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.
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.