Basis of a Tensor Product - Cooperstein - Theorem 10.2

In summary, the basis of a tensor product is a set of vectors that span the entire space and can be calculated by taking the tensor product of the basis vectors from the individual vector spaces. The Cooperstein Theorem 10.2 states that the basis of a tensor product is equal to the tensor product of the basis vectors from the individual vector spaces. The basis of a tensor product is important because it allows for easier calculations and understanding of properties. The basis can change if the basis vectors from the individual vector spaces are changed, but it remains unique if the basis vectors are unchanged.
  • #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 an aspect of Theorem 10.2 regarding the basis of a tensor product ... ... Theorem 10.2 reads as follows:View attachment 5437I do not follow the proof of this Theorem as it appears to me that \(\displaystyle Z'\), the space spanned by \(\displaystyle X'\) is actually equal to \(\displaystyle Z\) ... but the Theorem implies it is a proper subset ... otherwise why mention \(\displaystyle Z'\) at all ...I will explain my confusion by taking a simple example involving two vector spaces \(\displaystyle V\) and \(\displaystyle W\) where \(\displaystyle V\) has basis \(\displaystyle B_1 = \{ v_1, v_2, v_3 \}\) and \(\displaystyle W\) has basis \(\displaystyle B_2 = \{ w_1, w_2 \} \) ... ...So ... following the proof we set \(\displaystyle B = \{ v \otimes w \ | \ v \in B_1, w \in B_2 \}\) ... ...Elements of \(\displaystyle B\) then are ... ... \(\displaystyle v_1 \otimes w_1, \ v_1 \otimes w_2, \ v_2 \otimes w_1, \ v_2 \otimes w_2, \ v_3 \otimes w_1, \ v_3 \otimes w_2\) ... ...and we have\(\displaystyle X' = B_1 \times B_2 \)

\(\displaystyle = \{ (v_1, w_1) , \ (v_1, w_2) , \ (v_2, w_1) , \ (v_2, w_2) , \ (v_3, w_1) , \ (v_3, w_2) \} \)Now Z' is the subspace of \(\displaystyle Z\) spanned by \(\displaystyle X'\) ... ...But \(\displaystyle B_1\) spans \(\displaystyle V\) and \(\displaystyle B_2\) spans \(\displaystyle W\) ... ... so surely \(\displaystyle B_1 \times B_2\) spans \(\displaystyle V \times W\) ... ...

... BUT ... this does not seem to be what the Theorem implies (although it is possible under the proof ...)

Can someone please clarify the above for me and critique my example ... how is \(\displaystyle Z'\) a proper subset of \(\displaystyle Z\) ...?Hope someone can help ...

Peter
 
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  • #2
Dear Peter,

Thank you for bringing up your confusion regarding Theorem 10.2 in Cooperstein's book. I can understand how the proof may seem contradictory to the statement of the theorem.

Firstly, let me clarify that Z' is indeed a proper subset of Z. This is because Z' is the subspace of Z spanned by the set X', while Z is the full tensor product space of V and W. In other words, Z' is a smaller space compared to Z.

Now, in your example, you have correctly identified the elements of B, which are the basis elements of the tensor product space V ⊗ W. However, when we define X' as the set B_1 × B_2, we are essentially taking the Cartesian product of the two basis sets. This means that we are considering ordered pairs of basis elements, rather than the individual basis elements themselves.

For example, in your set X', the element (v_1, w_1) is not the same as the basis element v_1 ⊗ w_1. The former is an ordered pair, while the latter is a single element in the tensor product space. So, while B_1 × B_2 may span V ⊗ W, it is not the same as the basis set of V ⊗ W.

Therefore, in the proof of Theorem 10.2, we are showing that the set X' is a basis for the subspace Z' of Z. This means that Z' is spanned by the elements in X', but it is not the full tensor product space Z. This is why Z' is a proper subset of Z.

I hope this explanation helps to clarify your confusion. If you have any further questions, please do not hesitate to ask.
 

Related to Basis of a Tensor Product - Cooperstein - Theorem 10.2

What is the Basis of a Tensor Product?

The basis of a tensor product is a set of vectors that span the entire space of the tensor product. This means that any vector in the tensor product can be expressed as a linear combination of the basis vectors.

How is the Basis of a Tensor Product Calculated?

The basis of a tensor product can be calculated by taking the tensor product of the basis vectors from the individual vector spaces that are being multiplied together. This results in a set of basis vectors for the tensor product space.

What is the Cooperstein Theorem 10.2?

The Cooperstein Theorem 10.2 is a mathematical theorem that states that the basis of a tensor product is equal to the tensor product of the basis vectors from the individual vector spaces. This theorem is an important result in linear algebra and is used in many applications.

Why is the Basis of a Tensor Product Important?

The basis of a tensor product is important because it allows us to represent any vector in the tensor product space using a linear combination of the basis vectors. This makes it easier to perform calculations and understand the properties of the tensor product.

Can the Basis of a Tensor Product Change?

Yes, the basis of a tensor product can change if the basis vectors from the individual vector spaces are changed. However, the basis of a tensor product is unique and remains the same as long as the basis vectors from the individual vector spaces are unchanged.

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