matrix: the order of the index, \(J\), can be chosen arbitrary, for example, \((1, ..., 1), (1, ..., 2), ..., (dim V_1, ..., dim V_k)\), which is the most natural one.
Also each of \({b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k}\) and \({b_1}^{l_1} \otimes ... \otimes {b_k}^{l_k}\) can be regarded to be a column vector (a kind of matrix) with the chosen order of \(J\).
Then, \({b'_1}^{j_1} \otimes ... \otimes {b'_k}^{j_k} = M^{j_1, ..., j_k}_{l_1, ..., l_k} {b_1}^{l_1} \otimes ... \otimes {b_k}^{l_k}\) is the usual multiplication of the square matrix and the column vector.
In fact, that is natural, because it is a transition of bases for a vectors space, and although we denote the basis, \(B^*\), as \(\{{b_1}^{j_1} \otimes ... \otimes {b_k}^{j_k}\}\), just because that is somehow convenient for clarifying what each element is, in fact, the basis can be denoted also like \(\{e^1, ..., e^{dim V_1 * ... * dim V_k}\}\).
For any another matrix, \(N^{m_1, ..., m_k}_{j_1, ..., j_k}\), with the chosen order of \(J\), \(N^{m_1, ..., m_k}_{j_1, ..., j_k} M^{j_1, ..., j_k}_{l_1, ..., l_k}\) is the usual multiplication of the square matrices.
Step 3:
The reason why we want to regard \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) as a square matrix is that we want to take the inverse of it, while the reason why we want to take the inverse of it is that the inverse represents the transition of the components, by the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this: it certainly has the inverse, because it is a transition of bases.
The inverse of \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) is the matrix, \(N^{m_1, ..., m_k}_{j_1, ..., j_k}\), such that \(N^{m_1, ..., m_k}_{j_1, ..., j_k} M^{j_1, ..., j_k}_{l_1, ..., l_k} = \delta^{m_1}_{l_1} ... \delta^{m_k}_{l_k}\): the product of the reverse order is automatically guaranteed to be \(I\), because we know that \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) is invertible: from \(N M = I\), \(M N = M N M M^{-1} = M I M^{-1} = I\).
There is \({M_1}^{m_1}_{j_1} ... {M_k}^{m_k}_{j_k}\), which is a \((dim V_1 * ... * dim V_k) \times (dim V_1 * ... * dim V_k)\) matrix.
\({M_1}^{m_1}_{j_1} ... {M_k}^{m_k}_{j_k} M^{j_1, ..., j_k}_{l_1, ..., l_k} = {M_1}^{m_1}_{j_1} ... {M_k}^{m_k}_{j_k} {{M_1}^{-1}}^{j_1}_{l_1} ... {{M_k}^{-1}}^{j_k}_{l_k} = {M_1}^{m_1}_{j_1} {{M_1}^{-1}}^{j_1}_{l_1} ... {M_k}^{m_k}_{j_k} ... {{M_k}^{-1}}^{j_k}_{l_k} = \delta^{m_1}_{l_1} ... \delta^{m_k}_{l_k}\), which means that \({M_1}^{m_1}_{j_1} ... {M_k}^{m_k}_{j_k}\) is the inverse of \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\).
So, \({M^{-1}}^{j_1, ..., j_k}_{l_1, ..., l_k} = {M_1}^{j_1}_{l_1} ... {M_k}^{j_k}_{l_k}\).
\({M^{-1}}^{j_1, ..., j_k}_{l_1, ..., l_k}\) represents the transition of tensor components, by the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
Step 4:
\(B\) and \(B'\) are indeed some bases for \(V_1 \otimes ... \otimes V_k\), by the proposition that the tensor product of any \(k\) finite-dimensional vectors spaces has the basis that consists of the classes induced by any basis elements.
\([(({b'_1}_{j_1}, ..., {b'_k}_{j_k}))] = [(({b_1}_{l_1}, ..., {b_k}_{l_k}))] {M_1}^{l_1}_{j_1} ... {M_k}^{l_k}_{j_k}\) holds, by the proposition that for the tensor product of any \(k\) finite-dimensional vectors spaces over any field, the transition of the standard bases with respect to any bases for the vector spaces is this.
Step 5:
\(M^{l_1, ..., l_k}_{j_1, ..., j_k}\) may not look like any matrix unless \(k = 1\), because it is a multi-dimensional array.
But the set of the combinations, \(J := \{(j_1, ..., j_k) \vert 1 \le j_1 \le dim V_1, ..., 1 \le j_k \le dim V_k\}\), whose order is \(dim V_1 * ... * dim V_k\), can be regarded as a single index set. And \(J = \{(l_1, ..., l_k) \vert 1 \le l_1 \le dim V_1, ..., 1 \le l_k \le dim V_k\}\) can be regarded as a single index set.
So, \(M^{l_1, ..., l_k}_{j_1, ..., j_k}\) can be regarded to be a \((dim V_1 * ... * dim V_k) \times (dim V_1 * ... * dim V_k)\) square matrix: the order of the index, \(J\), can be chosen arbitrary, for example, \((1, ..., 1), (1, ..., 2), ..., (dim V_1, ..., dim V_k)\), which is the most natural one.
Also each of \([(({b'_1}_{j_1}, ..., {b'_k}_{j_k}))]\) and \([(({b_1}_{l_1}, ..., {b_k}_{l_k}))]\) can be regarded to be a row vector (a kind of matrix) with the chosen order of \(J\).
Then, \([(({b'_1}_{j_1}, ..., {b'_k}_{j_k}))] = [(({b_1}_{l_1}, ..., {b_k}_{l_k}))] M^{l_1, ..., l_k}_{j_1, ..., j_k}\) is the usual multiplication of the row vector and the square matrix.
