showing global min point

Question

let's $F(x)= x'Ax + a'x $

where $x'Ax$ is a quadratic form and $a'$ is defined as a vector.

$$A:= \left[ \begin{matrix} 6 &1&1 \\ 1&2&0 \\ 1&0&4\end{matrix}\right] $$

Does there exist a global minimum point (absolute min) for this function $F(x)$?

note: I know that if the function is given as a linear structure, whether this is convex or concave is not important. we only exemine $x'Ax$.

but i cannot perfectly prove this question. thank you for helping.

Answer 1

Isn't minimum or maximum were the gradient is zero (vector)?

$$ \frac{{\rm d}F(x)}{{\rm d}x} = 2 A x + a' =0 $$

So the unique solution

$$ x = - \frac{1}{2} A^{-1} a' $$

if it exists must be a minimum or maximum.

Edit 1

So for your example suppose $a'=(a_1,a_2,a_3)$ and $x=(x_1,x_2,x_3)$ then

$$ \frac{{\rm d}F(x)}{{\rm d}x} = 2 \begin{pmatrix} 6 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 4 \end{pmatrix} \begin{pmatrix} x_1 \\x_2 \\x_3 \end{pmatrix}+ \begin{pmatrix} a_1 \\a_2 \\a_3 \end{pmatrix} = \begin{pmatrix} 12 x_1+2 x_2+2 x_3 + a_1 \\ 2 x_1+4 x_2 + a_2 \\ 2 x_1 + 8 x_4 + a_3\end{pmatrix}$$

To make this the zero vector you need

$$ \begin{pmatrix} x_1 \\x_2 \\x_3 \end{pmatrix} =-\frac{1}{2} \begin{pmatrix} 6 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 4 \end{pmatrix}^{-1} \begin{pmatrix} a_1 \\a_2 \\a_3 \end{pmatrix} = \begin{pmatrix} -\frac{4 a_1-2 a_2-a_3}{42} \\ \frac{4 a_1-23 a_3-a_3}{84} \\ \frac{2 a_1-a_2 -11 a_3}{84} \end{pmatrix} $$

This makes $${\rm extrema}(F(x)) = \frac{-8 a_1^2+8 a_1 a_2+4 a_1 a_3-23 a_2^2-2 a_2 a_3 -11 a_3}{168} $$

Answer 2

A is symmetric and we can make it diagonal using an orthogonal matrix. Then we convert our problem into the form: $$f(x)=\lambda_1 x^2+\lambda_2 y^2+\lambda_3 z^2+ a x+by +c z$$, which is high school math. We find the eigenvalues of A are about $\{6.6,3.6,1.7\}$, therefore there exists minimum value of f.