Let V phi be a nondegenerate finitedimensional orthogonal sp

Let (V, phi) be a non-degenerate finite-dimensional orthogonal space over a field F and T: V rightarrow V an isometry. Prove that det(T) = plusminus 1.

Solution

Given that T a transformation from V to V is an isometry.

Let T(a) = A

and T(b) = B

Then since T is an isometry, d(a,b) = d(A,B) for all a, b

Or |T(b)-T(a) |= |b-a|

|T(b-a)| = |b-a|

This is possible only if T is having a determinant with absolute value 1.

In other words T can be isometry only if

det (T) = 1 or -1