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: 
It's rather awkward to define positive-definiteness using "regularity" and pivots. In particular, I'm not even sure all positive-definite matrices (as normally defined) fit under this category. But for this particular proof, you can form an $LU$ decomposition in which $L$ is a lower triangular matrix composed with only row addition matrices. Then $\det(A) = \det(U)$. Since $U$ is upper triangular with the pivot entries of $A$ as it's diagonal entries, it follows that $\det(U)$ is a product of $A$'s pivots. But all of $A$'s pivots are positive, so it follows that $\det(U) = \det(A)$ is also positive.

The trace part is not even true. So the matrices as you've defined them cannot actually be positive-definite. The below matrix is "regular" and has positive pivots
$$\begin{pmatrix}1 & -2 & 2 \\ 2 & -1 & 1 \\2 & 2 & -1\end{pmatrix} \rightarrow \begin{pmatrix}1 & -2 & 2 \\ 0 & 3 & -3 \\0 & 6 & -5\end{pmatrix}\rightarrow \begin{pmatrix}1 & -2 & 2 \\ 0 & 3 & -3 \\0 & 0 & 1\end{pmatrix}$$
Incidentally the trace is negative.
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.