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Consider the following $(2d+1)\times (2d+1)$ matrix:

$$ A = \begin{pmatrix} 0 &\frac{2d}{2} & 0 &0 & \cdots &0 & 0 \\ \frac{1}{2} & 0 & \frac{2d-1}{2} &0& \cdots & 0 & 0\\ 0 & \frac{2}{2} &0 & \frac{2d-2}{2} &\cdots & 0 & 0\\ 0 & 0 & \frac{3}{2} & 0 &\cdots &0 & 0\\ \vdots &\vdots &\vdots &\vdots & \ddots &\vdots &\vdots\\ 0 &0 &0 &0 &\cdots &0 & 1/2 \\ 0 &0 &0 &0 &\cdots &\frac{2d}{2} & 0 \end{pmatrix} $$

How can one show that $A$ has eigenvalues $-d, -d+1,...,d$?

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  • $\begingroup$ Also, it looks like $A$ is of size $(2d+1)×(2d+1).$ So, it defines a linear transformation from $\Bbb R^{2d+1}$ to itself, say. $\endgroup$
    – no upstairs
    Commented Apr 15 at 2:10
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    $\begingroup$ This question was on the Alibaba 2024 qualifying round. The qualifying round has now ended. I think it is okay that this question is now posted to MO, though I'm not sure. $\endgroup$
    – mathworker21
    Commented Apr 15 at 2:51
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    $\begingroup$ This is known as the Sylvester-Kac matrix and related forms. $\endgroup$
    – Toni Mhax
    Commented Apr 15 at 6:09
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    $\begingroup$ This and many similar matrices are discussed in Muir's book "A Treatise on the Theory of Determinants". $\endgroup$
    – Nemo
    Commented Apr 15 at 9:23
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    $\begingroup$ This looks like the matrix representing the sum of a simple root vector and its opposite in an irreducible representation of $\mathfrak{sl}_2$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 17 at 5:18

2 Answers 2

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You hope to show for each $n$ that the $(n+1)\times(n+1)$ matrix $$ B_n = \begin{pmatrix} 0 &n & 0 &0 & \cdots &0 & 0 \\ 1 & 0 & n-1 &0& \cdots & 0 & 0\\ 0 & 2&0 & n-2 &\cdots & 0 & 0\\ 0 & 0 & 3 & 0 &\cdots &0 & 0\\ \vdots &\vdots &\vdots &\vdots & \ddots &\vdots &\vdots\\ 0 &0 &0 &0 &\cdots &0 & 1 \\ 0 &0 &0 &0 &\cdots &n & 0 \end{pmatrix} $$ has eigenvalues $-n,2-n,\dots,n-2,n$ (your matrix is related by $A=\frac12B_{2d}$).

Recall that there is a $(n+1)$-dimensional representation $\varphi\colon\mathfrak{sl}_2\to \mathfrak{gl}_{n+1}(\mathbb C)$ of $\mathfrak{sl}_2$ given by $$\begin{align*} \varphi(e)&=\begin{pmatrix} 0 &n & 0 &0 & \cdots &0 & 0 \\ 0 & 0 & n-1 &0& \cdots & 0 & 0\\ 0 & 0&0 & n-2 &\cdots & 0 & 0\\ 0 & 0 & 0 & 0 &\cdots &0 & 0\\ \vdots &\vdots &\vdots &\vdots & \ddots &\vdots &\vdots\\ 0 &0 &0 &0 &\cdots &0 & 1 \\ 0 &0 &0 &0 &\cdots &0 & 0 \end{pmatrix}\\ \varphi(h)&=\begin{pmatrix} n &0 & 0 &0 & \cdots &0 & 0 \\ 0 & n-2 &0 &0& \cdots & 0 & 0\\ 0 & 0&n-4 & 0 &\cdots & 0 & 0\\ 0 & 0 & 0 & 0 &\cdots &0 & 0\\ \vdots &\vdots &\vdots &\vdots & \ddots &\vdots &\vdots\\ 0 &0 &0 &0 &\cdots &2-n & 0 \\ 0 &0 &0 &0 &\cdots &0 & -n \end{pmatrix}\\ \varphi(f)&=\begin{pmatrix} 0 &0 & 0 &0 & \cdots &0 & 0 \\ 1 & 0 & 0 &0& \cdots & 0 & 0\\ 0 & 2&0 & 0 &\cdots & 0 & 0\\ 0 & 0 & 3 & 0 &\cdots &0 & 0\\ \vdots &\vdots &\vdots &\vdots & \ddots &\vdots &\vdots\\ 0 &0 &0 &0 &\cdots &0 & 0 \\ 0 &0 &0 &0 &\cdots &n & 0 \end{pmatrix} \end{align*}$$ Thus, $B_n$ is simply $\varphi(e+f)$. But $e+f$ and $h$ are conjugate under the adjoint action of $\mathrm{SL}_2$ on $\mathfrak{sl}_2$. Thus $B_n$ is conjugate to $\varphi(h)$. But the eigenvalues of $\varphi(h)$ are clearly $-n,2-n,\dots,n-2,n$.

