1026: For Tensors Space or Tensor Product of Vectors Spaces, Transition of Standard Bases or Components Is Square Matrix, and Inverse Is Product of Inverses

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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

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof
• 3: Note

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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 $k$ vectors spaces over field.
• 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 the tensor product of any $k$ finite-dimensional vectors spaces has the basis that consists of the classes induced by any basis elements.
• The reader admits 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.
• The reader admits 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.
• The reader admits 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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$\{V_1,...,V_k\}$: $\subseteq \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$L(V_1,...,V_k:F)$: $=\text{ the tensors space }$
$V_1\otimes ...\otimes V_k$: $=\text{ the tensor product }$
$\{B_1,...,B_k\}$: $B_j\in \{\text{ the bases of }{V}_j\}=\{{b_j}_l|1\le l\le dim{V}_j\}$
$\{B_1^{\prime },...,B_k^{\prime }\}$: $B_j^{\prime }\in \{\text{ the bases of }{V}_j\}=\{{b_j^{\prime }}_l|1\le l\le dim{V}_j\}$
$\{B_1^{\ast },...,B_k^{\ast }\}$: $B_j^{\ast }=\text{ the dual basis of }{B}_j=\{{b_j}^l|1\le l\le dim{V}_j\}$
$\{B_1^{\prime \ast },...,B_k^{\prime \ast }\}$: $B_j^{\prime \ast }=\text{ the dual basis of }{B}_j^{\prime }=\{{b_j^{\prime }}^l|1\le l\le dim{V}_j\}$
$B^{\ast }$: $=\{{b_1}^{j_1}\otimes ...\otimes {b_k}^{j_k}|{b_l}^{j_l}\in B_l^{\ast }\}$, $\in \{\text{ the bases for }L(V_1,...,V_k:F)\}$
$B^{\prime \ast }$: $=\{{b_1^{\prime }}^{j_1}\otimes ...\otimes {b_k^{\prime }}^{j_k}|{b_l^{\prime }}^{j_l}\in B_l^{\prime \ast }\}$, $\in \{\text{ the bases for }L(V_1,...,V_k:F)\}$
$B$: $=\{[(({b_1}_{j_1},...,{b_k}_{j_k}))]|{b_l}_{j_l}\in B_l\}$, $\in \{\text{ the bases for }{V}_1\otimes ...\otimes V_k\}$
$B^{\prime }$: $=\{[(({b_1^{\prime }}_{j_1},...,{b_k^{\prime }}_{j_k}))]|{b_l^{\prime }}_{j_l}\in B_l^{\prime }\}$, $\in \{\text{ the bases for }{V}_1\otimes ...\otimes V_k\}$
//

Statements:
${b_j^{\prime }}_l={b_j}_m{M_j}_l^m$
$⟹$
(
(
${b_1^{\prime }}^{j_1}\otimes ...\otimes {b_k^{\prime }}^{j_k}={{M_1}^{-1}}_{l_1}^{j_1}...{{M_k}^{-1}}_{l_k}^{j_k}{b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k}$
$\wedge$
$M_{l_1,...,l_k}^{j_1,...,j_k}:={{M_1}^{-1}}_{l_1}^{j_1}...{{M_k}^{-1}}_{l_k}^{j_k}$ is a square matrix
$\wedge$
${M^{-1}}_{l_1,...,l_k}^{j_1,...,j_k}={M_1}_{l_1}^{j_1}...{M_k}_{l_k}^{j_k}$
)
$\wedge$
(
$[(({b_1^{\prime }}_{j_1},...,{b_k^{\prime }}_{j_k}))]=[(({b_1}_{l_1},...,{b_k}_{l_k}))]{M_1}_{j_1}^{l_1}...{M_k}_{j_k}^{l_k}$
$\wedge$
$M_{j_1,...,j_k}^{l_1,...,l_k}:={M_1}_{j_1}^{l_1}...{M_k}_{j_k}^{l_k}$ is a square matrix
$\wedge$
${M^{-1}}_{j_1,...,j_k}^{l_1,...,l_k}={{M_1}^{-1}}_{j_1}^{l_1}...{{M_k}^{-1}}_{j_k}^{l_k}$
)
)
//

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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 $M_{l_1,...,l_k}^{j_1,...,j_k}$ is a square matrix; Step 3: see that the inverse of $M_{l_1,...,l_k}^{j_1,...,j_k}$ 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 $M_{j_1,...,j_k}^{l_1,...,l_k}$ is a square matrix; Step 6: see that the inverse of $M_{j_1,...,j_k}^{l_1,...,l_k}$ is as is claimed; Step 7: see that also the components transitions are some square matrices.

