Is the set $ {\rm SL }(n, \mathbb {R}) $ connected in ${\rm M }(n, \mathbb {R}) $

Is the set $ {\rm SL }(n, \mathbb {R}) $ connected in ${\rm M }(n, \mathbb {R}) $?

Can you give me hints? I have only concept on following tags.

The answer is yes, and it is even pathwise connected. But it requires some minimal knowledge in linear algebra. An easy way is to use the fact that the group $SL(n,\mathbb{R})$ is generated by transvections=elementary matrices. Then each such matrix can easily be connected to $I_n$ within $SL(n,\mathbb{R})$.

Indeed, let $T=I_n+\lambda E_{i,j}$ be an elementary matrix (diagonal of $1$'s and a unique possibly nonzero off-diagonal coefficient $\lambda$ in $(i,j)$ position for some $i\neq j$. Then $$ T_t:=I_n+t\lambda E_{i,j} $$ defines a path connecting $T$ and $I_n$ with $\det T_t=1$ for every $t$.

Related: see this thread for many proofs of the fact that the set of real square matrices with positive determinant is pathwise connected, hence connected. Note that it implies the case of $SL(n,\mathbb{R})$. Indeed, if $S$ has determinant and if $S_t$ is a path connecting $S$ to $I_n$ within the matrices of positive determinant, then $$ \widetilde{S_t}:=\frac{S_t}{\sqrt[n]{\det S_t}} $$ connects $S$ and $I_n$ within $SL(n,\mathbb{R})$.

Here is an alternative proof of connectedness if you don't mind using something less elementary.

Let $A=USV^T$ be a singular value decomposition over $\mathbb{R}$. As $\det A=1$, we may assume WLOG that $\det U=\det V=1$ and hence $U=e^{K_1}$ and $V=e^{K_2}$ for some skew symmetric matrices $K_1,K_2$. Let $S=\operatorname{diag}(\sigma_1,\sigma_2,\ldots,\sigma_n)$. Then $I$ is pathwise connected to $A$ via the path $$f(t)=e^{tK_1}\operatorname{diag}(\sigma_1^t,\sigma_2^t,\ldots,\sigma_n^t)e^{-tK_2}.$$