# Find the geometry of the curves of the contour lines of $f(x) = \frac{1}{2}x^tAx + b^tx + c$

by Guerlando OCs   Last Updated August 15, 2018 02:20 AM

Find the geometry of the curves of the contour lines of a quadratic function

$f(x) = \frac{1}{2}x^tAx + b^tx + c$ for $A\in \mathbb{R}^{2\times 2}$, $b\in \mathbb{R}^2, c\in \mathbb{R}$ 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 $A = \begin{bmatrix} a & b \\ b & a\\ \end{bmatrix}$ then the function becomes

$$f((x_1,x_2)) = \frac{1}{2}\begin{bmatrix} x_1 & x_2 \end{bmatrix}\begin{bmatrix} a & b \\ b & a\\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2\\ \end{bmatrix} + \begin{bmatrix} b_1 & b_2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2\\ \end{bmatrix} + c = \\ \frac{1}{2}(ax_1^2 + 2bx_1x_2 + ax_2^2) + b_1x_1 + b_2x_2 + c$$

I don't know if I can simplify $(ax_1^2 + 2bx_1x_2 + ax_2^2)$. 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 $\frac{1}{2}(ax_1^2 + 2bx_1x_2 + ax_2^2)>0$ right?

And $A\ge 0 \implies \frac{1}{2}(ax_1^2 + 2bx_1x_2 + ax_2^2)\ge 0$, 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.

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