Find the geometry of the curves of the contour lines of $f(x) = \frac{1}{2}x^tAx + b^tx + c$ 
Find the geometry of the curves of the contour lines of a quadratic
  function $$f(x) = \frac{1}{2}x^tAx + b^tx + c$$ where $A \in \mathbb{R}^{2 \times 2}$, $b\in \mathbb{R}^2$ and $c\in \mathbb{R}$ in the following cases:
  
  
*
  
*$A>0$
  
*$A\ge 0$ and there exists $x$ such that $Ax+b = 0$
  
*$A\ge 0$ and there is no $x$ such that $Ax + b = 0$
  
*$A$ is undefined and non-singular.

I suppose $^t$ is the transpose. What is $A>0$? 
I'm trying to develop a technique to see this. If we write $A = \begin{bmatrix} 
     a & b \\
     b & c\\  
     \end{bmatrix}$ then the function becomes
$$f((x_1,x_2)) = \frac{1}{2}\begin{bmatrix} 
     x_1  & x_2  
     \end{bmatrix}\begin{bmatrix} 
     a & b \\
     b & c\\  
     \end{bmatrix}\begin{bmatrix} 
     x_1 \\
     x_2\\  
     \end{bmatrix} + \begin{bmatrix} 
     b_1  & b_2  
     \end{bmatrix}\begin{bmatrix} 
     x_1 \\
     x_2\\  
     \end{bmatrix} + d = \\ \frac{1}{2}(ax_1^2 + 2bx_1x_2 + cx_2^2) + b_1x_1 + b_2x_2 + d $$
I don't know if I can simplify $(ax_1^2 + 2bx_1x_2 + cx_2^2)$. I think not. Maybe it's a shape of its own and I should recognize. I thin I can see it as approximate $x^2 + y^2$ everywhere since they grow much faster than $xy$. So the contour here would be circles? Is there a more accurate way of drawing the contour? I cannot suppose they're circles, I need to find what they truly are.
Anyways, $A>0$ implies that $\frac{1}{2}(ax_1^2 + 2bx_1x_2 + cx_2^2)>0$ right? 
And $A\ge 0 \implies \frac{1}{2}(ax_1^2 + 2bx_1x_2 + cx_2^2)\ge 0$, and the condition there exists $x$ such that $Ax+b=0$ means there exists $x$ such that $\begin{bmatrix} 
     ax_1 + bx_2 + b_1 \\
     ax_2 + bx_1 + b_2\\  
     \end{bmatrix} = \begin{bmatrix} 
     0 \\
     0\\  
     \end{bmatrix}\implies ax_1 + bx_2 + b_1 = 0, ax_2 + bx_1 + b_2 = 0$
What this should tell me? 
$A$ being undefined and non singular means an arbitrary nonsingular matrix, I suppose. So the invertibility or the determinant of $A$ plays a role here. 
 A: In the optimization community, $A > 0$ (where $A$ is a matrix), or more commonly written as $A \succ 0$, usually means $A$ is positive definite. Also, $A \succeq 0 $ means positive semidefinite.
Notice that $\nabla f(x) = Ax+b$ and $\nabla^2 f(x) = A$.
If $A \succ 0$, we have a strongly convex quadratic function.  The curves should be ellipses. They can be very elongated if the condition number is large.
If $A \succeq 0$ but $A \not \succ 0$, then at least one of the eigenvalues is $0$.


*

*If there is an $x$ such that  $\nabla f(x)=0$, there are multiple  solutions to the system $$\nabla f(x)=Ax+b=0$$ It is like a parabolic sheet with multiple minimum points. It is also possible for the function to be constant everywhere.

*Suppose there is no $x$ such that $\nabla f(x)=0$, for example when $A=0, b \ne 0$, there is no minimum point, the problem is unbounded.
The most ambiguous part would be the word "undefined". I would guess it to be indefinite. If we have a positive eigenvalues and a negative eigenvalue, we have a saddle point. 
Edit: 
Usually, positive definite and positive semidefinite matrices refer to the case where the matrices are symmetric and we know that there exist orthogonal matrices $U$ such that $A=UDU^T$ where $D=\mbox{diag}(\lambda_1, \lambda_2)$ is a diagonal matrix which consists of the eigenvalues. 
$$f(x) = x^TAx+b^Tx+c=x^TUDU^Tx+b^TUU^Tx + c$$
Let $y= U^Tx$ and $p = U^Tb$. We might want to study 
\begin{align}g(y)&=y^TDy+p^Ty+c \\
&= \sum_{i=1}^2 (\lambda_i y_i^2 + p_i y_i)+c\end{align}
If $A \succ 0 $, then all the eigenvalues are positive.
\begin{align}g(y) &= \sum_{i=1}^2 \lambda_i \left( y_i^2 + \frac{p_i}{\lambda_i} y_i\right)+c \\
&= \sum_{i=1}^2 \left[ \lambda_i\left( y_i + \frac{p_i}{2\lambda_i} \right)^2 - \lambda_i\left( \frac{p_i^2}{4\lambda_i^2}\right)^2\right]+c\end{align}
Let $z_i = y_i + \frac{p_i}{2\lambda_i}$ and $q = c - \sum_{i=1}^2\lambda_i\left( \frac{p_i^2}{4\lambda_i^2}\right)^2$
then we have $$h(z) = \sum_{i=1}^2 \lambda_i z_i^2 + q$$
It should be easy to recognize the contour of $h$ should be consists of ellipses. From variable $y$ to $z$ what happens was a translation. From variable $x$ to $y$, the transformation is an orthogonal matrix multiplication, it is just a rotation / reflection. Hence $f$ contours consists of ellipses.

Similar analysis can be repeated as long as none of the $\lambda_i$ are zero. 
In the indefinite and non-singular case, one of the eigenvalue is positive and one of the eigenvalue is positive. By the same argument, we notice that the contour lines consists of hyperbola.

If $A=0$ and $b=0$ then $f$ is a constant. 
If $A=0$ and $b \ne 0$, then $f$ is linear and the contour consists of parallel straight lines.
If we consider the case where $A$ is non-zero, positive semidefinite but not positive definite, then one of the eigenvalue, $\lambda_1$ is positive and the other is $\lambda_2=0$.
Then $$g(y) = \lambda_1y_1^2+p_1y_1+p_2y_2+c$$
Also, $Ax+b=0$ become $UDU^Tx + b=0$, which is equivalent to $DU^Tx +U^Tb=0$ which is $\lambda_1 y_1 + p_1 =0$ and $0=p_2$. Hence whether the system $Ax+b=0$ is consistent and only if $p_2=0$.
If $Ax+b=0$ is consistent, that is if $p_2=0$, then our function $g$ is $$g(y) = \lambda_1y_1^2+p_1y_1+c$$
It is independent of $y_2$ and notice that it has multiple global minimum of the form of $(y_1^*, y_2)$. The contour lines are straight lines parallel to the $y_2$-axis.
If $Ax+b=0$ is not consistent, that is if $p_2 \ne 0$, then our function is $$g(y) = \lambda_1y_1^2+p_1y_1+p_2y_2+c$$
and since $p_2 \ne 0$, the contour lines consists of parabolas.

Picture source can be found here
