Left & Right Multiplication Maps on Algebras .... Bresar, Lemma 1.25 .... ....

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SUMMARY

The discussion focuses on the proof of Lemma 1.25 from Matej Bresar's "Introduction to Noncommutative Algebra," specifically regarding the dimensions of multiplication algebras and endomorphism algebras. Participants clarify that the dimension of the multiplication algebra, denoted as $$M(A)$$, is greater than or equal to $$d^2$$ due to the linear independence of operators $$L_{u_{i}}R_{u_{j}}$$. Furthermore, it is established that since $$M(A)$$ is a subspace of $$\text{End}_{F}(A)$$, its dimension must equal $$d^2$$, leading to the conclusion that $$M(A) = \text{End}_{F}(A)$$.

PREREQUISITES
  • Understanding of finite dimensional division algebras
  • Familiarity with linear algebra concepts, particularly linear independence
  • Knowledge of endomorphism algebras, specifically $$\text{End}_{F}(A)$$
  • Basic comprehension of algebraic structures and dimensions in vector spaces
NEXT STEPS
  • Study the implications of Lemma 1.24 in Bresar's text regarding linear independence
  • Explore the properties of multiplication algebras in noncommutative algebra
  • Investigate the relationship between dimensions of subspaces and their parent spaces
  • Review examples of finite dimensional division algebras to solidify understanding
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Mathematicians, algebraists, and students studying noncommutative algebra, particularly those interested in the structure and properties of multiplication algebras and endomorphism algebras.

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I am reading Matej Bresar's book, "Introduction to Noncommutative Algebra" and am currently focussed on Chapter 1: Finite Dimensional Division Algebras ... ...

I need help with the proof of Lemma 1.25 ...

Lemma 1.25 reads as follows:View attachment 6261
View attachment 6262
My questions on the proof of Lemma 1.25 are as follows:Question 1

In the above text from Bresar we read the following:

" ... ... Therefore $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$ ... ... "Can someone please explain exactly why Bresar is concluding that $$[ M(A) \ : \ F ] \gt d^2$$ ... ... ?Question 2

In the above text from Bresar we read the following:

" ... ... Therefore $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$

and so $$M(A) = [ \text{ End}_F (A) \ : \ F ]$$. ... ... "Can someone please explain exactly why $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$ ... ...

... implies that ... $$M(A) = [ \text{ End}_F (A) \ : \ F ]$$ ...
Hope someone can help ...

Peter
===========================================================*** NOTE ***

So that readers of the above post will be able to understand the context and notation of the post ... I am providing Bresar's first two pages on Multiplication Algebras ... ... as follows:View attachment 6263
https://www.physicsforums.com/attachments/6264
 
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Hi Peter,

A little short on time at the moment, but hopefully I can push you in the right direction.

Peter said:
Question 1

In the above text from Bresar we read the following:

" ... ... Therefore $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$ ... ... "Can someone please explain exactly why Bresar is concluding that $$[ M(A) \ : \ F ] \gt d^2$$ ... ... ?

Because we can conclude from Lemma 1.24 that the operators $L_{u_{i}}R_{u_{j}}$ are linearly independent and there are a total of $d^{2}$ of such operators. Thus the space $M(A)$ that contains all linear combinations of such operators must have at least this dimension (in general if a linear space contains $k$ linearly independent vectors, then its dimension must be greater than or equal to $k$). Note: You wrote a strict inequality in your original post when it should really include equality.

Peter said:
Question 2

In the above text from Bresar we read the following:

" ... ... Therefore $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$

and so $$M(A) = [ \text{ End}_F (A) \ : \ F ]$$. ... ... "Can someone please explain exactly why $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$ ... ...

... implies that ... $$M(A) = [ \text{ End}_F (A) \ : \ F ]$$ ...

We know the dimension of $\text{End}_{F}(A)$ over $F$ is $d^{2}$. Since $M(A)$ is a linear subspace of $\text{End}_{F}(A)$, its dimension is at most $d^{2}$. Thus we have $d^{2}\leq M(A)\leq d^{2}$.

I hope that helps. Let me know if anything needs clarification.
 
GJA said:
Hi Peter,

A little short on time at the moment, but hopefully I can push you in the right direction.
Because we can conclude from Lemma 1.24 that the operators $L_{u_{i}}R_{u_{j}}$ are linearly independent and there are a total of $d^{2}$ of such operators. Thus the space $M(A)$ that contains all linear combinations of such operators must have at least this dimension (in general if a linear space contains $k$ linearly independent vectors, then its dimension must be greater than or equal to $k$). Note: You wrote a strict inequality in your original post when it should really include equality.
We know the dimension of $\text{End}_{F}(A)$ over $F$ is $d^{2}$. Since $M(A)$ is a linear subspace of $\text{End}_{F}(A)$, its dimension is at most $d^{2}$. Thus we have $d^{2}\leq M(A)\leq d^{2}$.

I hope that helps. Let me know if anything needs clarification.
Thanks for the help GJA ... really appreciate it ...

Just reflecting on what you have said ...

Peter
 

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