Tensor Product - Knapp - Theorem 6.10 .... Further Question

[Math Amateur]

I am reading Anthony W. Knapp's book: Basic Algebra in order to understand tensor products ... ...

I need some help with a further aspect of the proof of Theorem 6.10 in Section 6 of Chapter VI: Multilinear Algebra ...

The text of Theorem 6.10 reads as follows:

The above proof mentions Figure 6.1 which is provided below ... as follows:

In the above text, in the proof of Theorem 6.10 under "PROOF OF EXISTENCE" we read:

" ... ... The bilinearity of $b$ shows that $B_1$ maps $V_0$ to $0$. By Proposition 2.25, $B_1$ descends to a linear map $B \ : \ V_1/V_0 \longrightarrow U$, and we have $Bi = b$. "
My questions are as follows:Question 1

Can someone please give a detailed demonstration of how the bilinearity of $b$ shows that $B_1$ maps $V_0$ to $0$?Question 2

Can someone please explain what is meant by "$B_1$ descends to a linear map $B \ : \ V_1/V_0 \longrightarrow U$" and show why this is the case ... also showing why/how $Bi = b$ ... ... ?

Hope someone can help ...

Peter===========================================================*** EDIT ***

The above post mentions Proposition 2.25 so I am providing the text ... as follows:

============================================================*** EDIT 2 ***

After a little reflection it appears that the answer to my Question 2 above should "fall out" or result from matching the situation in Theorem 6.10 to that in Proposition 2.25 ... also I have noticed a remark of Knapp's following the statement of Proposition 2.25 which reads as follows:

So that explains the language: "$B_1$ descends to a linear map $B \ : \ V_1/V_0 \longrightarrow U$" ... ...

BUT ... I remain perplexed over question 1 ...

Peter

 

[Math Amateur]

I am reading Anthony W. Knapp's book: Basic Algebra in order to understand tensor products ... ...

I need some help with a further aspect of the proof of Theorem 6.10 in Section 6 of Chapter VI: Multilinear Algebra ...

The text of Theorem 6.10 reads as follows:

The above proof mentions Figure 6.1 which is provided below ... as follows:

In the above text, in the proof of Theorem 6.10 under "PROOF OF EXISTENCE" we read:

" ... ... The bilinearity of $b$ shows that $B_1$ maps $V_0$ to $0$. By Proposition 2.25, $B_1$ descends to a linear map $B \ : \ V_1/V_0 \longrightarrow U$, and we have $Bi = b$. "
My questions are as follows:Question 1

Can someone please give a detailed demonstration of how the bilinearity of $b$ shows that $B_1$ maps $V_0$ to $0$?Question 2

Can someone please explain what is meant by "$B_1$ descends to a linear map $B \ : \ V_1/V_0 \longrightarrow U$" and show why this is the case ... also showing why/how $Bi = b$ ... ... ?

Hope someone can help ...

Peter===========================================================*** EDIT ***

The above post mentions Proposition 2.25 so I am providing the text ... as follows:

============================================================*** EDIT 2 ***

After a little reflection it appears that the answer to my Question 2 above should "fall out" or result from matching the situation in Theorem 6.10 to that in Proposition 2.25 ... also I have noticed a remark of Knapp's following the statement of Proposition 2.25 which reads as follows:

So that explains the language: "$B_1$ descends to a linear map $B \ : \ V_1/V_0 \longrightarrow U$" ... ...

BUT ... I remain perplexed over question 1 ...

Peter

... ... BUT NOTE ... after further reflection and work ...

I am having trouble applying Proposition 2.25 to Theorem 6.10 ... SO ... Question 2 remains a problem ... hope someone can help ...AND ... I remain perplexed over question 1 ...

Peter

 

[lavinia]

Here is a way of thinking that may help with tensor products.

When one multiplies two numbers $xy$ ,for instance real or complex numbers, then the rules of ordinary multiplication and addition say that

$(ax)y = a(xy) = x(ay)$, $(x_1 + x_2)y = x_1y + x_2y$ and $x(y_1 + y_2) = xy_1 + xy_2$

One would like to have exactly the same rules apply when $x$ and $y$ are vectors in two(usually different) vector spaces and $a$ is a scalar. Since it makes no sense to multiply vectors one needs to decide what a multiplication would mean. Your book defines the tensor product as the solution to a universal mapping problem. andrewkirk defines it in another way.

Whatever the construction one gets "products" $x⊗y$ in a new vector space, $V⊗W$, that satisfy the rules of multiplication and addition. That is:

* $a(x⊗y) = (ax)⊗y = x⊗(ay)$, $(x_1+x_2)⊗y = x_1⊗y + x_2⊗y$ and $x⊗(y_1+y_2) = x⊗y_1+x⊗y_2$

This new vector space will be all linear combinations $Σ_{i}a_{i}x_{i}⊗y_{i}$ subject to the required rules of multiplication and addition. That is: the tensor product is just the vectors space of all symbols $Σ_{i}a_{i}x_{i}⊗y_{i}$ subject to the equivalence relations *

One can think of the tensor product of two vector spaces completely formally in this way. You tell yourself, " However it was constructed, this is what is must look like."

It is easy to check that a linear map from $V⊗W$ into another vector space,$U$, determines a bilinear map from $V$x$W$ into $U$ and visa versa.

Here is an exercise: Let $V$ be a vector space over the real numbers and view the complex numbers $C$ as a vectors space over the real numbers as well. What is the space $V⊗C$? Is it a complex vector space?

 

[lavinia]

Inner products can also be defined on free abelian groups. $g(x,y)$ is a symmetric positive definite bilinear form that takes values in the integers. There is the additional condition that for every linear map $L$ of $V$ into the integers there is a unique $v_0$ such that
$L(w) = g(w,v_0)$ for all $w$ in $L$.

- Show that there is only one inner product on the integers.