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I have the following tensor (takes a vector of length $m$ and returns a matrix $m \times m$):

$C(y) = A \operatorname{diag}(A^T y ) A^{-1}$

for some invertible matrix $A$ of size $m \times m$ ($y$ is of size $m \times 1$).

Let's say I have $C(y)$ (a way to compute it, etc.) for any $y$ I want - is there a way to solve for $A$? (i.e. identify what $A$ is.)

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    @anon: the problem is that $\mathbf y$ is fixed, and that the same similarity matrix $A$ must both diagonalize $C(\mathbf y)$ and produce the spectal-vector $\lambda$ from $\mathbf y$.2011-09-24
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    @anon: that's a clearer remark to me than your previous one, and upon reflection it seems essentially complete. I spent some time writing a full solution analogous to your comment; but you should turn your comment into an answer to claim the credit.2011-09-24
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    Hm, just because $C=A^{-1}LA$ doesn't mean $A$'s columns are eigenvectors. [Thus I've deleted my comments and answer.] I'll have to see if the original idea can be augmented a bit. (And I do realize $C$ is diagonalizable precisely when $y\ne0$.)2011-09-25
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    Note that $\operatorname{Tr} C(y)=\sum_i(A^Ty)_i$ and $\det C(y)=\prod_i(A^Ty)_i$. For instance, by using the canonical basis vectors for $y$, you can get the sums and products of the rows of $A$.2011-09-25
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    I think it would be helpful to further discuss the answers that have been deleted. Since not everyone can see the deleted answers, it might help if you undelete them. @Niel, I assume you deleted your answer because the argument for dispensing with the permutations was wrong? Still it solves the problem up to permutations. dotproduct, I don't think your argument against using a single $y$ is valid. The fact that $y=0$ obviously doesn't help doesn't imply that $A$ can't be reconstructed using a single $y$ -- we do get a full matrix for a single $y$.2011-09-25
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    @anon: I think the fact that $A$ diagonalizes $C$ does mean that the columns of $A$ are eigenvectors of $C$. Denoting the $i$-th column of $A$ by $a_i$ and the $i$-th canonical basis vector by $e_i$, we have $Ca_i=A\Gamma A^{-1}a_i=A\Gamma e_i=A\Gamma_ie_i=\Gamma_ia_i$. I do think the solution lies in using this; it's just that we have to deal with the permutations somehow. For a given $y$, there is one solution for $A$ for every permutation of the eigenvalues; we might be able to identify the right one by substituting a second $y$, but that might take $O(m!)$ time.2011-09-25
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    @joriki: Ah, I guess it is made of eigenvectors then. I've undeleted my answer; I'll try to elaborate on it later for OP but I'm not feeling well right now so I'll just leave it for discussion. As for the permutations, it might be the case that the permuting of the eigenvectors and the permuting of the eigenvalues cancel each other out in seeking out $A$...2011-09-25
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    @anon: Thanks. About the permutations: I don't think there's that kind of cancellation. If $U^{-1}CU=\Lambda$, then $A$ could be any matrix of the form $U\Pi D$, where $\Pi$ is a permutation matrix and $D$ is diagonal, and substituting into the condition yields $\Lambda=\operatorname{diag}(D^T\Pi^TU^Ty)$. Unless $U^T=y$ contains zeroes, we can find a $D$ for each possible $\Pi$ to solve this, and these will generally lead to different $A$s.2011-09-25
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    @joriki could it be that solving this equation for a given $y$ gives only necessary conditions that $A$ has to satisfy? this sounds like it could be true - however, on the other hand, if there is only a single solution for $A$ given a certain $y$, then that /has/ to be right the $A$, isn't it.2011-09-25
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    @dotproduct: I don't understand how your comment relates to what I'd written. I wrote that for one $y$ we may get different possible $A$s corresponding to different permutations, so yes, that means that for a given $y$ we only get necessary conditions on $A$, namely being one of those possibilities -- I don't understand why you write "could it be". I don't understand the second sentence, either -- yes, if there's only a single solution for $A$, then that has to be the right $A$ -- that sounds tautological to me.2011-09-25
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    @joriki sorry, that was just a general remark about trying different $y$s, I wasn't necessarily addressing directly your previous comment. What's surprising me is that for a certain $y$ we could potentially get a single $A$ that satisfies the equation with that $y$ and with another $y$ we could get a different $y$. I don't think that could actually happen (because it is a contradiction), but I need to convince myself about it.2011-09-25
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    "we could get a different *$A$*..."2011-09-25
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    @dotproduct: Note that you can edit your comments (i.e. you could have changed the $y$ to an $A$). If we get two single but different $A$s for two different $y$s, we must have made a mistake, since as you say that would be a contradiction. I don't think there's anything mysterious there, it's the same as in simpler cases -- we can plug some $y$ into $(y-a)^2$ and deduce two different possible values of $a$; then if we plug in a different $y$ we again get two possible values of $a$, and only one of them matches one of the other two. We never get values none of which matches one of the others.2011-09-25
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    @joriki: about your comment of my argument of using $y = 0$. It could be that a single $y$ gives the full answer, but we would have to find a $y$ such that $A^T y$ does not contain a 0, is that correct? also, doesn't the $y=0$ argument mean that $A$ does not consist of the eigenvectors of *any* $C(y)$?2011-09-25
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    @dotproduct: We seem to be talking past each other. I'm sorry, I can't make any sense of your comments. Again you said "it could be" even though I said in my comments that I don't think a single $y$ gives the full answer -- if you disagree, please say why. (I only said that your argument why it shouldn't give the full answer was invalid.) "Finding" a $y$ such that $A^T$ doesn't contain a $0$ isn't difficult, since this will be the case for all random $y$s except for a set of measure $0$; but I'm not sure why we would want such a $y$.2011-09-25
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    @joriki, about the first part of the comment, sorry, I guess I have misunderstood you. About the second comment: I thought there might be problems with cases where the diagonal matrix $\mathrm{diag}(A^t y)$ has 0 on the diagonal somewhere (just like $y=0$ gives a problem) - we could be losing some information because the rank of $C(y)$ in this case is less than the rank of $A$. I am not sure if it is true.2011-09-25
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    (in any case, I suspect that there is no single solution for this. for example, if we consider $m=1$, then we have $C(y) = a^2 y$. In this case, it seems like there is no single solution! if $a = c$ is a solution, then so is $a = -c$. I am not sure how it generalizes to arbitrary $m$.)2011-09-25

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