How do you show a variance matrix is nonnegative definite?

Answer 1

I assume you mean covariance matrix and I assume that you are familiar with the definition:

C = E[(X-u)(X-u)T]

where X is a random vector and u = E(X) is the mean

The definition of non-negative definite is:

xTCx ≥ 0 for any vector x Є R

So is xTE[(X-u)(X-u)T]x ≥ 0?

Then, from one of the covariance properties:

E[(xT(X-u))((X-u)Tx)] = E[xT([(X-u)(X-u)T]x)] = E[((xTI)x)] = E[xTx]

Finally, since we've already defined x to have only real values, xTx is therefore non-negative definite by definition.