TITLE: What is an 'independent parameter' of a matrix?
QUESTION [5 upvotes]: I saw the following question a book of mine:

Show that an $n\times n$ orthogonal matrix has $n(n-1)/2$ independent parameters.

I have no idea what an independent parameter is. Could you explain it to me?

REPLY [3 votes]: @ Bernard , you did not show the algebraic independence of the $n(n+1)/2$ relations. For instance, if $U$ is a fixed square matrix, the commutativity relation $UX-XU=0$ gives $n^2$ relations between the entries ; yet the previous relations are not independent. In fact, generically, there are $n^2-n$ independent relations. More generally, if $X$ satisfies a matrix equation $f(X)=0$ the number of independent relations between the entries is given by the rank of $Df(X)$, the derivative of $f$ in $X$. Here $f(X)=A^TA-I$ and $Df(X):H\rightarrow H^TA+A^TH$. Note that $Df(X)=0$ iff $A^TH\in \mathcal{K}$, the set of skew symmetric matrices. Then $Df(X)(H)=0$  iff $H\in A^{-T}\mathcal{K}$. Since $A$ is invertible the vector space $ A^{-T}\mathcal{K}$ has the dimension of $\mathcal{K}$, that is $n(n-1)/2$ and we are done.