What is an 'independent parameter' of a matrix? 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?
 A: @ 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.
A: It means that an $n\times n$ matrix is just an element of $\mathbf R^{n^2}$ since it has $n^2$ coefficients $a_{ij}$, which are considered as parameters. These parameters are, informally, independent since you can choose any of them as you please, whatever the other coefficients are.
In the vector space (of dimension $n^2$) of all matrices, it asserts the orthogonal matrices
satisfy $\dfrac{n(n+1)}2$ (quadratic) independent relations, hence a number of independent coefficients equal to $$n^2-\frac{n(n+1)}2=\frac{n(n-1)}2.$$
These relations result from $A\,^{\mathrm t}\! A=I$: as the product of these matrices is symmetric, it is enough to write:
$$\sum_{j=1}^na_{ij}a_{kj}=\delta_{ik},\quad 1\le i\le k\le n.$$
There are as many relations as pairs $(i,k)$ satisfying the above relationsinequalities.
