# How the smallest eigenvalue change in a structured matrix?

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!

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

• 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 – jack Sep 15 '16 at 15:38
• Could you answer my question please? @RobertIsrael – jack Sep 15 '16 at 22:12
• 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. – Robert Israel Sep 15 '16 at 22:34