How can I prove that the dimensions of $SO(n)$ and $SL(n)$ are $n(n-1)/2$ and $n^2-1$ respectively?

Question

One of the explanations I found mentions that for $SO(n)$ we start with the dimension of $n \times n$ matrices, which is $n^2$. Then we substract the number of constraints. $n$ for the normalization and $n(n-1)/2$ for the orthogonality. Why are these the number of constraints and how can I use this for $\mathrm{dim} \ SL(n)$? :)

Answer 1

$SL(n)$ is the submanifold of $M(n,\mathbb{R})$ defined by the submersion (around the identity)$f(X)=det(X)$ and $SL(n)=f^{-1}(1)$. The dimension of $SL(n)$ is the dimension of the tangent space of $SL(n)$ at $I_n$ which is defined by the matrices such that $df_I(A)=tr(A)=0$, the dimension of this space is $n^2-1$.

For $SO(n)$, let $Sym(n)$ be the space of symmetric matrices, show that $f:M(n,\mathbb{R})\rightarrow Sym(n)$ defined by $f(X)=XX^T$ is a submersion around the identity, $O(n)=f^{-1}(I_n)$, show that the tangent space of $O(n)$ ($df_I(A)=0$) at $I_n$ is the space od skew symmetric matrices and deduce that $dim(O(n))=dim(SO(n))={{n(n-1)}\over 2}$.