How to find bounds on the eigenvalues of a matrix?

Question

Given this matrix

$M=\begin{bmatrix} 0 & m-1 & 2 & n-1\\ 1 & m-2& 1 & n-1\\ 2 & m-1 & 0 & 2(n-1)\\ 1 & m-1 & 2 & 2(n-2)\end{bmatrix},$ show that if $\alpha_1,\alpha_2$ are two negative eigenvalues of $M$, then $\alpha_1+\alpha_2<-3$. Alsoo $m,n\ge 4$ are integers.

Using Gershgorin circles theorem, the eigenvalues would lie in the intervals $B(0,m+n)$, $B(m-2,n-1)$, $B(0,2n+m-1)$,$ B(2n-4,m+2)$ where $B(x,y)$ denotes the interval with centre $x$ and radius $y$. But it's not helping to find the required bound?

Does there exist any theorem/trick to find a better upper bound of the sum of two negative eigenvalues of a matrix?

If someone can help, I will be grateful.

Answer 1

I am able to accurately calculate the eigenvalues in perturbation theory, as follows:

I perform a similarity transformation on the matrix $M$, given by $M'=S^{-1}MS$ with _{$$S=\left( \begin{array}{cccc} 1 & 1 & \frac{-\sqrt{m^2+4 n^2}+m-2 n}{2 m} & \frac{\sqrt{m^2+4 n^2}+m-2 n}{2 m} \\ 0 & 0 & \frac{-\sqrt{m^2+4 n^2}+m-2 n}{2 m} & \frac{\sqrt{m^2+4 n^2}+m-2 n}{2 m} \\ \frac{1}{2} \left(-\sqrt{5}-1\right) & \frac{1}{2} \left(\sqrt{5}-1\right) & 1 & 1 \\ 0 & 0 & 1 & 1 \\ \end{array} \right).$$} The matrix $M'$ has the same eigenvalues as $M$ and is given by $M'=M_0+M_1$, with _{$$M_0=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m+2 n\right) & 0 \\ 0 & 0 & 0 & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) \\ \end{array} \right),$$ $$M_1=\left( \begin{array}{cccc} \frac{1}{2} \left(-\sqrt{5}-3\right) & 0 & \frac{\sqrt{m^2+4 n^2}+\left(\sqrt{5}-2\right) m+2 n}{2 \sqrt{5} m} & \frac{-\sqrt{m^2+4 n^2}+\left(\sqrt{5}-2\right) m+2 n}{2 \sqrt{5} m} \\ 0 & \frac{1}{2} \left(\sqrt{5}-3\right) & \frac{-\sqrt{m^2+4 n^2}+\left(\sqrt{5}+2\right) m-2 n}{2 \sqrt{5} m} & \frac{\sqrt{m^2+4 n^2}+\left(\sqrt{5}+2\right) m-2 n}{2 \sqrt{5} m} \\ \frac{-\sqrt{5} \left(\sqrt{m^2+4 n^2}-2 n\right)-m}{2 \sqrt{m^2+4 n^2}} & \frac{\sqrt{5} \left(\sqrt{m^2+4 n^2}-2 n\right)-m}{2 \sqrt{m^2+4 n^2}} & -\frac{3 \sqrt{m^2+4 n^2}+m-2 n}{2 \sqrt{m^2+4 n^2}} & -\frac{\sqrt{m^2+4 n^2}+m-2 n}{2 \sqrt{m^2+4 n^2}} \\ \frac{m-\sqrt{5} \left(\sqrt{m^2+4 n^2}+2 n\right)}{2 \sqrt{m^2+4 n^2}} & \frac{\sqrt{5} \left(\sqrt{m^2+4 n^2}+2 n\right)+m}{2 \sqrt{m^2+4 n^2}} & \frac{-\sqrt{m^2+4 n^2}+m-2 n}{2 \sqrt{m^2+4 n^2}} & \frac{-3 \sqrt{m^2+4 n^2}+m-2 n}{2 \sqrt{m^2+4 n^2}} \\ \end{array} \right).$$}

Now we note that for large $m,n$ of order $1/\epsilon$, the matrix $M_0$ is of order $1/\epsilon$ while the matrix $M_1$ is of order unity. We can therefore perform perturbation theory in the small parameter $\epsilon$.

Let me first look at the positive eigenvalues. To zeroth order these are given by $\beta_\pm=\frac{1}{2} \left(\pm \sqrt{m^2+4 n^2}+m+2 n\right)$. To first order these are the $(3,3)$ and $(4,4)$ diagonal elements of $M_0+M_1$, which gives the two positive eigenvalues $$\beta_\pm=\frac{1}{2} \left(m+2 n\pm\sqrt{m^2+4 n^2}\right)\pm\frac{m-2 n \mp 3 \sqrt{m^2+4 n^2}}{2 \sqrt{m^2+4 n^2}}.$$ This expression for the positive eigenvalues is quite accurate, see the plot for $m=n$, where the exact expressions (red and green lines) are almost indistinguishable from the perturbative result (blue and orange).

Now for the negative eigenvalues. To first order these are given by the $(1,1)$ and $(2,2)$ diagonal elements of $M_0+M_1$, equal to $$\alpha_1=\frac{1}{2} \left(-\sqrt{5}-3\right)=-2.618,\;\;\alpha_2=\frac{1}{2} \left(\sqrt{5}-3\right)=-0.382.$$

Here is the plot for $m=n$, again red and green is exact, blue and orange the perturbative result.

So to first order in $\epsilon$ one has $\alpha_1+\alpha_2=-3$. The challenge is to prove that higher order corrections are negative. Incidentally, the restriction to integer $m,n$ does not seem to play any role (in the plot $m=n$ is varied continuously).

Answer 2

This can be done with the help of Mathematica. First, the function $f(m,n)$ finds the sum of the negative eigenvalues of the matrix under consideration. Second, we numerically find the maximum value of $f(x,y)$ over the integers greater than or equal to $4$ with relative error $10^{-3}$. The results are as follows.

f[m_, n_] := Total[Map[Min[0, #] &, 
Eigenvalues[{{0, m - 1, 2, n - 1}, {1, m - 2, 1, n - 1}, {2, m - 1,
0, 2 n - 2}, {1, m - 1, 2, 2 n - 4}}] // N]]
NMaximize[{f[x, y],   x >= 4 && y >= 4 && x \[Element] Integers && y \[Element] Integers}, 
{x, y}, Method -> {"DifferentialEvolution", "ScalingFactor" -> 1}, 
AccuracyGoal -> 3, PrecisionGoal -> 3, MaxIterations -> 1000]

$$\{-3.,\{x\to 12854143321479777526031848555457428151289960383333269504,y\to 1051298338250582241682620486779844006807195121763745792\}\} $$ Edit. A typo in the code for the matrix under consideration.