Determinant Question

[Corribus]

Hello all -
I have encountered a mathematics problem which is beyond my capability to solve. I need a general solution to the following equation:

\(\displaystyle M_N(x) = 0\)

where \(\displaystyle M_N(x)\) is the determinant of an N x N matrix defined with the following matrix elements \(\displaystyle A_{i,j}\):

\(\displaystyle A_{i,j} = \left\{ \begin{array}{ll}
x & \mbox{ if } i = j \\
\frac{1}{\left|(i-j)^3 \right|} & \mbox{otherwise} \end{array} \right.\)

I need a general expression for x as a function of N, the matrix dimension. I can solve these matrices for individual cases (i.e., N = 3), but I am unable to determine an expression for the general solution.

The N = 3 matrix by the way should look something like this:

\(\displaystyle M_{3}(x) = \left| \begin{array}{ccc}
x & 1 & \frac{1}{8} \\
1 & x & 1 \\
\frac{1}{8} & 1 & x \end{array} \right|.\)

And the N = 4 matrix would be:

\(\displaystyle M_{N}(x) = \left| \begin{array}{cccc}
x & 1 & \frac{1}{8} & \frac{1}{27} \\
1 & x & 1 & \frac{1}{8} \\
\frac{1}{8} & 1 & x & 1 \\
\frac{1}{27} & \frac{1}{8} & 1 & x \end{array} \right|.\)

Just as an example of what I'm looking for, a simpler problem is when \(\displaystyle M_N(x)\) is defined as

\(\displaystyle A_{i,j} = \left\{ \begin{array}{ccc}
x & \mbox{ if } i = j \\
1 & \mbox{ if } i = j \pm 1 \\
0 & \mbox{, otherwise} \end{array} \right. \]\)

In this case the general solution for \(\displaystyle M_N(x) = 0\) of the kind I'm looking for is

\(\displaystyle x = -2cos{\frac{k\pi}{N+1}}\) where k is allowed values of 1, 2, ..., N.

(Although I can't derive this result either.) Finally, I realize that the solution to this problem may be a very complicated function of N. While a general solution would be wonderful, I'm primarily interested in the maximum and minimum allowed values of this function as N approaches infinity. For instance, in the above "simpler" problem, these limits are \(\displaystyle 2 \mbox{for} (k \rightarrow \infty)\) and - \(\displaystyle 2 \mbox{ for} (k = 1)\). Thus the total "dispersion" of values (i.e., the maximum possible value minus the minimum possible value, what I'm REALLY interested in) is \(\displaystyle 4\).

If anyone can solve this or point me in the right direction, I'd very much appreciate it. Just FYI I am a physical chemist - this isn't a homework problem or anything. I'm not a mathematician for forgive me is some of my formalism is incorrect.

 

[Corribus]

No takers, huh? I simplified the problem a little bit in form. Oh well...

 

[jolly]

No takers, huh? I simplified the problem a little bit in form. Oh well...

It's just hard to follow, that's all. Sometimes, on forums like this, certain math problems don't come out as nice.

 

[Corribus]

Understood. Although it appears really complciated, the problem actually is very simple to write down. And it's also very simple (in principle) to solve for a given value of N. But to come up with a general expression for N is quite difficult it seems. The latter problem (the simplified tridiagonal matrix, with many elements going to zero) is a very often used matrix in physical chemistry, and the result (which I gave) is fairly well known. It describes some of the properties of conducting polymers, although the "hard" matrix would describe them better, which is why I'm interested. Unfortunately, I cannot find a derivation for the easier problem and I can't even find a solution to the harder problem, anywhere. Anyway - I know sometimes math problems are harder to follow by reading it rather than have someone just tell you in person.

 

[Gene]

I have been fussing with the original problem with very limited success. Am I correct that you would be satisfied with a general equation based on

\(\displaystyle M_{3}(x) = \left| \begin{array}{ccc}
x & 1 & 0 \\
1 & x & 1 \\
0 & 1 & x \end{array} \right|.\)

= x³-2x

And:

\(\displaystyle M_{4}(x) = \left| \begin{array}{cccc}
x & 1 & 0 & 0 \\
1 & x & 1 & 0 \\
0 & 1 & x & 1 \\
0 & 0 & 1 & x \end{array} \right|.\)

=x^4-3x²+1

Or am I not even playing in the right ballpark?

 

[Corribus]

Gene -
These tridiagonal matrices (where everything but the three diagonals are zero) do give the expansions as you wrote them. The general solution for an N x N tridiagonal matrix of this form I gave in my original post (although I cannot derive it - so if you can do that, it would also be helpful). The solution for the roots is x = -2 cos (k*pi/(N+1)) where k can take values of any integer from 1 to N (so that there are k roots).

