I have a problem to solve and I have no idea how to do. I need to find the maximum and minimum of $G=2x²+2y²-15z²+8xy-12yz-12xz$ with $x²+y²+z²=1$ I'm supposed to use the diagonalization but I don't know with which matrix started... Thanks for your help :)
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You need to find the points (x,y,z) such that the partial derivatives are all equal to zero. Also when $x^2+y^2+z^2=1$ then how can $G$ be written? – Aggelos Bessis Nov 13 '16 at 23:35
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I can write G just in term of x and y ? – Mathieu Hassenforder Nov 13 '16 at 23:40
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Not quite, but it would be easier to compute the derivatives this way: $G=2(x^2+y^2+z^2)-17z^2+8xy-12yz-12xz$. So now you find the points : $f_x=0,f_y=0,f_z=0$ After that I'm assuming you'll need the Hessian. – Aggelos Bessis Nov 13 '16 at 23:46
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Thanks, but is it normal that I find x=y=z=0 when I compute the partital derivative ? – Mathieu Hassenforder Nov 14 '16 at 00:53
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hint: Lagrange Multiplier will work for this question ! Let $H(x,y,z) = x^2+y^2+z^2-1$, and solve $\nabla G = \alpha \nabla H$. You solve for $x,y,z, \alpha$.
DeepSea
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I know I can do it with Lagrange multiplier but is it possible to solve it with matrix diagonalization ? – Mathieu Hassenforder Nov 13 '16 at 23:39
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Hint: You can write G = $x' \cdot Q \cdot x$ where $x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $Q = \begin{bmatrix} 2 & 4 & -6 \\ 4 & 2 & -6 \\ -6 & -6 & -15 \end{bmatrix} $
Emil Kerimov
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Thanks, now I found a diagonal matrix : J = [ -18 0 0, 0 -2 0, 0 0 9] How can I deduce the maximum and the minimum now ? – Mathieu Hassenforder Nov 14 '16 at 00:44
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The minimum would be -18, and the maximum 9. With the solutions being the normalized eigenvectors corresponding to those eigenvalues. – Emil Kerimov Nov 14 '16 at 01:08
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Thanks a lot, and now, how can i find the coordinates of the maximum and the minimum ? – Mathieu Hassenforder Nov 14 '16 at 01:11
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Because after finding the value of the maximum and the minimum, I have to find the values of 3x²+2y²+z² when G is maximum and minimum – Mathieu Hassenforder Nov 14 '16 at 01:14
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So for example the minimum eigenvalue -18, corresponded with the eigenvector $\begin{bmatrix} 1 & 1 & 4 \end{bmatrix}$, when normalized it becomes $\begin{bmatrix} \frac{1}{3 \sqrt{2}} & \frac{1}{3 \sqrt{2}} & \frac{4}{3 \sqrt{2}} \end{bmatrix}$ – Emil Kerimov Nov 14 '16 at 01:15
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@MathieuHassenforder: note that $1=x^T x=x^T R^{-1} R x = x^T R^T R x=(Rx)^T (Rx)$, where $Q=R^{-1} D R= R^T D R$. – G Cab Nov 14 '16 at 01:20
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Oh yeah !! I solved the problem ! Thanks a lot for your help :) – Mathieu Hassenforder Nov 14 '16 at 03:35
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@MathieuHassenforder if this answered your question, you can accept this answer. – Emil Kerimov Nov 14 '16 at 03:38