What is the Limit of the Hankel Determinant in a Matrix Challenge Problem?

[MountEvariste]

Challenge Problem: Let $A$ be an $r \times r$ matrix with distinct eigenvalues $λ_1, . . . , λ_r$. For $n \ge 0$, let $a(n)$ be
the trace of $A^n$. Let $H(n)$ be the $r \times r$ the Hankel matrix with $(i, j)$ entry $a(i + j + n - 2)$. Show that

$ \displaystyle
\lim_{n \to \infty}
\lvert \det H(n) \rvert ^{1/n} =\prod_{k=1}^{r} \lvert \lambda_k \rvert
$.​

 

[Opalg]

Challenge Problem: Let $A$ be an $r \times r$ matrix with distinct eigenvalues $λ_1, . . . , λ_r$. For $n \ge 0$, let $a(n)$ be
the trace of $A_n$. Let $H(n)$ be the $r \times r$ the Hankel matrix with $(i, j)$ entry $a(i + j + n - 2)$. Show that
$$\lim_{n \to \infty} \lvert \det H(n) \rvert ^{1/n} =\prod_{k=1}^{r} \lvert \lambda_k \rvert .$$

I'm assuming that $A_n$ in the statement of the problem should be $A^n$.
[sp]Since $A$ has distinct eigenvalues, it is diagonalizable. So its trace is $\lambda_1 + \ldots + \lambda_r$ and the trace of $A^n$ is $a(n) = \lambda_1^n + \ldots + \lambda_r^n$.

Let $x^r + c_1x^{r-1} + \ldots + c_{r-1}x + c_r = 0$ be the equation whose roots are $\lambda_1, \ldots, \lambda_r$. Apply the Newton sum formula $a(k) + c_1a(k-1) + \ldots + c_ra(k-r) = 0$ to each element of the final column of $H(n)$ (where $n>r$) to get $$\begin{aligned}\det H(n) &= \begin{vmatrix}a(n) & a(n+1) & \ldots & a(n+r-2) & a(n+r-1) \\ a(n+1) & a(n+2) & \ldots & a(n+r-1) & a(n+r) \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a(n+r-1) & a(n+r) & \ldots & a(n+2r-3) & a(n + 2r-2) \end{vmatrix} \\ \\ &= \begin{vmatrix}a(n) & a(n+1) & \ldots & a(n+r-2) & -c_1a(n+r-2) - \ldots - c_{r-1}a(n) - c_ra(n-1) \\ a(n+1) & a(n+2) & \ldots & a(n+r-1) & -c_1a(n+r-1) - \ldots - c_{r-1}a(n+1) - c_ra(n) \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a(n+r-1) & a(n+r) & \ldots & a(n+2r-3) & -c_1a(n+2r-3) - \ldots - c_{r-1}a(n+r-1) - c_ra(n+r-2) \end{vmatrix}. \end{aligned}$$ For each $k$ with $1\leqslant k\leqslant r-1$, add $c_{r-k}$ times the $k$th column of that determinant to the last column, giving $$\det H(n) = -c_r\begin{vmatrix}a(n) & a(n+1) & \ldots & a(n+r-2) & a(n-1) \\ a(n+1) & a(n+2) & \ldots & a(n+r-1) & a(n) \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a(n+r-1) & a(n+r) & \ldots & a(n+2r-3) & a(n + r-2) \end{vmatrix}.$$ But that determinant is just $\det H(n-1)$ with the first column shifted all the way to the right. This shows that $\det H(n) = \pm c_r\det H(n-1)$. By induction, $\det H(n) = \pm c_r^{n-r}\det H(r)$.

Therefore \(\displaystyle |\det H(n)|^{1/n} = |c_r|\left|\frac{\det H(r)}{c_r^r}\right|^{1/n} \to |c_r|\) as $n\to\infty$. But $c_r$ is $\pm$ the product of the roots of the equation $x^r + c_1x^{r-1} + \ldots + c_{r-1}x + c_r = 0$, namely \(\displaystyle \prod_{k=1}^r\lambda_k.\) So \(\displaystyle \lim_{n\to\infty}|\det H(n)|^{1/n} = \prod_{k=1}^r|\lambda_k|\), as required.

[/sp]

 

[MountEvariste]

...

Thanks for such a beautiful solution! (Bow)

Yeah, indeed I meant $A^n$. Fixed now.