Show that a 2x2 matrix A is symmetric positive definite if and only if A is symmetric, trace(A) > 0 and det(A) > 0 I need to show two parts of the implication are true. 
First: if $A$ is $2\times 2$ and is symmetric positive definite then $trace(A)>0$ and $\det(A)>0$.
Second: if $trace(A)>0$ and $\det(A)>0$ then $A$ is symmetric positive definite.
For the first part I was thinking: $A$ is symmetric and positive definite then $A$ has its eigenvalues positive. If $A$ is $2\times 2$  then characteristic polynomial of $A$ is
 $x^2−x.trace(A)+\det(A)=0$  If we compute the discriminant we get $(tr(A))^2 ≥ 4.\det(A)$ 
Now $tr(A)$ is squared so it is positive. How do I know that $\det(A)$ is also positive?
After copper.hat's response we argue that the eigenvalues of A are all positive because $A$ is spsd and the $\det(A)$ is the product of its eigenvalues. Thus $\det(A)$ is strictly positive.
Now, how do I verify the second part?
Thanks.

What isn't quite proved here is 2nd part in number 3 of Jesus RS' answer, I think?
Anyhoo:
Proof:

*

*Real symmetric is orthogonally diagonalisable. (I think we use just diagonalisable._ Let $A=QDQ^T$.


*We want to show $x^TAx=v^TDv > 0$ if $v=Q^Tx \ne 0$ (iff $x \ne0$ since $Q^T$ is invertible (since $Q^T$ is orthogonal. again, I think just diagonalisable is fine))'.


*For $D=diag[d_1,d_2]$, we have $v^TDv=d_1v_1^2+d_2v_2^2$.


*We have $0 < tr(A)=tr(D) = d_1 + d_2$ and $0 < \det(A)=\det(D) = d_1 d_2$, i.e. $d_1,d_2>0$.


*By (3) and (4), we have $v^TDv > 0$ if $v \ne 0$. QED
