Here is an n-by-n structured matrix, $$ \begin{bmatrix} 1 & 1-c & 1-2c & ..... & 1-nc \\ 1-c & 1 & 1-c & ..... & 1-(n-1)c\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ 1-nc & 1-(n-1)c & 1-(n-2)c & ..... & 1\\ \end{bmatrix} $$ It's a symmetric toeplitz matrix. c is a constant less than $\frac{1}{n}$ which means all entries are positive. I have known that this matrix is positive definite.

What I want to do is to prove that the smallest eigenvalue decrease with c. I have simulated how the smallest eigenvalue change using MATLAB. Now I need to prove that using analysis. Could anyone give me any tip about that please? Thanks so much!

  • $\begingroup$ Do you mean "increase with $c$"? For $c=0$ the smallest eigenvalue is $0$. $\endgroup$ – Robert Israel Sep 14 '16 at 18:08
  • 1
    $\begingroup$ Also the matrix is $n+1$ by $n+1$. $\endgroup$ – Robert Israel Sep 14 '16 at 18:51

Let your matrix be $T $.

The characteristic polynomial of $T$ is of the form $$ P_n(\lambda,c) = \lambda^{n+1} - (n+1) \lambda^n + \sum_{k=0}^{n-1} (a_{n,k} c^{n+1-k} - b_{n,k} c^{n-k}) \lambda^k $$ for some constants $a_{n,k}$ and $b_{n,k}$. It may be possible to compute a closed form for these.

Differentiating $P_n(\lambda,c) = 0$ implicitly, we get

$$ \dfrac{d\lambda}{dc} = - \dfrac{\partial P/\partial c}{\partial P/\partial \lambda}$$

as long as the denominator is nonzero, and this will have constant sign as long as both numerator and denominator are nonzero.

For $c=0$, the eigenvalues are $n+1$ (with multiplicity $1$) and $0$ (with multiplicity $n$. For $0 < c < 1/n$, $T$ is positive definite so the eigenvalues are positive. Thus the lowest $n$ eigenvalues will be increasing as a function of $c$ for small $c$.

The problem (and I don't see immediately how to solve it) is to show that when $\lambda$ is the lowest eigenvalue, $\partial P/\partial c$ and $\partial P/\partial \lambda$ don't hit $0$ before $c= 1/n$.

Here's a plot in the case $n=5$. Note how the first and second eigenvalues cross at approximately $c = 0.33$, after which the lowest eigenvalue is decreasing. But before $c = 1/5$ the first $5$ eigenvalues are increasing as functions of $c$.

enter image description here

  • $\begingroup$ Thank you so much!! But I have two questions: 1. how can you get the characteristic polynomial of T? 2. why do you claim "this (dλdcdλdc) will have constant sign as long as the numerator is nonzero."? @robert $\endgroup$ – jack Sep 15 '16 at 15:38
  • $\begingroup$ Could you answer my question please? @RobertIsrael $\endgroup$ – jack Sep 15 '16 at 22:12
  • $\begingroup$ 1) Using software such as Maple. 2) Sorry: I meant "as long as both numerator and denominator are nonzero" (I edited). That is, a fraction can only change sign when the numerator or denominator changes sign, and here the numerator and denominator are continuous functions. $\endgroup$ – Robert Israel Sep 15 '16 at 22:34

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.