The special linear group is path connected (transvections)

Idea. Each generator lies on an explicit path through $I$; a product of generators can be shrunk to $I$ by shrinking all the parameters simultaneously.

Step 1: $SL_n(\mathbb{R})$ is generated by transvections. Let $A\in SL_n(\mathbb{R})$. Left multiplication by $T_{ij}(t)$ adds $t$ times row $j$ to row $i$. Using these operations we reduce $A$ to the identity:
- *Making the $(1,1)$ entry equal to $1$.* Some entry of the first column is non-zero (else $\det A=0$). If $a_{11}=0$, add that row to row $1$ so that $a_{11}\ne0$. If $a_{11}\ne1$ we want to make it $1$: if $a_{11}\ne 0$, add row $2$ times $\frac{1-a_{11}}{a_{21}}$ if $a_{21}\ne0$. If $a_{21}=0$ first add row $2$ to... (some entry $a_{k1}\neq0$ with $k\ge2$ exists unless the first column is $(a_{11},0,\dots,0)^T$; in that case add row $2$ to row $1$ after having made $a_{22}\ne 0$ or move on). In all cases one can reach $a_{11}=1$ using only transvections whenever $n\ge2$ (a standard fact).
- *Clearing the first column and first row.* Subtract multiples of row $1$ from the others to clear the first column; then use column operations (right multiplication by transvections) to clear the first row.
- *Induction.* We are left with $1\oplus A'$ with $\det A'=1$; repeat with $A'$ in $SL_{n-1}$. At the end we reach a diagonal matrix with determinant $1$ and last entry $1$, hence $I$ (the diagonal entries equal to $1$ except possibly the last; the determinant being $1$, it is $1$).

So $A$ is a product of transvections: $A=T_1T_2\cdots T_m$ with $T_k=T_{i_kj_k}(t_k)$.

Step 2: the path. Define $\gamma(s)=T_{i_1j_1}(st_1)\cdots T_{i_mj_m}(st_m)$, $s\in[0,1]$. Each factor has determinant $1$ ($T_{ij}(t)$ is unipotent triangular after permutation), so $\gamma(s)\in SL_n$. It is continuous with $\gamma(0)=I$ and $\gamma(1)=A$.

Conclusion. Every $A\in SL_n(\mathbb{R})$ is joined to $I$ in $SL_n(\mathbb{R})$.

Remark. This proof works over any topological field in which $SL_n$ is generated by transvections; it proves for instance that $SL_n(\mathbb{C})$ is connected. It does not by itself prove that $GL_n^+$ is connected: that needs one more step, scaling by $|\det|^{1/n}$.

Note. The source solution only states the generation fact; the elimination argument above was written out in more detail. For a complete treatment of the first bullet, see any proof that $SL_n(F)=E_n(F)$ for a field $F$.