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I am trying to work through some problems to find the minimizer and minimum value of a function. The book I am using doesn't have a clear cut example and I can't seem to find a good example online anywhere, so I have no idea where to begin. The function I am trying to do this for is $$ 3x^{2} + 3xy + 3y^{2} - 2x - 2y + 4 $$

The only thing I can find is that we have to rewrite the equation in the following compact matrix notation $$ p(\textbf{x}) = \textbf{x}^{T}K\textbf{x} - 2\textbf{x}^{T}\textbf{f} + c $$ where $K = (k_{ij})$ is a symmetric $n \times n$ matrix, $\textbf{f}$ is a constant vector, and $c$ is a constant scalar. However, it doesn't really explain how to come up with those values. I'm guessing that in the problem I am trying to work through that $c = 4$ and that $\textbf{x} = [x, y]^{T}$, but I have no idea how to determine $K$ and $\textbf{f}$.

If someone could show me how to find $K$ and $\textbf{f}$ I would greatly appreciate it.

1 Answers1

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By inspection

$K = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$

$$ \textbf{x}^{T}K\textbf{x} = a x^2 + 2 b x y + c y^2 = 3 x^2 + 3 x y + 3 y^2 $$

So $K = \begin{pmatrix} 3 & 3/2 \\ 3/2 & 3 \end{pmatrix}$.

Similarly

$\textbf{f} = \begin{pmatrix} f_1 \\ f_2 \end{pmatrix}$

$$ -2\textbf{x}^{T}\textbf{f} = -2 f_1 x - 2 f_2 y = -2 x -2 y $$

Thus $\textbf{f} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

Rol
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  • Okay, so given $\textbf{x}^{T} = \begin{bmatrix}x & y\end{bmatrix}$ and the quadratic equation $ax^{2} + bxy + cy^{2} + dx + ey + f$ is it safe to say the following? $$ K = \begin{bmatrix} a & \frac{b}{2} \ \frac{b}{2} & c \end{bmatrix} \quad\text{and}\quad \textbf{f} = \begin{bmatrix} -\frac{d}{2} \ -\frac{e}{2} \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} d \ e \end{bmatrix} $$ – user224534 Mar 18 '15 at 07:37
  • Yes, and it is not only safe, but the only possibility – Rol Mar 18 '15 at 07:44
  • Thank you, it would be really helpful if they just put that in the book. – user224534 Mar 18 '15 at 07:49
  • Indeed, I also learn better by looking at examples. Feel free to browse this page – Rol Mar 18 '15 at 07:54