Positive Definite Matrix Determinant 
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?

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.