This is essentially what I want for the non-tridiagonal matrices.

For instance:

\(\displaystyle M_{2}(x) = \left| \begin{array}{cc}
x & 1 \\
1 & x \\
\end{array} \right|.\)

= x^2-1

\(\displaystyle M_{3}(x) = \left| \begin{array}{ccc}
x & 1 & \frac{1}{8} \\
1 & x & 1 \\
\frac{1}{8} & 1 & x \end{array} \right|.\)

= x^3-(129/64)x+1/4

And:

\(\displaystyle M_{4}(x) = \left| \begin{array}{cccc}
x & 1 & \frac{1}{8} & \frac{1}{27} \\
1 & x & 1 & \frac{1}{8} \\
\frac{1}{8} & 1 & x & 1 \\
\frac{1}{27} & \frac{1}{8} & 1 & x \end{array} \right|.\)

=x^4-(70745/23328)x^2+(14/27)x+(2672857/2985984)

So I want a general solution to det(MNx).
Is that clearer?

 

[Corribus]

Actually, though, the more I look at this, the more I think this problem is going to be prohibitively difficult to solve.

Therefore maybe I could simplify it is a bit. The tridiagonal matrices, which Gene wrote out, have known solutions. They also use a lot of approximations that make them (and their solutions) undesirable. However, the pentadiagonal matrices may be a good compromise between the tridiagonal matrices and the infinite matrices which I had originally requested solutions for. The pentadiagonal matrices are:

\(\displaystyle M_{2}(x) = \left| \begin{array}{cc}
x & 1 \\
1 & x \\
\end{array} \right|.\)

\(\displaystyle M_{3}(x) = \left| \begin{array}{ccc}
x & 1 & \frac{1}{8} \\
1 & x & 1 \\
\frac{1}{8} & 1 & x \end{array} \right|.\)

\(\displaystyle M_{4}(x) = \left| \begin{array}{cccc}
x & 1 & \frac{1}{8} & 0 \\
1 & x & 1 & \frac{1}{8} \\
\frac{1}{8} & 1 & x & 1 \\
0 & \frac{1}{8} & 1 & x \end{array} \right|.\)

\(\displaystyle M_{5}(x) = \left| \begin{array}{ccccc}
x & 1 & \frac{1}{8} & 0 & 0 \\
1 & x & 1 & \frac{1}{8} & 0 \\
\frac{1}{8} & 1 & x & 1 & \frac{1}{8} \\
0 & \frac{1}{8} & 1 & x & 1 \\
0 & 0 & \frac{1}{8} & 1 & x \end{array} \right|.\)

\(\displaystyle M_{6}(x) = \left| \begin{array}{ccccc}
x & 1 & \frac{1}{8} & 0 & 0 & 0 \\
1 & x & 1 & \frac{1}{8} & 0 & 0 \\
\frac{1}{8} & 1 & x & 1 & \frac{1}{8} & 0 \\
0 & \frac{1}{8} & 1 & x & 1 & \frac{1}{8} \\
0 & 0 & \frac{1}{8} & 1 & x & 1 \\
0 & 0 & 0 & \frac{1}{8} & 1 & x \end{array} \right|.\)

, etc.

If a general solution to the determinants of these matrices could be found, that, I think, would be a good compromise.

 

[Corribus]

Assuming I copied everything over right from Maple, the solutions to those determinants are, respectively:

\(\displaystyle M_{1}(x)=x\)

\(\displaystyle M_{2}(x)=x^{2}-1\)

\(\displaystyle M_{3}(x)=x^{3}-\frac{129}{64}x+\frac{1}{4}\)

\(\displaystyle M_{4}(x)=x^{4}-\frac{97}{32}x^{2}+\frac{1}{2}x+\frac{3969}{4096}\)

\(\displaystyle M_{5}(x)=x^{5}-\frac{259}{64}x^{3}+\frac{3}{4}x^{2}+\frac{6081}{2048}x-\frac{127}{256}\)

\(\displaystyle M_{6}(x)=x^{6}-\frac{81}{16}x^{4}+x^3+\frac{6145}{1024}x^{2}-\frac{3}{2}x-\frac{897}{1024}\)

\(\displaystyle M_{7}(x)=x^{7}-\frac{389}{64}x^{5}+\frac{5}{4}x^{4}+\frac{41223}{4096}x^{3}-\frac{771}{256}x^{2}-\frac{481473}{131072}x+\frac{11908}{16384}\)


Is there a pattern there? I could probably solve for the roots as well.