1023: Tensor Product of k Finite-Dimensional Vectors Spaces Has Basis That Consists of Classes Induced by Basis Elements

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description/proof of that tensor product of $k$ finite-dimensional vectors spaces has basis that consists of classes induced by basis elements

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of tensor product of $k$ vectors spaces.
• The reader knows a definition of basis of module.
• 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 tensor product of tensors.
• The reader admits the proposition that for any multilinear map from any finite product vectors space, there is the unique linear map from the tensor product of the finite number of vectors spaces such that the multilinear map is the linear map after the canonical map from the product vectors space into the tensor product.

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Target Context

• The reader will have a description and a proof of 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.

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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:
$F$: $\in \{\text{ the fields }\}$
$\{V_1,...,V_k\}$: $\subseteq \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$\{B_1,...,B_k\}$: $B_l\in \{\text{ the bases for }{V}_l\}$
$V_1\otimes ...\otimes V_k$: $=\text{ the tensor product }$
$B$: $=\{[((b_{1,j_1},...,b_{k,j_k}))]|b_{l,j_l}\in B_l\}$
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Statements:
$B\in \{\text{ the bases for }{V}_1\otimes ...\otimes V_k\}$
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Let us call $B$ "the standard basis with respect to $\{B_1,...,B_k\}$": it is not determined unless $\{B_1,...,B_k\}$ is specified.

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2: Note

Each $V_j$ needs to be finite-dimensional, because Proof uses the dual of $B_j$: the definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space requires $V_j$ to be finite-dimensional.

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3: Proof

Whole Strategy: Step 1: see that $B$ spans $V_1\otimes ...\otimes V_k$; Step 2: see that $B$ is linearly independent.

Step 1:

Let us see that $B$ spans $V_1\otimes ...\otimes V_k$.

Each element of $V_1\otimes ...\otimes V_k$ is $[r^1((v_{1,1},...,v_{1,k}))+...+r^l((v_{l,1},...,v_{l,k}))]$ by the definition of tensor product of $k$ vectors spaces.

$[r^1((v_{1,1},...,v_{1,k}))+...+r^l((v_{l,1},...,v_{l,k}))]=r^1[((v_{1,1},...,v_{1,k}))]+...+r^l[((v_{l,1},...,v_{l,k}))]$, which is by the definition of quotient vectors space of vectors space by sub-'vectors space'.

Let us look at each $r^j[((v_{j,1},...,v_{j,k}))]$.

$r^j[((v_{j,1},...,v_{j,k}))]=r^j[((\sum _{m_1\in S_{1,j}}{v}_{j,1}^{m_1}{b}_{1,m_1},...,\sum _{m_k\in S_{k,j}}{v}_{j,k}^{m_k}{b}_{k,m_k}))]$, where $S_{l,j}$ is a finite index set, $=r^j\sum _{m_1\in S_{1,j}}{v}_{j,1}^{m_1}[((b_{1,m_1},...,\sum _{m_k\in S_{k,j}}{v}_{j,k}^{m_k}{b}_{k,m_k}))]$, which is by "the 2 legitimate rules", $=...=r^j\sum _{m_1\in S_{1,j}}{v}_{j,1}^{m_1}...\sum _{m_k\in S_{k,j}}{v}_{j,k}^{m_k}[((b_{1,m_1},...,b_{k,m_k}))]$, which means that $r^j[((v_{j,1},...,v_{j,k}))]$ is a linear combination of $B$.

So, $[r^1((v_{1,1},...,v_{1,k}))+...+r^l((v_{l,1},...,v_{l,k}))]=r^1[((v_{1,1},...,v_{1,k}))]+...+r^l[((v_{l,1},...,v_{l,k}))]$ is a linear combination of $B$.

So, $B$ spans $V_1\otimes ...\otimes V_k$.

Note that this step did not use finite-dimensional-ness of $V_j$ s, so, $B$ spans $V_1\otimes ...\otimes V_k$ even when $V_j$ s are infinite-dimensional.

Step 2:

Let us see that $B$ is linearly independent.

Let $r^1[((b_{1,j_{1,1}},...,b_{k,j_{1,k}}))]+...+r^l[((b_{1,j_{l,1}},...,b_{k,j_{l,k}}))]=0$.

Let us take the dual basis of each $B_j$, $B_j^{\ast }=\{b_j^l\}$.

Let us take the multilinear map, $f_{j_{m,1},...,j_{m,k}}:V_1\times ...\times V_k\to F=b_1^{j_{m,1}}\otimes ...\otimes b_k^{j_{m,k}}$.

By the proposition that for any multilinear map from any finite product vectors space, there is the unique linear map from the tensor product of the finite number of vectors spaces such that the multilinear map is the linear map after the canonical map from the product vectors space into the tensor product, there is the unique linear map, $f_{j_{m,1},...,j_{m,k}}^{\prime }:V_1\otimes ...\otimes V_k\to F$, such that $f_{j_{m,1},...,j_{m,k}}=f_{j_{m,1},...,j_{m,k}}^{\prime }\circ g$, where $g:V_1\times ...\times V_k\to V_1\otimes ...\otimes V_k,(v_1,...,v_k)\mapsto [((v_1,...,v_k))]$.

$0=f_{j_{m,1},...,j_{m,k}}^{\prime }(0)=f_{j_{m,1},...,j_{m,k}}^{\prime }(r^1[((b_{1,j_{1,1}},...,b_{k,j_{1,k}}))]+...+r^l[((b_{1,j_{l,1}},...,b_{k,j_{l,k}}))])=r^1{f}_{j_{m,1},...,j_{m,k}}^{\prime }([((b_{1,j_{1,1}},...,b_{k,j_{1,k}}))])+...+r^l{f}_{j_{m,1},...,j_{m,k}}^{\prime }([((b_{1,j_{l,1}},...,b_{k,j_{l,k}}))])$, because $f_{j_{m,1},...,j_{m,k}}^{\prime }$ is linear, but $f_{j_{m,1},...,j_{m,k}}^{\prime }([((b_{1,j_{n,1}},...,b_{k,j_{n,k}}))])=f_{j_{m,1},...,j_{m,k}}^{\prime }\circ g((b_{1,j_{n,1}},...,b_{k,j_{n,k}}))=f_{j_{m,1},...,j_{m,k}}((b_{1,j_{n,1}},...,b_{k,j_{n,k}}))=b_1^{j_{m,1}}\otimes ...\otimes b_k^{j_{m,k}}((b_{1,j_{n,1}},...,b_{k,j_{n,k}}))=b_1^{j_{m,1}}(b_{1,j_{n,1}})...b_k^{j_{m,k}}(b_{k,j_{n,k}})={\delta }_{m,n}$, so, $0=r^m$.

So, $B$ is linearly independent.

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References

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