MHB Theorem 6.10 in Knapp's Basic Algebra: Exploring Bilinearity & Descending Maps

[Math Amateur]

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

I need some help with an aspect of Theorem 6.10 in Section 6 of Chapter VI: Multilinear Algebra ...

The text of Theorem 6.10 reads as follows:
View attachment 5405
View attachment 5406

The above proof mentions Figure 6.1 which is provided below ... as follows:View attachment 5407
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:
View attachment 5408

============================================================*** 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:

View attachment 5409So that explains the language: "$$B_1$$ descends to a linear map $$B \ : \ V_1/V_0 \longrightarrow U$$" ... ... BUT NOTE ...

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

 

[GJA]

Hi Peter,

Question 1

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

To show that $B_{1}$ maps $V_{0}$ to 0, it's enough to show that it sends basis elements to zero. Thus we need only compute $B_{1}$ on each of the four types of basis elements for $V_{0}$; these are the four expressions set apart in the text. For example, the first computation would be

$B_{1}((e_{1}+e_{2},f)-(e_{1},f)-(e_{2},f))=b(e_{1}+e_{2},f)-b(e_{1},f)-b(e_{2},f)=0,$

where the second equality follows from the bilinearity of $b$.

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 $$Bi = b$$ ... ... ?

In Proposition 2.25 take $V=V_{1}$ and $W=U$, and the quotient space to be $V_{1}/V_{0}$; i.e. $U=V_{0}$ (I am trying to match the notation in Proposition 2.25, so please do not take the above to mean $W=V_{0}$). Then we have $Bq=B_{1}$. Now a computation will give you that $B\iota = b$; i.e. start with $B\iota (e,f)$, then see if you can use the result of Proposition 2.25 together with the definition of $\iota$, $q$, and $B_{1}$ to obtain $B\iota(e,f)=b(e,f).$

Let me know if this is still unclear. Good luck!

 

[Math Amateur]

Hi Peter,
To show that $B_{1}$ maps $V_{0}$ to 0, it's enough to show that it sends basis elements to zero. Thus we need only compute $B_{1}$ on each of the four types of basis elements for $V_{0}$; these are the four expressions set apart in the text. For example, the first computation would be

$B_{1}((e_{1}+e_{2},f)-(e_{1},f)-(e_{2},f))=b(e_{1}+e_{2},f)-b(e_{1},f)-b(e_{2},f)=0,$

where the second equality follows from the bilinearity of $b$.
In Proposition 2.25 take $V=V_{1}$ and $W=U$, and the quotient space to be $V_{1}/V_{0}$; i.e. $U=V_{0}$ (I am trying to match the notation in Proposition 2.25, so please do not take the above to mean $W=V_{0}$). Then we have $Bq=B_{1}$. Now a computation will give you that $B\iota = b$; i.e. start with $B\iota (e,f)$, then see if you can use the result of Proposition 2.25 together with the definition of $\iota$, $q$, and $B_{1}$ to obtain $B\iota(e,f)=b(e,f).$

Let me know if this is still unclear. Good luck!

Thanks GJA ... most helpful ...

BUT ... I am still a bit perplexed with showing why/how $$Bi = b$$ ...I will try to explain my issues ...

Now Proposition 2.25 has a situation as shown below in Figure 1 ...View attachment 5414At this point you suggest putting $$V = V_1$$ and $$W = U$$ giving us the following situation ...
View attachment 5415Now ... for us to get $$Bq = B_1$$ we require Figure 2 to look like Figure 3 below ... not quite sure how the assignment of maps in this way is justified, mind ...
...

https://www.physicsforums.com/attachments/5416
I am bothered by what is the exact nature of U ... (what is the form of the elements of U ... and why are they a particular form) ... .. and I am also worried and unsure about justifying that $$B_1$$ is a map between $$V_1$$ and $$U$$ ... ...Can you elaborate and clarify ... ...


To further indicate my worries ... consider the following ...Knapp indicates that $$U$$ is the codomain of a map $$b$$ given by:$$b \ : \ E \times F \longrightarrow U$$ ... ... as in Figure 6.1 shown below:https://www.physicsforums.com/attachments/5417
But Knapp also defines $$U$$ as the codomain of the map $$B_1$$ which Knapp defines as follows:$$B_1 ( \sum_{ (finite) } c_i ( e_i , f_i ) ) = \sum_{ (finite) } c_i b ( e_i , f_i )$$
Now, for the above to 'work', the codomain of $$b$$ must be the same as the codomain of $$B_1$$ ... ... BUT ... ... is this the case? ... ... and if it is how do we show this rigorously ...

Note also that when you write:

" ... ... Then we have $Bq=B_{1}$. Now a computation will give you that $B\iota = b$; i.e. start with $B\iota (e,f)$, then see if you can use the result of Proposition 2.25 together with the definition of $\iota$, $q$, and $B_{1}$ to obtain $B\iota(e,f)=b(e,f).$ ... ... "I am unsure of exactly how we get from $$Bq = b$$ to $$B \iota = b$$ ... can you be more explicit ... ...


