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I made some number of calculations with doubly centrosymmetric matrices $ 4 \times 4$ with positive integer entries (by doubly centrosymmetric I mean matrix which stays the same after rotation its entries about central point by $90^\circ$ - example below) and I've received interesting results for its eigenvalues:

matrix has always (if inverse exists) two real positive eigenvalues and two pure imaginary. Additionally these pure imaginary are always integer whereas real ones usually are not integer.

  • How these facts can be explained ?

Example: $\begin{bmatrix} 23 & 15 & 17 & 23 \\ 17 & 53 & 53 & 15 \\ 15 & 53 & 53 & 17 \\ 23 & 17 & 15 & 23 \end{bmatrix}$

Eigenvalues: $(119.863, 0.000i), ( 32.137, 0.000i), ( 0.000, 2.000i), ( 0.000, -2.000i)$

Especially, it is interesting why pure imaginary are integer in contrast to real ones?

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  • $\begingroup$ Interestingly enough pattern seems to be preserved also for $5 \times 5$ doubly centrosymmetric matrices, In this case we have 3 real eigenvalues and 2 integer pure imaginary. $\endgroup$
    – Widawensen
    Mar 16, 2017 at 16:10

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Nice observation! We can confirm that your observation holds (except that one of the real roots may be negative) for all $4\times 4$ matrices of the form you specified by analyzing the characteristic polynomial of the matrix.

Let's denote the elements of the matrix satisfying your form as

$A=\begin{bmatrix} a & b & c & a \\ c & d & d & b \\ b & d & d & c \\ a & c & b & a \end{bmatrix}.$

The matrix has only four degrees of freedom, so it's no surprise that the characteristic polynomial $p(\lambda)=\det(A-\lambda I)$ can be written down fairly compactly. After computing, simplifying, and factoring the characteristic polynomial, we get

$$p(\lambda)=\{\lambda^2-2(a+d)\lambda+4ad-(b+c)^2\}\times\{\lambda + (b-c)^2\}.$$

For $a,b,c,d\in\mathbb{Z^+},$ this gives pure imaginary solutions $\pm|(b-c)|i$ and real solutions $(a+d)\pm\sqrt{(a-d)^2+(b+c)^2}.$

As a check, we can look at your example with $a=23,b=15,c=17,$ and $d=53.$ Plugging in these values into the roots we found, we see that your example matrix has eigenvalues $\pm 2i,$ $76+2\sqrt{481},$ and $76-2\sqrt{481}.$

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  • $\begingroup$ Excellent that you have found so clear form of the characteristic polynomial, it explains everything. One more mystery of LA unveiled. Thank you.. $\endgroup$
    – Widawensen
    Mar 31, 2017 at 8:09
  • $\begingroup$ One more interesting fact I observed from the form of the given by you characteristic polynomial, for integer $a,b,c,d$ when $a,d$ are different primes then a matrix can't be singular. Generally matrix consisting of different primes can't be singular. However there are some cases for composite $a,d$ when a matrix has $0$ eigenvalue - the necessary condition is of course $4ad=(b+c)(b+c)$. Example $a=9, d=4, b=5, c=7$ $\endgroup$
    – Widawensen
    Mar 31, 2017 at 12:52

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