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Jul 22, 2020 at 16:16 comment added Rodrigo de Azevedo X-posted: math.stackexchange.com/q/3764350/339790
Jul 22, 2020 at 0:48 history edited heng CC BY-SA 4.0
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Jul 22, 2020 at 0:35 comment added heng @lambda I agree, from $\mathbf{y}$ to $\mathbf{x}$ there's a sign ambiguity even if $\mathbf{A}$ is full rank. Actually in my case $\mathbf{x} > 0$. I want to know whether there are other methods to solve quadratic equations?
Jul 22, 2020 at 0:31 comment added heng @LSpice Let's say $\mathbf{A}$ is always full rank in this case. I want to know how many quadratic equations are required to obtain $\mathbf{x} \in \mathbb{R}^3$, is p=3 enough?
Jul 21, 2020 at 16:06 comment added Rodrigo de Azevedo Related: mathoverflow.net/q/308163
Jul 21, 2020 at 15:48 comment added Rodrigo de Azevedo Have you tried in $\mathbb{R}^2$ first?
Jul 21, 2020 at 15:40 history edited Rodrigo de Azevedo
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Jul 21, 2020 at 15:30 history edited Rodrigo de Azevedo CC BY-SA 4.0
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Jul 21, 2020 at 13:42 comment added lambda If $\mathbf x$ is a solution then so is $-\mathbf x$, so you can't ever have uniqueness. Up to sign you can certainly recover $\mathbf x$ from $\mathbf y$ so the condition for having only two solutions is indeed just that the matrix has full rank.
Jul 21, 2020 at 13:42 review Close votes
Jul 27, 2020 at 3:03
Jul 21, 2020 at 13:23 comment added LSpice What do you mean by "What is the minimal value of $p$ to guarantee a unique solution"? Do you mean the minimal value for which a solution can be unique? No matter how big $p$ is, $\mathbf A$ could be, say, the $0$ matrix, in which case there are no solutions, or some low-rank matrix, in which case there could be multiple solutions.
Jul 21, 2020 at 13:22 history edited LSpice CC BY-SA 4.0
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Jul 21, 2020 at 13:06 history edited heng CC BY-SA 4.0
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Jul 21, 2020 at 12:05 review First posts
Jul 21, 2020 at 12:23
Jul 21, 2020 at 11:56 history asked heng CC BY-SA 4.0