 Solution

Regard the space of real $n\times n$ matrices as ${\mathbb{R}}^{n^2}$, and use the smooth determinant map. For invertible $A$, factor $A+tH=A(I+t{A}^{-1}H)$. Expanding the determinant to first order givessince the linear coefficient of $\det (I+tK)$ is the sum of the diagonal entries of $K$. At every $A$ with determinant one, taking $H=A$ makes this derivative $n\ne 0$. Thus one is a regular value, and part (ii) makes $\mathrm{SL}(n,\mathbb{R})$ a submanifold of dimension $n^2-1$.

Its tangent space is the kernel of this differential. At the identity,Indeed tangent derivatives of curves with constant determinant lie in that kernel, and the local regular-value coordinates show every kernel vector is tangent to the level set.