Basis of a Tensor Product - Theorem 10.2 - Another Question .... ....

[Math Amateur]

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:View attachment 5438

A diagram involving the mappings \(\displaystyle \iota\) and \(\displaystyle \gamma'\) is as follows:View attachment 5439My questions are as follows:Question 1

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

Peter

 

[jvicens]

Dear Peter,

Thank you for your post and for sharing your questions on Theorem 10.2 in Bruce N. Cooperstein's book on Advanced Linear Algebra. The section on Tensor Products is indeed an important and complex topic in linear algebra, and I am happy to assist you in understanding the proof of Theorem 10.2.

To answer your first question, we know that there exists a multilinear map \gamma' \ : \ X \longrightarrow Z' because of the universal property of tensor products. The universal property states that for any multilinear map \gamma \ : \ X \times Y \longrightarrow Z, there exists a unique linear map \gamma' \ : \ X \otimes Y \longrightarrow Z' such that \gamma = \gamma' \circ \iota. In other words, for any multilinear map \gamma, there exists a unique linear map \gamma' that factors through the tensor product \iota and is thus a map from X \otimes Y to Z'.

As for your second question, the mechanics under the mapping \gamma' involve taking the elements in X \otimes X' and mapping them to elements in Z'. This is done by defining the linear map \gamma' on the basis elements of X \otimes X' and extending it to all elements in X \otimes X' using linearity. Since X \otimes X' is a basis for X \otimes X', the linear map \gamma' is uniquely determined by its values on the basis elements. Therefore, the elements in X \otimes X' will end up in Z' under the mapping \gamma'.

I hope this helps clarify the proof of Theorem 10.2 for you. If you have any further questions or need further clarification, please do not hesitate to ask.