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$$A=\left[ \begin{array}{cc} a & b \\ c & d \end{array}\right] $$

$$B=\left[ \begin{array}{ccc} a^{2} & ab & b^{2} \\ 2ac & ad+bc & 2bd \\ c^{2} & cd & d^{2} \end{array}\right] $$

Let eigenvalues of $A$ be $a_1,a_2$. Then, eigenvalues of $B$ are $a_1^2$, $a_1a_2$, $a_2^2$. Why does this happen? Is this related to spectral analysis of boolean function or spectral property of Cayley graph?

Giovanni Resta
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kswim
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  • I'm not sure why you picked this matrix, but notice that if you interpret the columns of $A$ as $v=ax+c$ and $w=bx+d$, then the columns of $B$ are $v^2, vw,$ and $w^2$ (multiplication of polynomials and then extracting coefficients as coordinates). My stupid guess would then be that the eigenvalues should be $a_1^2, a_1a_2,$ and $a_2^2$, but this doesn't seem to be the case :/ – Eric Stucky Feb 15 '16 at 02:56
  • I asked this question because of your guess. I observed that it is generalized as $a_1^{i^1}\cdots a_n^{i^n}$ where $i^1,\dots,i^n$ are nonnegative integer solution and their summation is $n-1$. – kswim Feb 15 '16 at 03:34
  • Upon further analysis, my stupid guess seems to be correct: compare these eigenvalues and these eigenvalues, for instance. Since symmetric polynomials are showing up so much, I asked some of the combinatorialists in my department; they suspect (and now I do too) that this has something to do with Schur-Weyl duality, since the operation that sends $A$ to $B$ is a polynomial representation of $GL(n)$. – Eric Stucky Feb 15 '16 at 22:13
  • In particular, my notes say that (since the representation is homogeneous) this guarantees the existence of some multiset of monomials $x^ny^m$ so that if the eigenvalues of $A$ are $a,b$, then the eigenvalues of $B$ are all corresponding $a^nb^m$. But my notes do not have the proof (nor a reference :P) so I'm not sure how to guarantee that our monomials are $x^2, xy$, and $y^2$. – Eric Stucky Feb 15 '16 at 22:26

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