Step 1:

$B^{\ast }$ and $B^{\prime \ast }$ 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.

${b_1^{\prime }}^{j_1}\otimes ...\otimes {b_k^{\prime }}^{j_k}={{M_1}^{-1}}_{l_1}^{j_1}...{{M_k}^{-1}}_{l_k}^{j_k}{b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k}$ 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:

$M_{l_1,...,l_k}^{j_1,...,j_k}$ may not look like any matrix unless $k=1$, because it is a multi-dimensional array.

But the set of the combinations, $J:=\{(j_1,...,j_k)|1\le j_1\le dim{V}_1,...,1\le j_k\le dim{V}_k\}$, whose order is $dim{V}_1\ast ...\ast dim{V}_k$, can be regarded as a single index set. And $J=\{(l_1,...,l_k)|1\le l_1\le dim{V}_1,...,1\le l_k\le dim{V}_k\}$ can be regarded as a single index set.

So, $M_{l_1,...,l_k}^{j_1,...,j_k}$ can be regarded to be a $(dim{V}_1\ast ...\ast dim{V}_k)\times (dim{V}_1\ast ...\ast dim{V}_k)$ 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 ${b_1^{\prime }}^{j_1}\otimes ...\otimes {b_k^{\prime }}^{j_k}$ 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, ${b_1^{\prime }}^{j_1}\otimes ...\otimes {b_k^{\prime }}^{j_k}=M_{l_1,...,l_k}^{j_1,...,j_k}{b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k}$ 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, $B^{\ast }$, 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 $\{e^1,...,e^{dim{V}_1\ast ...\ast dim{V}_k}\}$.

For any another matrix, $N_{j_1,...,j_k}^{m_1,...,m_k}$, with the chosen order of $J$, $N_{j_1,...,j_k}^{m_1,...,m_k}{M}_{l_1,...,l_k}^{j_1,...,j_k}$ is the usual multiplication of the square matrices.

Step 3:

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, 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 $M_{l_1,...,l_k}^{j_1,...,j_k}$ is the matrix, $N_{j_1,...,j_k}^{m_1,...,m_k}$, such that $N_{j_1,...,j_k}^{m_1,...,m_k}{M}_{l_1,...,l_k}^{j_1,...,j_k}={\delta }_{l_1}^{m_1}...{\delta }_{l_k}^{m_k}$: 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: from $NM=I$, $MN=MNM{M}^{-1}=MI{M}^{-1}=I$.

There is ${M_1}_{j_1}^{m_1}...{M_k}_{j_k}^{m_k}$, which is a $(dim{V}_1\ast ...\ast dim{V}_k)\times (dim{V}_1\ast ...\ast dim{V}_k)$ matrix.

${M_1}_{j_1}^{m_1}...{M_k}_{j_k}^{m_k}{M}_{l_1,...,l_k}^{j_1,...,j_k}={M_1}_{j_1}^{m_1}...{M_k}_{j_k}^{m_k}{{M_1}^{-1}}_{l_1}^{j_1}...{{M_k}^{-1}}_{l_k}^{j_k}={M_1}_{j_1}^{m_1}{{M_1}^{-1}}_{l_1}^{j_1}...{M_k}_{j_k}^{m_k}...{{M_k}^{-1}}_{l_k}^{j_k}={\delta }_{l_1}^{m_1}...{\delta }_{l_k}^{m_k}$, which means that ${M_1}_{j_1}^{m_1}...{M_k}_{j_k}^{m_k}$ is the inverse of $M_{l_1,...,l_k}^{j_1,...,j_k}$.

So, ${M^{-1}}_{l_1,...,l_k}^{j_1,...,j_k}={M_1}_{l_1}^{j_1}...{M_k}_{l_k}^{j_k}$.

${M^{-1}}_{l_1,...,l_k}^{j_1,...,j_k}$ 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 4:

$B$ and $B^{\prime }$ 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.

$[(({b_1^{\prime }}_{j_1},...,{b_k^{\prime }}_{j_k}))]=[(({b_1}_{l_1},...,{b_k}_{l_k}))]{M_1}_{j_1}^{l_1}...{M_k}_{j_k}^{l_k}$ 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:

$M_{j_1,...,j_k}^{l_1,...,l_k}$ may not look like any matrix unless $k=1$, because it is a multi-dimensional array.

But the set of the combinations, $J:=\{(j_1,...,j_k)|1\le j_1\le dim{V}_1,...,1\le j_k\le dim{V}_k\}$, whose order is $dim{V}_1\ast ...\ast dim{V}_k$, can be regarded as a single index set. And $J=\{(l_1,...,l_k)|1\le l_1\le dim{V}_1,...,1\le l_k\le dim{V}_k\}$ can be regarded as a single index set.

