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.
We will solve it by assuming a function which is +ve definite and then using continuity definition of ϵ-δ.

So,if you know ϵ-δ continuity definition of a function,then only consider this solution,otherwise skip it.

A is +ve definite.lets define a function :

f(x)= |xA + (1-x)I| for 0<=x<=1 -----(1), here I is the identity matrix.

clearly,f(x) is never 0.

Now we can easily see [xA+(1-x)I] > 0 for 0<=x<=1. 

Also,f is continuous on our assumed interval of x.

let us choose a point b in [0,1].

then |x-b| < δ (where δ>0) (assume) -----(2).

now, |f(x)-f(b)| = |xA+(1-x)I - bA-(1-b)I| = |(x-b)A+(b-x)I|.

|f(x)-f(b)| = |(x-b)A+(b-x)I| <= |(x-b)A| +|(b-x)I| (triangle inequality).

|f(x)-f(b)| <= |(x-b)A| +|(b-x)I| < |x-b| |A| |b-x| |I|.

|f(x)-f(b)| <= |x-b| |A| |b-x| |I| < $δ^2 |A|$ (using (2) ) [note: |I| = 1].

|f(x)-f(b)| <$ δ^2 |A|$ ----(3).

Now,since f is continuous ,thus |A| > 0 is a must as then only the definition of continuity holds.

thus we get |A| > 0 . and then we can choose δ = $[ϵ/|A|]^{1/2}$ and get :

|f(x)-f(b)| < ϵ .

Hence proved |A| > 0.