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The general linear group is generated by elementary matrices that add a multiple of row $j$ to row $i$ and elementary matrices that multiply row $i$ by a scalar. This is because you can write an invertible matrix as the product of elementary matrices, and a row swap matrix can be written as a product of the other two types of elementary matrix.

So far, I've worked out that all matrices that are products of this type of elementary matrix are in the special linear group, but I don't know how to proceed.

user26857
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user124398
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    The proof involves the Euclidean algorithm and is a little messy. I once typed it out in detail so I wouldn't have to do it again. You can find the proof here: https://drive.google.com/file/d/0B1rY67lS_DqlZW0zRmczN21oemhod0dZSnIydEY2VVladmVB/edit?usp=sharing – Elchanan Solomon Jan 28 '14 at 06:35
  • @IsaacSolomon: In your proof, can you actually take $ac=1$ to mean that $c=1/a$? We are only in a Euclidean ring. Perhaps it is better to multiply the first row by $c$ and add the first row to the second. –  Feb 28 '14 at 09:45
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    Since $ac = 1$, we know that $a$ and $c$ are invertible. The notation $1/a$ is (common?) shorthand to mean "that element which is the inverse of $a$", namely $c$. – Elchanan Solomon Feb 28 '14 at 12:31

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Exercise 2.4.8(b) (Artin's Algebra, 2nd edition). Prove that the elementary matrices of the first type generate $SL_n(\mathbb{R})$. Do the $2 \times 2$ case first.

If $E$ is an elementary matrix, then $X\mapsto EX$ can be described as follows: if $E$ is Type 1, with the off-diagonal entry $a$ at $(i,j)$, we do $r_i\mapsto r_i+ar_j$; if $E$ is Type 3, with the modified diagonal entry $c\ne0$ at index $i$, then $r_i\mapsto cr_i$.

Note that Type 1 matrices have determinant $1$ and Type 3 matrices have determinant $c$.

In order to show$$M = E_1E_2\dots E_k$$for some (permitted) elementary matrices $E_i$, it suffices to show $$I_n = F_kF_{k-1}\dots F_1M$$for some elementary $F_i$, since then$$M = F_1^{-1}\dots F_k^{-1},$$as elementary matrices are invertible, and their inverses are elementary as well.

Now, we consider $M\in SL_n(\mathbb{R})$. Using the row operations corresponding to Type 1 elementary matrices, we turn column $i$ into $e_i$ ($1$ at position $i$, $0$ elsewhere) from left to right.

Take the leftmost column $i$ with $c_i \ne e_i$, if it exists (otherwise, we are done). Since $\det(M) = 1 \ne 0$, we can not have $c_i$ written as a linear combination of$$c_1=e_1,\dots,\,c_{i-1}=e_{i-1};$$hence one of entries $i,i+1,\dots,n$ must be nonzero, say $j$.

Subtracting $r_j$ from the other rows as necessary, we first clear out all column $i$ (except for row $j$). Note that none of this affects columns $1$ through $i-1$. If $i=n$, we have a diagonal matrix with determinant $1$ and the first $n-1$ entries all $1$'s, so we are done. Otherwise, if $i<n$, pick an arbitrary row $k\ne j$ from $i$ to $n$, and add a suitable multiple of $r_j$ to $r_k$ so that the $(k,i)$ entry becomes $1$. Now subtract a suitable multiple of $r_k$ from $r_j$ so the $(j,i)$ entry becomes $0$. If $k=i$, we can proceed to column $i+1$; otherwise, add $r_k$ to $r_i$ and subtract $r_i$ from $r_k$, and then proceed to column $i+1$.

We can slightly simplify the proof by using elementary column operations as well (corresponding to right multiplication: it suffices to construct $F_1\cdots F_k M F_{k+j}\dots F_{k+1} = I_n$).