So, $M_{j_1,...,j_k}^{l_1,...,l_k}$ can be regarded to be a $(dim{V}_1\ast ...\ast dim{V}_k)\times (dim{V}_1\ast ...\ast dim{V}_k)$ 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 $[(({b_1^{\prime }}_{j_1},...,{b_k^{\prime }}_{j_k}))]$ 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, $[(({b_1^{\prime }}_{j_1},...,{b_k^{\prime }}_{j_k}))]=[(({b_1}_{l_1},...,{b_k}_{l_k}))]M_{j_1,...,j_k}^{l_1,...,l_k}$ 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 $\{e_1,...,e_{dim{V}_1\ast ...\ast dim{V}_k}\}$.

For any another matrix, $N_{l_1,...,l_k}^{m_1,...,m_k}$, with the chosen order of $J$, $N_{l_1,...,l_k}^{m_1,...,m_k}{M}_{j_1,...,j_k}^{l_1,...,l_k}$ is the usual multiplication of the square matrices.

Step 6:

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: 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, $N_{l_1,...,l_k}^{m_1,...,m_k}$, such that $N_{l_1,...,l_k}^{m_1,...,m_k}{M}_{j_1,...,j_k}^{l_1,...,l_k}={\delta }_{j_1}^{m_1}...{\delta }_{j_k}^{m_k}$: 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.

There is ${{M_1}^{-1}}_{l_1}^{m_1}...{{M_k}^{-1}}_{l_k}^{m_k}$, which is a $(dim{V}_1\ast ...\ast dim{V}_k)\times (dim{V}_1\ast ...\ast dim{V}_k)$ matrix.

${{M_1}^{-1}}_{l_1}^{m_1}...{{M_k}^{-1}}_{l_k}^{m_k}{M}_{j_1,...,j_k}^{l_1,...,l_k}={{M_1}^{-1}}_{l_1}^{m_1}...{{M_k}^{-1}}_{l_k}^{m_k}{M_1}_{j_1}^{l_1}...{M_k}_{j_k}^{l_k}={{M_1}^{-1}}_{l_1}^{m_1}{M_1}_{j_1}^{l_1}...{{M_k}^{-1}}_{l_k}^{m_k}...{M_k}_{j_k}^{l_k}={\delta }_{j_1}^{m_1}...{\delta }_{j_k}^{m_k}$, which means that ${{M_1}^{-1}}_{l_1}^{m_1}...{{M_k}^{-1}}_{l_k}^{m_k}$ is the inverse of $M_{j_1,...,j_k}^{l_1,...,l_k}$.

So, ${M^{-1}}_{j_1,...,j_k}^{l_1,...,l_k}={{M_1}^{-1}}_{j_1}^{l_1}...{{M_k}^{-1}}_{j_k}^{l_k}$.

${M^{-1}}_{j_1,...,j_k}^{l_1,...,l_k}$ 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 7:

So, we have gotten the components transitions, ${M^{-1}}_{l_1,...,l_k}^{j_1,...,j_k}$ and ${M^{-1}}_{j_1,...,j_k}^{l_1,...,l_k}$, also which are some matrices likewise, and the inverses are $M_{l_1,...,l_k}^{j_1,...,j_k}$ and $M_{j_1,...,j_k}^{l_1,...,l_k}$.

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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, $t\in L({V_1}^{\ast },...,{V_k}^{\ast },V_1,...,V_k:F)$, $t$ can be expressed with the components with respect to a standard basis as $t_{l_1,...,l_k}^{j_1,...,j_k}$, which resembles $M_{l_1,...,l_k}^{j_1,...,j_k}$ in form, but it is not so natural to regard it as a matrix: thinking of $t(v^1,...v^k,v_1,...,v_k)=t_{l_1,...,l_k}^{j_1,...,j_k}{v^1}_{j_1}...{v^k}_{j_k}{v_1}^{l_1}...{v_k}^{l_k}$, 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 $:V_1^{\ast }\times ...\times V_k^{\ast }\times V_1\times ...\times V_k\to F$, a non-linear map from a $(2(dim{V}_1+...+dim{V}_k))$-dimensional vectors space into $F$, and thinking of the $(dim{V}_1\ast ...\ast dim{V}_k{)}^2\times (dim{V}_1\ast ...\ast dim{V}_k{)}^2$ matrix is not meaningful in general.

Of course, as any matrix is just an arrangement of some ring elements, you can always regard $t_{l_1,...,l_k}^{j_1,...,j_k}$ as a matrix, if you want to.

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References

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