In fact, that is natural, because it is a transition of bases for a vectors space, and although we denote the basis, \(B\), as \(\{[(({b_1}_{j_1}, ..., {b_k}_{j_k}))]\}\), just because that is somehow convenient for clarifying what each element is, in fact, the basis can be denoted also like \(\{e_1, ..., e_{dim V_1 * ... * dim V_k}\}\).
For any another matrix, \(N^{m_1, ..., m_k}_{l_1, ..., l_k}\), with the chosen order of \(J\), \(N^{m_1, ..., m_k}_{l_1, ..., l_k} M^{l_1, ..., l_k}_{j_1, ..., j_k}\) is the usual multiplication of the square matrices.
Step 6:
The reason why we want to regard \(M^{l_1, ..., l_k}_{j_1, ..., j_k}\) as a square matrix is that we want to take the inverse of it, while the reason why we want to take the inverse of it is that the inverse represents the transition of the components: it certainly has the inverse, because it is a transition of bases.
The inverse of \(M^{l_1, ..., l_k}_{j_1, ..., j_k}\) is the matrix, \(N^{m_1, ..., m_k}_{l_1, ..., l_k}\), such that \(N^{m_1, ..., m_k}_{l_1, ..., l_k} M^{l_1, ..., l_k}_{j_1, ..., j_k} = \delta^{m_1}_{j_1} ... \delta^{m_k}_{j_k}\): the product of the reverse order is automatically guaranteed to be \(I\), because we know that \(M^{l_1, ..., l_k}_{j_1, ..., j_k}\) is invertible.
There is \({{M_1}^{-1}}^{m_1}_{l_1} ... {{M_k}^{-1}}^{m_k}_{l_k}\), which is a \((dim V_1 * ... * dim V_k) \times (dim V_1 * ... * dim V_k)\) matrix.
\({{M_1}^{-1}}^{m_1}_{l_1} ... {{M_k}^{-1}}^{m_k}_{l_k} M^{l_1, ..., l_k}_{j_1, ..., j_k} = {{M_1}^{-1}}^{m_1}_{l_1} ... {{M_k}^{-1}}^{m_k}_{l_k} {M_1}^{l_1}_{j_1} ... {M_k}^{l_k}_{j_k} = {{M_1}^{-1}}^{m_1}_{l_1} {M_1}^{l_1}_{j_1} ... {{M_k}^{-1}}^{m_k}_{l_k} ... {M_k}^{l_k}_{j_k} = \delta^{m_1}_{j_1} ... \delta^{m_k}_{j_k}\), which means that \({{M_1}^{-1}}^{m_1}_{l_1} ... {{M_k}^{-1}}^{m_k}_{l_k}\) is the inverse of \(M^{l_1, ..., l_k}_{j_1, ..., j_k}\).
So, \({M^{-1}}^{l_1, ..., l_k}_{j_1, ..., j_k} = {{M_1}^{-1}}^{l_1}_{j_1} ... {{M_k}^{-1}}^{l_k}_{j_k}\).
\({M^{-1}}^{l_1, ..., l_k}_{j_1, ..., j_k}\) represents the transition of tensor components, by the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
Step 7:
So, we have gotten the components transitions, \({M^{-1}}^{j_1, ..., j_k}_{l_1, ..., l_k}\) and \({M^{-1}}^{l_1, ..., l_k}_{j_1, ..., j_k}\), also which are some matrices likewise, and the inverses are \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) and \(M^{l_1, ..., l_k}_{j_1, ..., j_k}\).
3: Note
We are talking about the transition of bases or the transition of components, not about the tensor components themselves: for example, for a tensor, \(t \in L ({V_1}^*, ..., {V_k}^*, V_1, ..., V_k: F)\), \(t\) can be expressed with the components with respect to a standard basis as \(t^{j_1, ..., j_k}_{l_1, ..., l_k}\), which resembles \(M^{j_1, ..., j_k}_{l_1, ..., l_k}\) in form, but it is not so natural to regard it as a matrix: thinking of \(t (v^1, ... v^k, v_1, ..., v_k) = t^{j_1, ..., j_k}_{l_1, ..., l_k} {v^1}_{j_1} ... {v^k}_{j_k} {v_1}^{l_1} ... {v_k}^{l_k}\), in order to regard it as the multiplication of a matrix and a column vector, the column vector would be like \(({v^1}_1 ... {v^k}_1 {v_1}^1 ... {v_k}^1, {v^1}_1 ... {v^k}_1 {v_1}^1 ... {v_k}^2, ..., {v^1}_{dim V_1} ... {v^k}_{dim V_k} {v_1}^{dim V_1} ... {v_k}^{dim V_k})^t\) instead of like \(({v_1}_1, ..., {v_1}_{dim V_1}, ..., {v_k}^1, ..., {v_k}^{dim V_k})\), which might not be particularly meaningful (if it is meaningful for your situation, of course, it is fine).
The reason why that might not be meaningful is that \(t\) is not any linear map in general (refer to the proposition that a multilinear map is not necessarily linear): \(t\) is \(: V_1^{*} \times ... \times V_k^{*} \times V_1 \times ... \times V_k \to F\), a non-linear map from a \((2 (dim V_1 + ... + dim V_k))\)-dimensional vectors space into \(F\), and thinking of the \((dim V_1 * ... * dim V_k)^2 \times (dim V_1 * ... * dim V_k)^2\) matrix is not meaningful in general.
Of course, as any matrix is just an arrangement of some ring elements, you can always regard \(t^{j_1, ..., j_k}_{l_1, ..., l_k}\) as a matrix, if you want to.
References
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