Hope you can help clarify the above issues ...

Peter

 

[GJA]

Hi Peter,

You're trying to prove a so-called "universal" property of the tensor product of two vector spaces (see Universal Property section of https://en.wikipedia.org/wiki/Tensor_product). What you're trying to prove is a semi-remarkable property of the tensor product - namely that there is a bilinear map from $E\times F$ to the tensor product such that if $U$ is any given vector space and $b$ is any bilinear map from $E\times F$ to $U$, then there is a unique linear map ($B$ in this author's notation) so that the diagram in Figure 6.1 commutes (i.e. $B\iota = b$). So $U$ is a given vector space and $b$ is a given bilinear map.

I am bothered by what is the exact nature of U ... (what is the form of the elements of U ... and why are they a particular form) ... .. and I am also worried and unsure about justifying that $$B_1$$ is a map between $$V_1$$ and $$U$$ ... ...

As mentioned above, $U$ is a given vector space, nothing more is known about it. Since $b$ is a map whose codomain is $U$ and $B_{1}$ is defined via $b$, so too is $B_{1}$'s codomain $U$.

BUT ... I am still a bit perplexed with showing why/how $$Bi = b$$ ...

To get you going on the computation, start with $B\iota(e,f) = B((e,f) + V_{0})$ where we have applied the definition of $\iota.$ There are a few more steps after this initial one, but not too many. Remember that using the definition of $q$ and $B_{1}$ will be needed as well as the result from Proposition 2.25.

Hopefully this helps shed some light on the matter.

 

[Math Amateur]

Hi Peter,

You're trying to prove a so-called "universal" property of the tensor product of two vector spaces (see Universal Property section of https://en.wikipedia.org/wiki/Tensor_product). What you're trying to prove is a semi-remarkable property of the tensor product - namely that there is a bilinear map from $E\times F$ to the tensor product such that if $U$ is any given vector space and $b$ is any bilinear map from $E\times F$ to $U$, then there is a unique linear map ($B$ in this author's notation) so that the diagram in Figure 6.1 commutes (i.e. $B\iota = b$). So $U$ is a given vector space and $b$ is a given bilinear map.
As mentioned above, $U$ is a given vector space, nothing more is known about it. Since $b$ is a map whose codomain is $U$ and $B_{1}$ is defined via $b$, so too is $B_{1}$'s codomain $U$.
To get you going on the computation, start with $B\iota(e,f) = B((e,f) + V_{0})$ where we have applied the definition of $\iota.$ There are a few more steps after this initial one, but not too many. Remember that using the definition of $q$ and $B_{1}$ will be needed as well as the result from Proposition 2.25.

Hopefully this helps shed some light on the matter.

Thanks yet again for help GJA ... but I need further help ...

Indeed, you write:

" ... ... To get you going on the computation, start with $B\iota(e,f) = B((e,f) + V_{0})$ where we have applied the definition of $\iota.$ There are a few more steps after this initial one, but not too many. Remember that using the definition of $q$ and $B_{1}$ will be needed as well as the result from Proposition 2.25. ... ... "


Sorry GJA ... cannot proceed anywhere from $B\iota(e,f) = B((e,f) + V_{0})$ ... unsure how definitions of $$q$$ and $$B_1$$ can be used to show $$B\iota(e,f) = b(e,f)$$ or in other words that $$B\iota = b$$ ... ...
But just to indicate my thinking ... here is the set of maps involved with $$B$$ showing as existing ... because ... given the existence of the maps $$B_1$$ and $$q$$ ... Proposition 2.2.5 tells us $$B$$ exists ... so fitting $$B$$ into a diagram showing the other maps defined by Knapp in Theorem 10.6 we have the following diagram ... (is it correct?)
View attachment 5418
As mentioned above the linear map $$B$$ exists as indicated in Figure 2:
View attachment 5419
So from Figure 3 below, given that $$B$$ exists we can deduce that $$Bi = b$$ ... as long as we can assume the commutativity of the diagram ... ... ( can we? why exactly?) ... ...
View attachment 5420

Now I know that arguing from diagrams is not a proof ... but it is an indication of how a proof might work ... is my 'diagrammatic argument' correct in as far as it goes ... ?


I would however be grateful if you could help me progress your argument ... that starts with $B\iota(e,f) = B((e,f) + V_{0})$ ... ...

Peter

 

[GJA]

Hi Peter,

The picture you drew in Figure 1 is correct and will get you where you want to go. You want to recognize that $(e,f)\in E\times F$ and $(e,f)\in V_{1}$. Then,

$B\iota(e,f)=B((e,f)+V_{0})=Bq(e,f)=B_{1}(e,f)=b(e,f)$

Since $(e,f)$ was arbitrary, $B\iota =b$. Hope this helps!