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Let $F$ be a field. Let $x_1,\ldots,x_k,y_1,\ldots,y_k\in F$ be distinct elements in the field. Consider the $k\times k$ matrix that in position $i$, $j$ has the element $\frac{\prod_{l\neq i}(y_i - x_l)}{\prod_{l\neq j}(x_j - x_l)}$.

What are its eigenvectors (as a function of $x_1,\ldots,x_k,y_1,\ldots,y_k$)?

Update: I had a typo in the above expression: the correct expression is $\frac{\prod_{l\neq j}(y_i - x_l)}{\prod_{l\neq j}(x_j - x_l)}$ ($j$ replaces $i$ in the numerator). It's indeed a nice exercise to compute the characteristic polynomial of the matrix I specified above, but the expression for the correct matrix is hairy. If it's known (which would make sense, since the transformation is important), please share!

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  • $\begingroup$ Why do you want to know? $\endgroup$
    – Steve Huntsman
    Commented Aug 13, 2011 at 0:10
  • $\begingroup$ The last line of the questions says: "What are its eigenvectors (as a function of x1,…,xk,y1,…,yk)?" $\endgroup$
    – user17119
    Commented Aug 13, 2011 at 0:11
  • $\begingroup$ Oh, that was a "why" rather than a "what"... I'm trying to construct small families of low degree curves that are "spread out" (in a certain concrete sense) yet have many intersections. $\endgroup$
    – user17119
    Commented Aug 13, 2011 at 0:15
  • $\begingroup$ I apologize for the elementary question. Just solved it myself without too much difficulty (for some reason I thought it would be hairy). It would actually make a nice problem for those of you who teach linear algebra. $\endgroup$
    – user17119
    Commented Aug 13, 2011 at 1:15
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    $\begingroup$ If you solved it, upload your solution as an answer so it can help someone else. If you are shy about picking up reputation for solving your own question, you can mark the answer CW, although I don't think you need to. $\endgroup$
    – David E Speyer
    Commented Aug 18, 2011 at 1:27

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