# Thread: Elementary row operation

1. ## Elementary row operation

Let
$e$ be an elementary row operation,
$A,B,C$ be a $n \times n$ matrices
By $e(A)$ we mean the resultant matrix after we perform $e$ on $A$

Does the the following hold true -
$e(A)B = Ae(B)$ ?

Also, we know the $e(A)=e(I_n)A$. One way to prove this is for each kind of elementary row operation establish this fact by performing the computation on the matrix elements.

I was wondering if there is a more conceptual/abstract argument, logic, reason behind this to work.

Thanks

2. Originally Posted by aman_cc
Let
$e$ be an elementary row operation,
$A,B,C$ be a $n \times n$ matrices
By $e(A)$ we mean the resultant matrix after we perform $e$ on $A$

Does the the following hold true -
$e(A)B = Ae(B)$ ?
no! for example in $M_n(\mathbb{C}),$ choose $e=e_{11} + 2e_{22}, \ A=e_{12}, \ B=e_{21}.$ then: $(eA)B=e_{11} \neq 2e_{11}=A(eB).$

3. Originally Posted by NonCommAlg
no! for example in $M_n(\mathbb{C}),$ choose $e=e_{11} + 2e_{22}, \ A=e_{12}, \ B=e_{21}.$ then: $(eA)B=e_{11} \neq 2e_{11}=A(eB).$
Thanks NonCommAlg.

Though I am fairly confident of the following still would check this:
if $C=AB$ then $e(C) = e(A)B$

Also is there another conceptual-argument (based on linear transformation etc) for $e(A)=e(I_n)A$ to be true? The proof I have at my disposal is based on actual computation.

4. Originally Posted by aman_cc
Thanks NonCommAlg.

Though I am fairly confident of the following still would check this:
if $C=AB$ then $e(C) = e(A)B$
this is true because, (at least) over a field, "matrix multiplication" is associative. so $e(AB)=(eA)B.$ you're making your job hard by looking at $e$ as a map rather than just a matrix.

Also is there another conceptual-argument (based on linear transformation etc) for $e(A)=e(I_n)A$ to be true? The proof I have at my disposal is based on actual computation.
i don't know what kind of "concept" or even "computation" you're talking about here? $I_n$ is the identity matrix and so $eI_n=e,$ which gives us $e(I_n)A=(eI_n)A=eA=e(A).$

5. Originally Posted by NonCommAlg
i don't know what kind of "concept" or even "computation" you're talking about here? $I_n$ is the identity matrix and so $eI_n=e,$ which gives us $e(I_n)A=(eI_n)A=eA=e(A).$
Thanks for confirming the part (1). After you said it - I realized I was asking a very trivial question.

For part(2) - Let me try to explain.
When I say 'e' is an elementary row operation. I look at it as some kind of a function on A. e(A) in my way of thinking is image of A under e.

Now it so happens that $e(A) = e.A$
In R.H.S. I guess $e$ is treated like a matrix ( $e=e(I_n)$). And, $e.A$ is a matrix multiplication.

I guess it is this multiplication that you refer when you write $eI_n=e$

My question is why should $e(A) = e.A$? Is there a logic to it - or this is a fact which just happens to be true based on how we define
1. Elementary Row Operation
2. Matrix Multiplication

Please ignore this question if it is really not relevant. Maybe I need to get more clear in my head.

6. Originally Posted by aman_cc

My question is why should $e(A) = e.A$? Is there a logic to it - or this is a fact which just happens to be true based on how we define
1. Elementary Row Operation
2. Matrix Multiplication
well, the only way to prove that is direct calculation, i.e. to check that $eA$ really does the required row operation.

here's a question for you to think about: can you write an elementary matrix in terms of $e_{ij}$?

( $e_{ij},$ as usual, is the square matrix with 1 in the $i$-th row and $j$-th column and 0 everywhere else)

7. Originally Posted by NonCommAlg
well, the only way to prove that is direct calculation, i.e. to check that $eA$ really does the required row operation.

here's a question for you to think about: can you write an elementary matrix in terms of $e_{ij}$?

( $e_{ij},$ as usual, is the square matrix with 1 in the $i$-th row and $j$-th column and 0 everywhere else)

Let me try

$I_n = \sum e_{ii}$ , where $i \in [1,n]$
In the equations below I have used the above mentioned expansion.

Interchange of row - $Row_a \leftrightharpoons Row_b$
$e = I_n - e_{aa} - e_{bb} + e_{ab} + e_{ba}$

Multiplication by a scalar $Row_a = x \times Row_a$
$e = I_n - e_{aa} + x \times e_{aa}$

Adding a scalar multiple of a row to another row $Row_a = Row_a + x \times Row_b$
$e = I_n + x \times e_{ab}$

Am I anywhere close to what you wanted me to try? If yes, any specific importance of this method?

Thanks very much !!

8. Originally Posted by aman_cc
Let me try

$I_n = \sum e_{ii}$ , where $i \in [1,n]$
In the equations below I have used the above mentioned expansion.

Interchange of row - $Row_a \leftrightharpoons Row_b$
$e = I_n - e_{aa} - e_{bb} + e_{ab} + e_{ba}$

Multiplication by a scalar $Row_a = x \times Row_a$
$e = I_n - e_{aa} + x \times e_{aa}$

Adding a scalar multiple of a row to another row $Row_a = Row_a + x \times Row_b$
$e = I_n + x \times e_{ab}$

Am I anywhere close to what you wanted me to try? If yes, any specific importance of this method?

Thanks very much !!
right! of course, "multiplying by a scalar" and "adding a scalar multiple of a row to another row" have the same form, which is $I + ce_{ij}.$ so we basically have two distinct forms of matrices.

the reason that i gave you this question is that using these forms makes the proof $e(A)=eA$ easier. that's because we can write the matrix $A=[a_{ij}]$ in the form $A=\sum_{i,j}a_{ij}e_{ij}$ and then

we can use this fact that $e_{ij}e_{rs}=\delta_{jr}e_{is},$ where $\delta_{jr}$ is, as usual, the Kronecker's delta.

9. I get the idea. Thanks very much !!