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This is probably a stupid question, but what makes the three magical elementary row operations, as taught in elementary linear algebra courses, special? In other words, in what way are they "natural" (as opposed to "arbitrary")?

It seems that they're always presented in a somewhat haphazard manner ("these are the three legendary elementary row operations, don't ask why, they just are"). From what I understand, they satisfy some nice properties, such as the inverse of each being an operation of the same type, etc. But is there something that characterizes them, i.e. is there some definition of what constitutes "elementary" that's only satisfied by the three types of elementary matrices, and no other matrix?

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4 Answers 4

up vote 12 down vote accepted

They're not special. They're just convenient. It's relatively easy to tell what happens to a matrix when you apply an elementary row operation to it, and this isn't quite as true for more complicated types of operations.

In the language of group theory, elementary matrices form a set of generators for the group of invertible square matrices. You could choose a different set of generators if you wanted to, but again, the elementary matrices are convenient.

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Thank you, this is the answer I needed. So in some sense, they're the "most natural" of all arbitrary generators one can come up with? –  user33661 Jul 19 '12 at 1:33
    
They just have a very simple form. –  ncmathsadist Jul 19 '12 at 1:34
    
@paul: well, again, they're convenient. One way of quantifying their convenience is that they can all be written as $I + M$ where $M$ has very few nonzero entries. But I would hesitate to call them natural (this has a somewhat technical meaning in mathematics and I do not think it is satisfied here). –  Qiaochu Yuan Jul 19 '12 at 1:34
    
Can it be (easily) proven that three is the fewest number of generators needed to obtain the entire group of invertible square matrices? Edit: It's obviously not cyclic, because multiplication is not commutative, so can we come up with two (probably more complicated) matrices that generate the three "canonical" elementary matrices? If not, why not? (I'm unfamiliar with the etiquette here, does this question scream "ask new question"?) –  user33661 Jul 19 '12 at 1:43
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@paul: ask a new question. Keep in mind that two of the elementary matrices depend on a parameter, so there are uncountably many matrices involved. The group of invertible square matrices (over $\mathbb{R}$, say) is not even countably generated, let alone finitely generated. –  Qiaochu Yuan Jul 19 '12 at 1:45

Take a system of linear equations and apply the three elementary row operations to them, you will get an equivalent system of equations.

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What's the downvote for? –  Mhenni Benghorbal Nov 17 '13 at 23:03

By the way, it may be amusing to note that you don't really need the row-interchange operation. Thus, instead of having a special operation to interchange rows $i$ and $j$, you could:

  1. add row $j$ to row $i$
  2. subtract row $i$ from row $j$
  3. multiply row $j$ by $-1$
  4. subtract row $j$ from row $i$

and it would have the same effect. In terms of the elementary matrices,

$$ \pmatrix{1 & -1\cr 0 & 1\cr} \pmatrix{1 & 0\cr 0 & -1\cr} \pmatrix{1 & 0\cr -1 & 1\cr} \pmatrix{1 & 1\cr 0 & 1\cr} = \pmatrix{0 & 1\cr 1 & 0\cr}$$

Similarly, you don't need to be able to add an arbitrary nonzero multiple of row $j$ to row $i$, as long as you can add row $j$ to row $i$: to add $t$ times row $j$ to row $i$, you first multiply row $j$ by $t$, then add row $j$ to row $i$, then multiply row $j$ by $1/t$.

So you could get by with just

  1. add a row to another row
  2. multiply a row by a nonzero scalar

Not that I'd want to do this in practice...

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;) I discovered this a while ago, too, only to have my high school math teacher tell me I was wasting time "using less tools when I had more." –  user33661 Jul 19 '12 at 1:51
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@paul- never think of it as a waste of time to play around with concepts to find different ways of doing things. I don't think we'd get very far by just accepting things the way they are. Different perspectives are always worth cultivating. –  Chris Leary Jul 19 '12 at 2:54

Performing them preserves the solution space of any associated linear system. They reflect three ways in which we can manipulate a linear system and not change its solution

  • Permute two equations.
  • Take one equation, add a multiple of another to it and replace it with that.
  • Multiply any row by a nonzero scalar.
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Yes. "Preserving the solution space" is a nice property they possess. But are these the only three operations that have this property? For one thing, we can compose them to get other such operations. My question is, what makes them so elementary? –  user33661 Jul 19 '12 at 1:23
    
All permissible operations are a composition of these three simple operatiosn. –  ncmathsadist Jul 19 '12 at 1:25
    
Just to be clear, are you saying that all linear systems with the same solution space can be obtained from one another after a finite number of elementary row operations? Or am I misunderstanding something here? –  user33661 Jul 19 '12 at 1:28
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No. What I am saying is that any finite composition of these three operations will preserve the solution space of the associated system. You can use them to place a matrix in row-reduced eschalon form and learn abou the linear span of its row space and its rank. –  ncmathsadist Jul 19 '12 at 1:30
    
What you write here is a property of any invertible matrix, and does not even begin to answer the OP's question of why particularly this subset of the invertible matrices gets a distinguished name. –  Henning Makholm Jul 19 '12 at 12:18

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