MHB How Do You Interpret Rowen's Notation in Matrix Rings?

Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Louis Rowen's book, "Ring Theory"(Student Edition) ...

I have a problem interpreting Rowen's notation in Section 1.1 Matrix Rings and Idempotents ...

The relevant section of Rowen's text reads as follows:View attachment 6069
View attachment 6070In the above text from Rowen, we read the following:
" ... ... We obtain a more explicit notation by defining the $$n \times n$$ matric unit $$e_{ij}$$ to be the matrix whose $$i-j$$ entry is $$1$$, with all other entries $$0$$.Thus $$( r_{ij} ) = \sum_{i,j =1}^n r_{ij} e_{ij}$$ ; addition is componentwise and multiplication is given according to the rule $$( r_1 e_{ij} ) ( r_2 e_{uv} ) = \delta_{ju} (r_1 r_2) e_{iv} $$

... ... ...
I am having trouble understanding the rule $$( r_1 e_{ij} ) ( r_2 e_{uv} ) = \delta_{ju} (r_1 r_2) e_{iv}$$ ... ...

What are $$r_1$$ and $$r_2$$ ... where exactly do they come from ... ?

Can someone please explain the rule to me ...?To take a specific example ... suppose we are dealing with $$M_2 ( \mathbb{Z} )$$ and we have two matrices ...$$P = \begin{pmatrix} 1 & 3 \\ 5 & 4 \end{pmatrix}$$

and

$$Q = \begin{pmatrix} 2 & 1 \\ 3 & 3 \end{pmatrix}$$ ...In this specific case, what are $$r_1$$ and $$r_2$$ ... ... and how would the rule in question work ...?
Hope someone can help ...

Peter
 
Physics news on Phys.org
Peter said:
I am having trouble understanding the rule $$( r_1 e_{ij} ) ( r_2 e_{uv} ) = \delta_{ju} (r_1 r_2) e_{iv}$$ ... ...

What are $$r_1$$ and $$r_2$$ ... where exactly do they come from ... ?
$r_1$ and $r_2$ are numbers, or, more precisely, elements of the ring $R$. They are also coefficients, or coordinates, of a matrix in the basis consisting of $e_{ij}$. If one matrix is $\displaystyle\sum_{i,j=1}^nr^{(1)}_{ij}e_{ij}$ and another is $\displaystyle\sum_{i,j=1}^nr^{(2)}_{ij}e_{ij}$, then when you multiply them, you apply distributivity and get $\displaystyle\sum_{i,j,u,v=1}^nr^{(1)}_{ij}e_{ij}r^{(2)}_{uv}e_{uv}$, and the formula in your quote says how to compute $r^{(1)}_{ij}e_{ij}r^{(2)}_{uv}e_{uv}$.
 
Evgeny.Makarov said:
$r_1$ and $r_2$ are numbers, or, more precisely, elements of the ring $R$. They are also coefficients, or coordinates, of a matrix in the basis consisting of $e_{ij}$. If one matrix is $\displaystyle\sum_{i,j=1}^nr^{(1)}_{ij}e_{ij}$ and another is $\displaystyle\sum_{i,j=1}^nr^{(2)}_{ij}e_{ij}$, then when you multiply them, you apply distributivity and get $\displaystyle\sum_{i,j,u,v=1}^nr^{(1)}_{ij}e_{ij}r^{(2)}_{uv}e_{uv}$, and the formula in your quote says how to compute $r^{(1)}_{ij}e_{ij}r^{(2)}_{uv}e_{uv}$.
Thanks Evgeny ... your post was very clear ... and most helpfu ...

Peter
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
8
Views
1K
Replies
2
Views
1K