Graduate Matrix multiplication, Orthogonal matrix, Independent parameters

[LagrangeEuler]

Matrix multiplication is defined by
\sum_{k}a_{ik}b_{kj} where $a_{ik}$ and $b_{kj}$ are entries of the matrices $A$ and $B$. In definition of orthogonal matrix I saw
\sum_{k=1}^n a_{ki}a_{kj}=\delta_{ij}
This is because $A^TA=I$. How to know how many independent parameters we have in the case of nxn orthogonal matrix? So how many parameters you need to give me in order to know the other ones?

 

[hutchphd]

The number of independent elements will equal the number of elements minus the number of (independent) equations of constraint.

 

[mathwonk]

Maybe think geometrically? You want to choose n vectors each one length one and perpendicular to all the previous ones. So in (real) n-space the first vector's head must be chosen on the unit sphere, which has dimension n-1. Then we want the next vector chosen from the unit sphere in the hyperplane orthogonal to the first vector, so that is a unit sphere in n-1 space, so has dimension n-2. So after choosing two vectors we have ranged over (n-1)+(n-2) dimensions. It seems by the principle of what else could it be, that the answer is the sum of the first n-1 positive integers, or n(n-1)/2 dimensions for the space of all such matrices. and the geometry of that matrix space seems to be a product of spheres of those dimensions. does this seem to make sense?

Here is a rough "check" on it: look at the map from all matrices to symmetric matrices, sending a matrix A to the product A*A, where A* is the transpose of A. (Do you see why A*A is symmetric?) Symmetric matrices have dimension 1+2+...+n = (n+1)n/2 since the diagonal terms and those above determine those below. Thus we have a (smooth) map from a space of dim n^2 to a space of dim (n+1)n/2 and we want the dimension of the preimage of the single point Id. Assuming this map is a nice one, say surjective and all fibers of the same dimension, we would get the dimension of a fiber by subtracting those dimensions, which gives n(n-1)/2. Of course this leaves out a lot of details, but is comforting nonetheless, at least to me. By the way, this second calculation is the one suggested by hutchphd.

 

[jbunniii]

by the principle of what else could it be