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?
 A: 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.
A: 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.
A: 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.
A: 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. 
