Bounded elementary generation of SLn(R) by transvections

Each $E_{ij}(t)$ has determinant $1$: in its Leibniz expansion the identity permutation contributes $1$, and every nonidentity permutation term vanishes because the only nonzero off-diagonal entry is at $(i,j)$. By [F4], its inverse is the transvection $E_{ij}(-t)$; also $E_{ij}(s)E_{ij}(t)=E_{ij}(s+t)$ by [F1]–[F2], so each $H_{ij}$ is a subgroup.

Start with $M=g$. At stage $k$, where $1\le k<n$, the first $k-1$ rows and columns have already been cleared off the diagonal, their pivots $d_1,\dots ,d_{k-1}$ are nonzero, and $\det M=1$. If $M_{kk}=0$, row $k$ is not zero because $\det M\mathrm{\ne }0$; its entries in columns $j<k$ are zero, so some $j>k$ has $M_{kj}\mathrm{\ne }0$. Choose the least such $j$ and replace $M$ by $M{E}_{j,k}(1)$, which adds column $j$ to column $k$ and makes the pivot $M_{kk}+M_{kj}=M_{kj}\mathrm{\ne }0$. The cleared earlier rows remain unchanged because their entries in columns $j$ and $k$ are zero. This uses at most one transvection for pivot repair.

With $p:=M_{kk}\mathrm{\ne }0$, for each $j>k$ right-multiply by $E_{k,j}(-M_{kj}\mathrm{/}p)$ to clear entry $(k,j)$, then for each $i>k$ left-multiply by $E_{i,k}(-M_{ik}\mathrm{/}p)$ to clear entry $(i,k)$. These operations leave the earlier rows and columns cleared: earlier rows have zero entries in columns $k,j$, and row $k$ has zero entries outside its pivot after the first set of operations. Thus row and column $k$ are isolated with a nonzero pivot $d_k=p$. This costs at most $2(n-k)$ further transvections.

The operations in steps 2.1–3.1 are transvections, so [F1]–[F3] preserve $\det M=1$. Associativity collects the left multiplications into a product $U$ and the right multiplications into a product $V$, giving $UgV=D:=\mathrm{diag}(d_1,\dots ,d_n)$. Summing the stage costs gives ${\sum }_{k=1}^{n-1}(2(n-k)+1)=n^2-1$, so each of $U,V$ has at most $n^2-1$ factors. Every pivot $d_i$ is nonzero, and ${\prod }_{i=1}^n{d}_i=\det D=1$.

For $1\le k<n$, put $s_k=d_1\cdots d_k$ and let $D_k$ be diagonal with $s_k$ at position $k$, $s_k^{-1}$ at position $k+1$, and $1$ elsewhere. Then $D={\prod }_{k=1}^{n-1}{D}_k$: the first diagonal entry is $s_1=d_1$, an interior entry $j$ is $s_{j-1}^{-1}{s}_j=d_j$, and the last is $s_{n-1}^{-1}=d_n$ because ${\prod }_i{d}_i=1$.

In the $(k,k+1)$ block write $T_{+}(u)=E_{k,k+1}(u)$ and $T_{-}(v)=E_{k+1,k}(v)$. For every nonzero $a$, direct multiplication gives $T_{+}(a)T_{-}(-a^{-1})T_{+}(a)=\left(\begin{array}{cc}0& a\\ -a^{-1}& 0\end{array}\right)$ and $T_{+}(-1)T_{-}(1)T_{+}(-1)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, so their product is $\mathrm{diag}(a,a^{-1})$. Taking $a=s_k$ shows that each $D_k$ is a product of six transvections; therefore $D$ is a product of at most $6(n-1)$ transvections.

From $UgV=D$ we have $g=U^{-1}D{V}^{-1}$. By [F4], the inverses of the transvections in $U$ and $V$ are transvections, so $U^{-1}$ and $V^{-1}$ each use at most $n^2-1$ factors. Together with step 6.1, this writes $g$ as a product of at most $2(n^2-1)+6(n-1)=2n^2+6n-8\le M(n)$ transvections. Thus the stated bounded-generation claim holds.