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Let $A$ be symmetric positive definite (or maybe not, see the edit below), $x\in \mathbb{R}^n$, and span$\{v_1,\ldots,v_n\} = \mathbb{R}^n$.

What is a necessary and sufficient condition for when the two following problems give equivalent answers?

Here's the first problem: $$\min_{x_n}\ldots\min_{x_1} \frac{1}{2}\left(u-\sum_{m=1}^Nx_mv_m\right)^T A\left(u-\sum_{m=1}^Nx_mv_m\right) - f^T\left(u-\sum_{m=1}^Nx_mv_m\right)$$

Second problem: $$\min_x \frac{1}{2}\left(u-\sum_{m=1}^Nx_mv_m\right)^T A\left(u-\sum_{m=1}^Nx_mv_m\right) - f^T\left(u-\sum_{m=1}^Nx_mv_m\right)$$

Idea I think the answer should be that the directions $\{v_i\}_{i=1}^n$ are conjugate, meaning $$v_i^TAv_j = 0, i\neq j$$ I think I can show that this is a sufficient condition, but it takes quite a bit of writing. Any ideas on how to show that it's also a necessary condition? Is there a concrete example that would show it's necessary? Also, if you have a nice proof that it's sufficient, it'd be good to see that too, since I'm not totally happy with mine.

Thanks.

Edit: The question isn't actually clear on whether or not $A$ is symmetric positive definite. I suspect it is, but you may want to keep that in mind.

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  • $\begingroup$ And what is $f$? $\endgroup$ – Paul Sinclair Jan 7 '17 at 2:07
  • $\begingroup$ Just some given vector in $\mathbb{R}^n$. $\endgroup$ – Kurt Jan 7 '17 at 2:09
  • $\begingroup$ The two problems are always equivalent. $\endgroup$ – LinAlg Jan 7 '17 at 10:18

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