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

[Math Amateur]

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

In the above text from Bresar we read the following:

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

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

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

 

[GJA]

Hi Peter,

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

Question 1

In the above text from Bresar we read the following:

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

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.

Question 2

In the above text from Bresar we read the following:

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

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

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

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.

 

[Math Amateur]

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