Basis of a Tensor Product - Theorem 10.2 - Another Question

In summary, the conversation discusses the proof of Theorem 10.2 in Section 10.1 of Bruce N. Cooperstein's book, Advanced Linear Algebra (Second Edition). The focus is on understanding the basis of a tensor product and the existence of a multilinear map \gamma' that maps X' to a subspace of Z. The conversation also touches on the mechanics and representation of the map \gamma', as well as the definition of Z provided by Cooperstein.
  • #1
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I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

I am focused on Section 10.1 Introduction to Tensor Products ... ...

I need help with another aspect of the proof of Theorem 10.2 regarding the basis of a tensor product ... ...Theorem 10.2 reads as follows:
?temp_hash=f5d3c296ba033b45e904bd30300693b1.png
A diagram involving the mappings [itex]\iota[/itex] and [itex]\gamma'[/itex] is as follows:
?temp_hash=f5d3c296ba033b45e904bd30300693b1.png


My questions are as follows:Question 1

How do we know that there exists a multilinear map [itex]\gamma' \ : \ X \longrightarrow Z'[/itex] ?
Question 2What happens (what are the 'mechanics') under the mapping [itex]\gamma'[/itex] ... ... to the elements in X\X' ( that is [itex]X - X'[/itex])? How can we be sure that these elements end up in [itex]Z'[/itex] and not in Z\Z'? (see Figure 1 above)
Hope someone can help ...

Peter
 

Attachments

  • Cooperstein - 3 - Theorem 10.2     ....        ....png
    Cooperstein - 3 - Theorem 10.2 .... ....png
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  • Figure 1 - Cooperstein - Theorem 10.2             ... ... .png
    Figure 1 - Cooperstein - Theorem 10.2 ... ... .png
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  • #2
Re question 1:
Let ##V^\dagger\equiv V_1\times,...,\times V_m## and let's use angle brackets [itex]\langle...\rangle[/itex] to enclose components of Cartesian product spaces. Let the components of [itex]\mathscr{B}_j[/itex] be [itex]v_{j1},...,v_{jn_j}[/itex] where [itex]n_j[/itex] is the dim of [itex]V_j[/itex].
You haven't said what [itex]Z[/itex] is but let's assume it's the infinite-dimensional vector space over field [itex]F[/itex] with base set
$$\mathscr{B}^Z\equiv \{\langle u_1,...,u_m\rangle\ |\ \forall k:\ u_k\in V_k\}$$
More formally, [itex]Z[/itex] is the set of all functions from [itex]\mathscr{B}^Z[/itex] to [itex]F[/itex] for which each such function has finite support (ie is nonzero on only finitely many input values).
[itex]Z'[/itex] is the subset of [itex]Z[/itex] containing only functions whose support lies in [itex]X'[/itex], and it is easily shown to be a subspace.
Define the map [itex]\xi:X'\to Z'[/itex] that maps each Cartesian product of basis vectors [itex]\langle v_{1i_1},...,v_{mi_m}\rangle[/itex] to the function that returns zero for every input except [itex]\langle v_{1i_1},...,v_{mi_m}\rangle[/itex], for which it returns [itex]1_F[/itex]. Note that the image of [itex]\xi[/itex] is a basis for [itex]Z'[/itex].
Then a map [itex]\gamma':V^\dagger\to Z'[/itex] is multilinear, and agrees with [itex]\xi[/itex] on [itex]X'[/itex], if and only if it satisfies:
\begin{align*}
\gamma'\left(\left\langle \sum_{i_1}a_{1i_1}v_{1i_1},\, ...\, ,\sum_{i_m}a_{mi_m}v_{mi_m}\right\rangle\right)
&=\sum_{i_1}\sum_{i_2}\ ...\ \sum_{i_m} \prod_{k=1}^m a_{ki_k}\gamma'\left(\left\langle
v_{1i_1},\, ...\, ,v_{mi_m}\right\rangle\right)
\\&=
\sum_{i_1}\sum_{i_2}\ ...\ \sum_{i_m} \prod_{k=1}^m a_{ki_k}
\xi\left(\langle v_{1i_1},...,v_{mi_m}\rangle\right)
\end{align*}
where the first equality implements multilinearity and the second implements the requirement to agree with [itex]\xi[/itex].
Since [itex]X'[/itex] is a basis for [itex]V^\dagger[/itex], it follows that such a map [itex]\gamma'[/itex] exists. It is unique because the representation [itex]\sum_{i_k}a_{ki_k}v_{ki_k}[/itex] of the [itex]k[/itex]th coordinate of the input to [itex]\gamma'[/itex] is unique.
 
  • #3
Andrew,

Going through your post carefully shortly ...

Cooperstein provides a definition of Z in the introduction to Section 10.1 including in the proof of Theorem 10.1 ...

Relevant text from Cooperstein is as follows:
?temp_hash=a66ee8cd377f7c6b3c284cd0f239d602.png

?temp_hash=a66ee8cd377f7c6b3c284cd0f239d602.png

?temp_hash=a66ee8cd377f7c6b3c284cd0f239d602.png

?temp_hash=a66ee8cd377f7c6b3c284cd0f239d602.png


Hope that helps ...

Peter
 

Attachments

  • Cooperstein - 1 - Section 10.1 - PART 1     ....png
    Cooperstein - 1 - Section 10.1 - PART 1 ....png
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  • Cooperstein - 2 - Section 10.1 - PART 2     ....png
    Cooperstein - 2 - Section 10.1 - PART 2 ....png
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  • Cooperstein - 3 - Section 10.1 - PART 3     ....png
    Cooperstein - 3 - Section 10.1 - PART 3 ....png
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  • Cooperstein - 4 - Section 10.1 - PART 4     ....png
    Cooperstein - 4 - Section 10.1 - PART 4 ....png
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Related to Basis of a Tensor Product - Theorem 10.2 - Another Question

1. What is a tensor product?

A tensor product is a mathematical operation that combines two vector spaces to create a new vector space. It is denoted by the symbol ⊗ and is used in various fields of mathematics, physics, and engineering.

2. What is the basis of a tensor product?

The basis of a tensor product is a set of vectors that span the new vector space created by the tensor product. These vectors are formed by taking the tensor product of basis vectors from the original vector spaces.

3. What is Theorem 10.2 regarding the basis of a tensor product?

Theorem 10.2 states that the dimension of the tensor product of two vector spaces is equal to the product of the dimensions of the individual vector spaces. It also provides a formula for constructing the basis of the tensor product based on the basis of the original vector spaces.

4. How is Theorem 10.2 useful?

Theorem 10.2 is useful in various fields such as linear algebra, quantum mechanics, and computer science. It allows for the efficient computation of the basis of a tensor product, which is essential in solving complex problems involving vector spaces.

5. Can Theorem 10.2 be extended to more than two vector spaces?

Yes, Theorem 10.2 can be extended to any number of vector spaces. The dimension of the tensor product will be equal to the product of the dimensions of all the vector spaces, and the basis can be constructed using a similar formula as in the case of two vector spaces.

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