How can I solve a multivariable equation using the matrix form?

Question

I have the following equation:

$$ f(x,y) = x^2 + cy^2 + 2xy $$

I want to find the values of c for the function $f(x,y)$ has a minimum. However I have to solve it by writing the function in the following form:

$$ f(x,y) = \begin{bmatrix} x \\ y \end{bmatrix} ^T A \begin{bmatrix} x \\ y \end{bmatrix}$$

I generally solve the equations by differentiating by x and y and equating to zero, then forming a Hessian matrix and then checking if my point(s) are a minimum or not. However, I am very unfamiliar to solving it in the form above. Could someone explain the relevance of the form to finding the minimum.

Answer 1

One direct way of doing this is to write $A = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$ and multiply $$ (x, y) A (x,y)^T = \alpha x^2 + (\beta + \gamma)xy + \delta y^2 $$ which in your case forces $\alpha = 1, \delta = c$ and $\beta + \gamma = 2$, since you have a choice, pick $\beta = \gamma = 1$ so $A = A^T$ and we get $A = \begin{pmatrix} 1 & 1 \\ 1 & c \end{pmatrix}$.

Can you finish? Here is a useful set of lecture notes on optimization of quadratic forms.

Answer 2

We have: $f(x,y) = (x+y)^2 + (c-1)y^2$. Thus if $c = 1$, clearly the minimum is $0$ when $x = -y$. If $c \neq 1$, then either $c < 1$ or $c > 1$. For $c < 1$, take $x = -y$ and thus $f(-y,y) = (c-1)y^2$ has no minimum. For $c > 1$, again $f(-y,y) = (c-1)y^2 \ge 0$ and has a minimum of $0$ when $y = 0$ and in this case $x = -y = 0$ as well. In brief, the minimum occurs for $f$ when $c \ge 1$.

Answer 3

The quadratic form is $$ x'Ax = x^2 a_{11} + xy (a_{12}+a_{21}) + a_{22} y^2 $$ and you have $$ x^2+cy^2 +2xy $$ So $a_{11} = 1$, $a_{22} = c$, and $a_{12} = a_{21} = 1$.

The Hessian is the matrix $A$.