Find the points of the surface defined by $-x_1^2 + x_2^2 - x_3^2 + 10x_1x_3 = 1$ closest to the origin

Question

For the surface

$$\{ (x_1, x_2, x_3) \in \mathbb R^3 \mid -x_1^2+x_2^2-x_3^2 + 10x_1x_3=1 \}$$

I want to find the points on it closest to the origin. I know how to do a substantial part of this. First I make the matrix

$$A = \begin{pmatrix}-1 & 0 & 5 \\ 0 & 1 & 0 \\ 5 & 0 & -1\end{pmatrix}$$

and then get the eigenvalues

$$|A-\lambda I| = \begin{vmatrix} -1-\lambda & 0 & 5 \\ 0 & 1-\lambda & 0 \\ 5 & 0 & -1-\lambda\end{vmatrix}$$

$$ = (-1-\lambda)(1-\lambda)(-1-\lambda) + 5(-5(1-\lambda))$$

$$= (1-\lambda)([-1-\lambda]^2-25)$$

$$= (1-\lambda)(-1-\lambda-5)(-1-\lambda+5)$$

$$= (1-\lambda)(-6-\lambda)(-4-\lambda) $$

$$= (1-\lambda)(6+\lambda)(4+\lambda)$$

So the eigenvalues are $\lambda = 1,-6,-4$ and therefore the quadratic form can be represented as

$$c_1^2 - 6c_2^2 - 4c_3^2=1$$

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The part I don't get is how to find the point closest to the origin from here. I've seen solutions to problems similar to this where they merely set two variables to 0 and out of all such choices of two variables, they pick the one with the least value for the third which satisfies the equation.

However, how do we know this process will select the point closest to the origin? If coefficients were made suitably extreme, at least on the face of it, you'd think we might be able to engineer and example where picking some very small mixture of non-zero values of $c_1, c_2, c_3$ would satisfy the equation and be smaller than what you'd get for setting some two of them to 0.

In this particular case we could set $c_2=c_3=0$ so that $c_1=1$. But what guarantee is there that we couldn't give $c_2$ just a little something, drastically reducing the amount we need to give to $c_1$? Even if we couldn't do that here, what if the coefficient of $c_2$ were negative a million, so that you'd think just a little bit in $c_2$ goes a very long way to reducing what we need from $c_1$?

[Edit: As pointed out, I should have said that the characteristic equation was

$$(1-\lambda)(-6-\lambda)(4-\lambda)$$

so that the eigenvalues are 1, -6, and 4. Then the quadratic can be expressed as

$$c_1^2-6c_2^2+4c_3^2=1$$]

Answer 1

If you start at $(c_1,c_2,c_3)=(1,0,0)$ and give a little something to $c_2$, then you would have to increase $c_1$ in order to keep $c_1^2 - 4 c_2^2 - 6 c_3^2$ equal to $1$. (I'm not quite sure I understand your worries correctly, but you seem to think that $c_1$ can be decreased? If so, you've got it backwards.)

And any point $(c_1,c_2,c_3)$ with $c_1>1$ is farther from the origin than $(c_1,c_2,c_3)=(1,0,0)$, so $(1,0,0)$ is closest (together with $(-1,0,0)$).

Answer 2

It might be easier to see what’s going on by switching to cylindrical coordinates. With $c_1=r\cos\phi$ and $c_3=r\sin\phi$, the transformed equation (with corrected sign) becomes $$r^2={1+6c_2^2 \over 1+3\sin^2\phi}.$$ The numerator has an absolute minimum of $1$ at $c_2=0$ and the denominator lies in the interval $[1,4]$. The minimum value of $r^2$ clearly occurs at $c_2=0$, $\phi = \pm\pi/2$, i.e., at $c_1=0$, $c_3=\pm1/2$.

Looking at this geometrically, the intersections of this surface with the planes $c_2=const$ are homothetic ellipses centered on the $c_2$-axis, with the smallest of the family occuring at $c_2=0$: making any changes to $c_1$, $c_2$ and $c_3$ that moves you off of this plane can only move you away from the minimum-distance point(s). On the other hand, if you hold $c_2$ fixed at zero, you can’t change $c_1$ and $c_3$ arbitrarily. They are constrained by the minimal ellipse.

Answer 3

According to some theorem I forgot name of, you can get away with just finding points having gradient point to origo.

$$g(x_1,x_2,x_3) = -{x_1}^2+{x_2}^2-{x_3}^2+10x_1x_3 - 1$$

$$\nabla g = [-2x_1 + 10x_3, 2x_2, -2x_3 + 10x_1]^T$$

so $$ {\bf x} - \lambda (\nabla g ({\bf x})) = 0, \lambda \in \mathbb R$$

$$ [x_1 - \lambda(10x_3-2x_1), x_2 - \lambda(2x_2), x_3 - \lambda(10x_1-2x_3)] = \bf 0$$

We first see either $\lambda = 1/2$ or $x_2=0$ must hold for component $2$ to fulfill.

if $x_2\neq 0$:

This means for component 1: $x_1 - 1/2(10x_3-2x_1) = 0$ giving $5x_3 = 2x_1$

And component 3: that $5x_1 = 2x_3$

Only solution can then be along $[0,1,0]^T$

if $x_2=0$:

$$x_1-\lambda(10x_3-2x_1) = 0\\x_3-\lambda(10x_1-2x_3) = 0$$

from first: $\lambda = \frac{x_1}{10x_3-2x_1}$

$$x_3 - \frac{x_1(10x_1-2x_3)}{10x_3-2x_1} = 0$$ $$\frac{x_3(10x_3-2x_1)}{10x_3-2x_1} - \frac{x_1(10x_1-2x_3)}{10x_3-2x_1} = 0$$ simplification gives: $$x_3^2 = x_1^2$$

so $$\lambda[\pm 1,0,\pm 1]$$ must be solution.

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We can verify soundness by these Octave commands

% First define level set:
g = @(a) -a(1).^2+a(2).^2-a(3).^2+10*a(1)*a(3) - 1;
% Now define cost function 1e3 factor indicates we want fit to manifold to be 1000 times more important than being close to origo:
f = @(a) 1e3*g(a).^2 + norm(a).^2;
% Now run Nelder-Mead numerical optimization algorithm:
x = fminsearch(f,randn(1,3))

This gives me: a typical solution [-3.5353e-01 1.5206e-05 -3.5353e-01]

which would indicate $x_1=x_3 \approx -0.3535, x_2 \approx 0$