Question: > 
> Prove that a positive definite matrix has positive determinant and
>  positive trace.
> 
> 
> 

In order to be a positive determinant the matrix must be regular and have pivots that are positive which is the definition. Its obvious that the determinant must be positive since that is what a positive definite is, so how can I prove that?

Answer: 
All eigenvalues of a positive definite matrix are real and positive.

The determinant is the product of the eigenvalues, hence real and positive.

The trace is the sum of the eigenvalues, hence real and positive.
One more proof to show that the determinant of a +ve definite matrix is +ve:

let A be a +ve definite matrix. Now we know that A can be written as :

A = $(V)D(V^T)$ ,which is the single value decomposition form of the +ve definite matrix A.Here $V^T$ and V are the orthogonal vectors such that $ (V^T)V $ = 1 .Thus we can also write :

$|V^T|$ |V| = 1 ---(1).

Also D is the diagonal matrix with eigen values of A as its diagonal elements.

Now, A = $(V)D(V^T)$ .

or $|A| = |V| |D| |V^T| $ .

or |A| = |D| (from (1) ) ---(2) .

|D| = product of eigen values ----(3).

Thus if we prove that all the eigen values of a +ve definite matrix are +ve then we are done.
$(x^T)A(x)$ > 0 .

Ax=bx ,where b is the eigen value. 

thus $(x^T)bx > 0 $.

$b(x^T)x > 0 $.

$(x^T)x > 0 $ (always) .

thus b>0 .

thus from (2) and (3) : 

|A| = |D| > 0 

or |A| > 0 .

hence proved.