= Solution
Apply the <regular level set theorem> to $F(A)=\det A-1$ on the $n^2$-dimensional <vector space> of real <matrices>. At a point with determinant one, part (b) gives $DF_A(H)=\operatorname{tr}(A^{-1}H)$. Choosing $H=A/n$ gives $DF_A(H)=1$, so this derivative is <surjective>.
Hence \b[the <special linear group> $\operatorname{SL}(n,\mathbb R)$ is a smooth <submanifold> of dimension $n^2-1$]. At the identity,
$$
\boxed{T_I\operatorname{SL}(n,\mathbb R)=\{H:\operatorname{tr}H=0\}.}
$$
This is the <special linear Lie algebra>. When $n=1$, the group is a single point and the tangent space is zero, consistent with the same formula.
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