How do the eigen-values of the following tridiagonal matrix ($A$) change when adding random numbers $R_i$ (with a normal distribution with the mean 0 and variance $m$) to its diagonal. A is a square matrix defined as follows:

For i = 1 to N $$ A(i,i) = -2 + R_i $$ $$ A(i,i+1) = 1 $$ $$ A(i,i-1) = 1 $$ where m and N are constants.

Notice that the eigen-values are known for the case $R_i=0$:

for j=1 to N; $$ Eig(j) = -2 -2 cos(j\pi/N) $$

Numerical solution reveals that the eigen values around the maximum eigenvalue have the largest deviation from the case without random numbers ($R_i=0$). For example, the following figure shows eigen values with and without having the random numbers for m=0.1. But how we can solve the problem analytically? (Maybe by using central limit theorem)

enter image description here


I don't think this question has an easy answer. See section 1.3, "Eigenvalues and sums of Hermitian matrices," of Terry Tao's book on random matrices.


This seems to fall under eigenvalue perturbation problem.

The following assumes that eigenvectors $x_{0j}$ of the unperturbed matrix are chosen such that $||x_{0j}||^2 = x_{0j}^{*} x_{0j} = 1$.

An eigenvalue of perturbed matrix is $\lambda_j = \lambda_{0j} + \mathbf{x}_{0j}^{*} \delta A \mathbf{x}_{0j} + o(||\delta A||)$, where $\lambda_{0j} = -2 - 2 \cos (2 \pi j/N)$ is $j$-th eigenvalue for unperturbed matrix, $\mathbf{x}_{0j}$ is the corresponding eigenvector and $\delta A$ is a diagonal matrix with $R_i$ on its diagonal, $||\delta A|| = \max |R(i)|$. As a result, up to higher-degree , the expectation of $\lambda_j$ is the non-perturbed value $\lambda_{0j}$ and, if $R_i$ are uncorrelated, then $$ \mathrm{Var}(\lambda_j) = \mathrm{Var} \left(\sum_{k=1}^N R_k |(x_{0j})_k|^2 \right)= \sum_{k=1}^N \mathrm{Var} (R_k |(x_{0j})_k|^2) = \sum_{k=1}^N |(x_{0j})_k|^4 m \leq m $$ (up to higher-degree terms). Since eigenvectors have the form $(x_{0j})_k = \sqrt{1/N}\exp(-i t_{jk})$ with real $t_{jk}$, $|(x_{0j})_k|^2 = 1/\sqrt{N}$ for all $k$ and thus, $\lambda_j = \lambda_{0 j} + \frac{1}{N} \sum_{k=1}^N R_k + o(||\delta A||)$ and $\mathrm{Var}(\lambda_j) = m/N$.

For reference, the second-order and third-order corrections are $$ \delta \lambda^{(2)} = \sum_{\substack{k=1\\ k \neq j}}^{N} \frac{|x_{0k}^{*} \delta A x_{0j}|^2}{\lambda_{0j} - \lambda_{0k}} $$ and $$ \delta \lambda^{(3)} = \sum_{\substack{k=1\\ k \neq j}}^{N} \sum_{\substack{m=1\\ m \neq j}}^{N} \frac{x_{0j}^{*} \delta A x_{0m} x_{0m}^{*} \delta A x_{0k} x_{0k}^{*} \delta A x_{0j}}{(\lambda_{0m} - \lambda_{0j})(\lambda_{0k} - \lambda_{0j})} - x_{0j}^{*} \delta A x_{0j} \sum_{\substack{m = 1\\m \neq j}} \frac{|x_{0j}^{*} \delta A x_{0m}|^2}{(\lambda_{0m} - \lambda_{0j})^2}. $$

The formula for mean value for $\max R_i$, which describes how accurate above estimations are, can be found in Expectation of the maximum of gaussian random variables.

As a result of Bauer–Fike theorem, for any eigenvalue of the perturbed matrix $\lambda$ there is an eigenvalue of unperturbed matrix $\lambda_0$ such that $|\lambda - \lambda_0| \leq \max R_i$.

In addition, it follows from Gershgorin circle theorem that all eigenvalues belong to $[-4, 0]$.

  • $\begingroup$ Thanks, $R_i$ has a normal distribution with the mean 0 and variance $\sigma$. $\endgroup$ – Hesam Jul 3 '15 at 17:39
  • 1
    $\begingroup$ 1) If eigenvectors are complex, conjugate transpose should be used instead of transpose in above formulas. 2) If eigenvectors correspond to vectors of Fourier basis, then the first-order correction is the diagonal element of $\delta A$ in that basis. So corrections are related to double Fourier transform of $\delta A$. $\endgroup$ – Budenn Jul 3 '15 at 18:57
  • 1
    $\begingroup$ Discrete Fourier transform of a vector $x$ can be considered its multiplication by a matrix $F$: $\tilde{x} = F x$. $\delta A$, however, is a matrix (even if only its diagonal elements are non-zero), so result of transformation would be $F A F^*$. See en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/…. Essentially, to calculate it one has to first perform (1-D) DFT on rows of $\delta A$ and then on columns of the result. $\endgroup$ – Budenn Jul 3 '15 at 21:15
  • 1
    $\begingroup$ @Hesam, I added that to the answer. $\endgroup$ – Budenn Jul 4 '15 at 8:14
  • 1
    $\begingroup$ I suspect that is because 1) the "density" of eigenvalues is larger for eigenvalues closer to $0$, so for second-order correction of eigenvalues close to the greatest there are many summands with low denominators and, thus, large value, 2) for eigenvalues close to the greatest one, the summands mostly have the same sign. $\endgroup$ – Budenn Jul 4 '15 at 18:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.