Number of independent elements of an orthogonal matrix

Question

I don't understand the following fact.

The $n^2$ elements of an orthogonal matrix $A$ of order $n \times n$ are not independent. This follows from the fact that $A′A = I_{n}$ which implies that the elements of $A$ are subject to $n (n + 1)∕2$ equality constraints. Hence, the number of independent elements is $n^2 − \frac{n(n + 1)}{2} = \frac{n(n − 1)}{2}$

Can someone please explain where that equality constraint does come from and why the number of independent elements is $n^2 − \frac{n(n + 1)}{2} = \frac{n(n − 1)}{2}$?

Answer 1

(1) Equality constraints. They are from $A_i \cdot A_j = 0, \forall i \neq j$ ($\binom{n}{2} = n(n-1)/2$ in total) and $A_i \cdot A_i = 1, \forall i$ ($n$ in total). And we have $n(n-1)/2 + n = n(n+1)∕2$.

(2) Dimension. For the dimension, you may understand this from the perspective of "degree of freedom". Each equation reduces $1$ "degree of freedom", i.e., an element can be uniquely determined when all the other elements in the equation are given.

Answer 2

The equality $A'A=\operatorname{Id}_n$ means that:

• the norm of each column of $A$ is $1$ ($n$ equalities);
• if $i\ne j$, the $i$^th and the $j$^th columns of $A$ are orthogonal ($\frac{n(n-1)}2$ equalities).

So, you have $n+\frac{n(n-1)}2$ equalities, that is, you have $\frac{n(n+1)}2$ equalities. And $n^2-\frac{n(n+1)}2=\frac{n(n-1)}2$