Find the minimum of
f(x,y) = ax^2 + 2bxy + cy^2

Question

Find the minimum of 
f(x,y) = ax^2 + 2bxy + cy^2

photon336 · Answer

interesting, how would we find a minimum?

baru · Answer

points where $f_x=f_y=0$ ?

ganeshie8 · Answer

how about completing the square ?

ganeshie8 · Answer

btw, there is a really nice way to factor this quadratic using matrices : 

$$f(x,y)=\begin{bmatrix} x&y\end{bmatrix}\begin{bmatrix} a&c\c&b\end{bmatrix}\begin{bmatrix} x\ y\end{bmatrix}$$

ganeshie8 · Answer

It seems the pivots of that matrix in the middle might reveal some info about the minima of f(x,y)

ganeshie8 · Answer

I'm trying to understand the relationship between the given quadratic and the properties of that matrix

kainui · Answer

Slight mistake there, in your matrix you've mixed up your c and b, they should be swapped around.

Critical thing for symmetric matrices is that their eigenvalues have orthogonal eigenvectors. This ends up meaning some magical stuff because it means we can always find a linear transformation to bring our quadratic form into something of the form:

$$f(u,v)=pu^2+qv^2$$

Every step of the process is pretty interesting to me, but I'll go ahead and sketch out what happens when you diagonalize your quadratic form matrix.

$$x^\top A x = x^\top (P^{-1} D P) x$$
It turns out that since A is symmetric, $P^{-1} = P^\top$ as long as you normalize the eigenvectors in the matrix P. So this ends up becoming:

$$x^\top P^\top D Px = (Px)^\top D (Px)$$
So we define $u = Px$ which is really our change from the x,y basis to the u,v basis I described above. and we have:

$$u^\top D u =pu^2+qv^2 $$