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  • $\begingroup$ Can you explain the notations $\mathfrak{sl}_2$ and $\mathfrak{gl}_{n+1}(\mathbb C)$, and also what "the adjoint action of $\mathrm{SL}_2$ on $\mathfrak{sl}_2$" is? $\endgroup$
    – Iosif Pinelis
    Commented Apr 15 at 16:25
  • $\begingroup$ The Lie group $\mathrm{SL}_2$ has the tautological $2$-dimensional representation $V$, so $\mathrm{Sym}^nV$ is a $(n+1)$-dimensional representation. This is a group homomorphism $\mathrm{SL}_2\to\mathrm{GL}_{n+1}$, whose derivative is the Lie algebra homomorphism $\mathfrak{sl}_2\to \mathfrak{gl}_{n+1}$ in my answer. $\endgroup$
    – Kenta Suzuki
    Commented Apr 15 at 21:01
  • $\begingroup$ Thank you for your response. However, I know next to nothing about Lie groups and algebras, and I don't know what $\text{Sym}^n V$ stands for. Can the eigenvectors here be simply described? $\endgroup$
    – Iosif Pinelis
    Commented Apr 15 at 21:30
  • $\begingroup$ @IosifPinelis Think of your matrices as operators on the space of all homogeneous polynomials $h(x,y)$ of degree $n$. Then, in the monomial basis, $B_n$ is the operator $x\frac\partial{\partial y}+y\frac\partial{\partial x}$ (this is one possible realization of the $\mathfrak{sl}_2$-irrep). The $\lambda$-eigenfunctions for this operator are $\left(\frac{x+y}{x-y}\right)^{\frac\lambda2}F(x^2-y^2)$, with $F$ any function. So one just examines for which $\lambda$ can one obtain polynomials. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 19 at 16:25
  • $\begingroup$ @მამუკაჯიბლაძე : Thank you for your comment. Can you expand its last sentence, maybe in a formal answer? I think it would be useful to know the eigenvectors more or less explicitly. $\endgroup$
    – Iosif Pinelis
    Commented Apr 19 at 16:39
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As requested by Iosif Pinelis, here is another version of the (perfectly correct, in my opinion) answer by Kenta Suzuki.

Let $V$ be the $d+1$-dimensional vector space of homogeneous polynomials of degree $d$ in variables $x$, $y$. Then the matrix $B_n$ from that answer is the matrix of the operator $T=x\frac\partial{\partial y}+y\frac\partial{\partial x}$ in the monomial basis $x^d$, $x^{d-1}y$, ..., $xy^{d-1}$, $y^d$. In variables $x+y=u$, $x-y=v$ the operator $T$ is $u\frac\partial{\partial u}-v\frac\partial{\partial v}$, with $T(u^iv^j)=(i-j)u^iv^j$, so monomials in $u$, $v$ form an eigenbasis, with eigenvalues $d$, $d-2$, ..., $2-d$, $-d$.

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