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got a two-part question here.

Given (symmetric, positive definite) matrix $A$ = $$\begin{bmatrix} 2 & -1 \\ -1 & 2 \\ \end{bmatrix} $$ solve the problem: $arg max$ $g(x) = (Ax,x)$ subject to $\lVert x\rVert^2 = 1$.

Use two methods:

(1) Use the fact that the unit circle $S_1$ $\equiv$ $\{x: \lVert x\rVert = 1\}$ can be parametrized using a single variable $t$. This gives a function of one variable to be maximized.

(2) Enforce the constraint using Lagrange multipliers.

I know that the parametrization of $S_1$ using $t$ should be $x(t) = [cos(t), sin(t)]^T$, but I'm not sure how to proceed from there to finding the value that maximizes $g(x)$. As for the Lagrange multipliers, I'm not clear on how to find them from the constraint, nor on how to apply them to find the argmax. How is this done?

Any help on how to proceed is greatly appreciated!

user0102
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BigT
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  • For method 1, you just need to write $g(x_1,x_2)$ in terms of $(x_1,x_2)=(\cos(t), \sin(t))$. So, first, write $g(\cos(t), \sin(t))$ and explicitly calculate that function in terms of $t$. That is, $$g(\cos(t), \sin(t)) = (\cos(t); \sin(t))^T A (\cos(t); \sin(t)) = ...$$ – Michael Sep 21 '18 at 22:42
  • For method 2, say $x=(x_1,x_2)$. Then $g(x)=2x_1^2 -2 x_1 x_2 + 2 x_2^2$ and the constraint is $x_1^2+x_2^2-1 =0$. Are you now able to set up the Lagrange function and find the multipliers? – Ernie060 Sep 21 '18 at 22:50
  • Ah, yes! Using these methods, I was able to find that $g(x)$ is maximized where $(x_1, x_2) = (-\sqrt{\frac 12}, \sqrt{\frac 12})$, and where $(x_1, x_2) = (\sqrt{\frac 12}, -\sqrt{\frac 12})$. Thanks to you both - the clarification was really helpful! – BigT Sep 21 '18 at 23:40

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