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In order to a problem related to optics, I have been led to try and diagonalize the following $N\times N$ real matrix:

$A_N=(\min(i,j))_{1\le i,j\le N}$.

I call this a "discrete Brownian matrix" because it is reminiscent of the covariance matrix of standard Brownian motion starting at 0, i.e. $E_0\{x(s)x(t)\}=\min(s,t)$.

It seems that some astute induction trick might do it, but I'm lacking time to find it and I suspect this might be a classical matrix diagonalize. In the scaling limit $N\to\infty$, the matrix $A_N$ becomes the Brownian kernel $A_N(i/N,j/N)\sim N \min(s,t)$, $0\le s,t\le 1$, and it is a nice exercise to show that the eigenfunctions are $u_n(t)=\sin((n+1/2)\pi t)$ with eigenvalues $\lambda_n=1/((n+1/2)^2\pi^2)$, $n\ge 0$.

It would be nice, however, to have a result for finite $N$.

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    $\begingroup$ What do you want to know exactly ? The matrix is orthogonaly diagonalizable because it is real symmetric. Would you like to know explicitly a system of eigenvectors and eigenvalues ? $\endgroup$ Commented Dec 20, 2024 at 14:38
  • $\begingroup$ Yes, ideally I would like to know an explicit basis of eigenvectors and the corresponding eigenvalues. Knowing only the spectrum would be useful enough however. $\endgroup$ Commented Dec 20, 2024 at 17:33
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    $\begingroup$ You may find that you want here : sciencedirect.com/science/article/pii/… $\endgroup$ Commented Dec 20, 2024 at 20:56
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    $\begingroup$ I am not sure whether the tag (linear) created in this question was just a typo - I have replaced it with (linear-algebra) and (matrices). Maybe (eigenvalues) could be suitable, too? $\endgroup$ Commented Dec 25, 2024 at 11:13
  • $\begingroup$ Thank you for the quick answer pointing to the solution, and for the details of Trench's solution below. The solution is indeed nice and simple, and readily gives the expression of the eigenvalues and eigenvectors deduced by calculus in the continuum limit. $\endgroup$ Commented Dec 28, 2024 at 15:39

2 Answers 2

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As shown in the citation in the comment of @jvc, the inverse of $A_N$ is tridiagonal, \begin{align}\tag{1}\label{eq:1} A_N^{-1}= \begin{pmatrix} 2&-1\\ -1&2&-1\\ &-1&\ddots&\ddots\\ &&\ddots&2&-1\\ &&&-1&1\\ \end{pmatrix} \, , \end{align} with characteristic polynomial \begin{align} \tag{2a}\label{eq:2a} P_N(\lambda)&=\det(A_N^{-1}-\lambda \,1) \\ \tag{2b}\label{eq:2b} &=\langle 1,0| \begin{pmatrix}2-\lambda&-1\\1&0\end{pmatrix}^{N-1} |1-\lambda,0\rangle \\ \tag{2c}\label{eq:2c} &=\langle 1,0| \begin{pmatrix}2\cos\varphi&-1\\1&0\end{pmatrix}^{N-1} |1-4\sin^2\tfrac\varphi 2,0\rangle \\ \tag{2d}\label{eq:2d} &=\frac 1{\sin\varphi}\, \langle 1,0| \begin{pmatrix}\sin[N\varphi]&\sin[(N-1)\varphi]\\ \sin[(N-1)\varphi]&\sin[(N-2)\varphi]\end{pmatrix} |1-4\sin^2\tfrac\varphi 2,0\rangle \\ \tag{2e}\label{eq:2e} &=\frac{\cos\big[(N+\tfrac 1 2)\varphi\big]}{\cos \frac \varphi 2}\,. \end{align} Here we used the parametrization \begin{align}\tag{3}\label{eq:3} 2\cos\varphi=2-\lambda \, . \end{align} The eigenvalues $\lambda_\mu$ of $A_N^{-1}$ are the zeroes of $P_N(\lambda)$ and therefore \begin{align} \tag{4}\label{eq:4} \lambda_\mu = 4\sin^2 \frac{\varphi_\mu}{2}\,, \quad \varphi_\mu=2\pi\frac{2\mu + 1}{2N + 1}\,, \quad \mu = 0,1,\ldots,N-1 \, , \end{align} and the eigenvalues of $A_N$ are $\lambda_\mu^{-1}$.

The construction of the eigenvectors is straightforward, see, e.g., the other answer which just appeared while I was typing... I'll cite the (unnormalized) eigenvectors from Federico's answer, \begin{align} \tag{5}\label{eq:5} (x_\mu)_j = \sin\left(j\pi\frac{2\mu+1}{2N+1}\right) \, . \end{align} Merry Christmas to you all!

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    $\begingroup$ Thanks! If you wish you can copypaste the eigenvectors in here and I can delete my answer, so we have a single answer with everything. Sorry, I would have left the answer to you if I knew you were writing one. $\endgroup$ Commented Dec 25, 2024 at 18:50
  • $\begingroup$ @FedericoPoloni No problem, please leave your answer as it is. I have added the eigenvectors. $\endgroup$ Commented Dec 25, 2024 at 20:20
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This question was posed as a problem by William Trench in the bulletin IMAGE of the International Linear Algebra Society, in 1999. A full solution by the proposer is on page 28 of https://ilasic.org/wp-content/uploads/IMAGE/image22.pdf . The eigenvalues of $A$ are $$ \lambda_k = \frac{1}{2\left(1-\cos\frac{(2k+1)\pi}{2N+1}\right)}, \quad k=0,1,\dots,N-1, $$ and the corresponding eigenvector $x_k$ has entries $$ (x_k)_j = \sin \frac{(2k+1)j\pi}{2N+1}. $$

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    $\begingroup$ Many thanks for the reference to Trench's solution, which I had not been able to find. $\endgroup$ Commented Dec 25, 2024 at 19:42

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