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If I am given the surface $$x_1x_3-x_2x_4=1$$ $$x_2x_3+x_1x_4=0$$ how would I find the point on this surface closest to the origin?

I first tried $f(x)=x_1^2+x_2^2+x_3^2+x_4^2$ and $g(x)=x_1x_3-x_2x_4-x_2x_3-x_1x_4-1$

Then I found the gradient of each and tried to use a lagrange multiplier and got $$2x_1=\lambda(x_3-x_4)$$ $$2x_2=\lambda(-x_4-x_3)$$ $$2x_3=\lambda(x_1-x_2)$$ $$2x_4=\lambda(-x_2-x_1)$$ I know this will result in a system of equation to solve for $x_1,x_2,x_3,x_4$ and $\lambda$ but I am not sure how to get to that point.

EDIT: So I tried it again and got these new equations:$$\lambda_1x_3+\lambda_2x_4-2x_1=0$$ $$-\lambda_1x_4+\lambda_2x_3-2x_2=0$$ $$\lambda_1x_1+\lambda_2x_2-2x_3=0$$ $$-\lambda_1x_2+\lambda_2x_1-2x_4=0$$

I did this by having $g_1(x)=x_1x_3-x_2x_4-1$ and $g_2(x)=x_2x_3+x_1x_4$ and having

grad$f(x)=\lambda_1$grad$g_1(x)+\lambda_2$grad$g_2(x)$

But after this I am still stuck. With these four equations, $g_1(x)$, and $g_2(x)$ I have six equations and six unknowns which should work out but I cannot figure out what the next step is.

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    $\begingroup$ Your distance function $f$ is good, but your restriction function isn't. Do you know how to use Lagrange multipliers with multiple constraint functions? $\endgroup$ – Fimpellizieri Apr 5 '17 at 21:22
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    $\begingroup$ As you have two restrictions, you need two multipliers. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 5 '17 at 21:24
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Hint: is a homogeneous linear system with two parameters (the multipliers). If the determinant of the matrix of the system is $\ne 0$, the only solution is (0,0,0,0) that $\not\in$ manifold. When the determinant is $= 0$... can you continue?

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