Is A a Killing field if A is an Antisymmetric Matrix?

[JasonJo]

Homework Statement

A linear vector field on R^n, defined by a matrix A, is a Killing field if and only if A is antisymmetric.

Homework Equations

The Attempt at a Solution

I took R^2. Let A be the following matrix:
0 -10
10 0

A is antisymmetric. However, how in the heck is A a Killing field? For example, let
v = (1, 10), then Av = (-100, 10)
u = (2, 3), then Au = (-30, 20)
and <u, v> = 32 while <Au, Av> = 3200, so the dot product after we apply A is a multiple of the original dot product by the determinant of A (in this case it's 100).

What exactly am I misreading? The only restriction on A is that it is antisymmetric. A Killing field is a vector field that preserves the metric, i.e. in particular it will preserve the dot products of vectors.

 

[Dick]

Ok, I finally figured out what you mean here. On R^2, they are referring to a vector field of the form F(x*i+y*j)=(ax+by)*i+(cx+dy)*j, where i and j are the usual basis for R^2. Using the usual metric, what do Killing's equations look like? The covariant derivatives are just the ordinary partial derivatives.