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I have noticed that eigenvectors for matrices $2\times{2}$ made from $4$ consecutive big prime numbers, for example

$\begin{bmatrix} 100003 & 100019 \\ 100043 &100049 \\ \end{bmatrix}$,

have "always" approximate eigenvectors presented below in the matrix

$R=\begin{bmatrix} v_1 & v_2 \\ \end{bmatrix} = \begin{bmatrix} \dfrac{\sqrt{2}}{2} &-\dfrac{\sqrt{2}}{2} \\ \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} = Rotation(\pi/4) $

I suppose it's not only characteristic for prime numbers but also for any big numbers with relatively small differences between consecutive numbers, but ....

  • how to prove that it holds just for 4 consecutive big prime numbers?

What are consequences of these approximate eigenvector forms?

P.S.
Please notice dear reader that I'm not asking whether property I have presented is valid exclusively for primes numbers (big ones), but .. whether it is valid also for big prime numbers in any situation what requires however a little analysis of prime numbers properties.
For example to use Legendre conjecture.

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2 Answers 2

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Dividing your matrix by the first entry won't change anything. When you do, it has the form $$ M = \begin{bmatrix} 1 & 1 + a \\ 1 + b & 1 + c \end{bmatrix} $$ where $a, b, c$ are all small numbers. Hence it's almost symmetric, hence has almost orthogonal eigenvectors.

Now look at

$$ M\cdot \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2+a \\ 2+(b+c) \end{bmatrix} $$ Since $a,b,c$ are small, this vector is almost an eigenvector of eigenvalue 2. Similarly, $$\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ will almost be an eigenvector with eigenvalue $0$.

And that explains what you're seeing. It turns out to have nothing to do with prime-ness --- just the fact that your four numbers are close to each other.

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  • $\begingroup$ Ok. I have taken this into account but how to prove that it is satisfied just for prime numbers. Can we prove that they have sufficiently "small" differences for this formula to hold? This is my question.. $\endgroup$
    – Widawensen
    Dec 9, 2016 at 14:15
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    $\begingroup$ The prime number theorem (en.wikipedia.org/wiki/Prime_number_theorem) says that the gap between successive primes is, on average, much smaller than the primes themselves. That average isn't universally true (there may be long spans with no primes), but then again, your formula probably isn't either (exactly in those places). See the facts on "prime gaps" (en.wikipedia.org/wiki/Prime_gap#Upper_bounds) for more details. Since prime gaps have asymptotic bounds that ensure that for large enough primes, $a,b,c$ are small, your conjecture is true for large enough primes. $\endgroup$
    – John Hughes
    Dec 9, 2016 at 14:32
  • $\begingroup$ To answer your last question about "what's the significance of this?", I cannot see any, since it appears to be a far weaker statement than the known results on the distribution of primes and the sizes of prime gaps. $\endgroup$
    – John Hughes
    Dec 9, 2016 at 14:34
  • $\begingroup$ Hmm, so you are skeptical whether it holds for the case where gap occurs... If it were we would have known that the big gaps can be only for much bigger numbers (how many times?) than the gap is. $\endgroup$
    – Widawensen
    Dec 9, 2016 at 14:42
  • $\begingroup$ I'm not skeptical about anything here. But even if I were, it wouldn't matter --- the claim is true or false, regardless of my state of mind. Look, you can plot, for various values of $a,b,c$, the angle between the dominant eigenvector and $[1,1]$. That'll tell you how large a gap is needed (and between which two primes) to make you feel that the eigenvector is no longer "approximately" $[1,1]$ (I can't do that part -- you're the only just of "approximately"!). That'll tell you "how many times". And the results on upper bounds on gaps will give you some sense of whether such gaps ever occur. $\endgroup$
    – John Hughes
    Dec 9, 2016 at 14:48
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May be it is just because: $$ \begin{bmatrix} a &b \\ c &d \end{bmatrix} $$ has roughly eigenvectors (not normalized) $\langle 1, 1 \rangle$ and $\langle 1, -1 \rangle$ if $$ a+b \approx c+d \\ b-a \approx c-d $$ The first eqn holds because, in your setting of 4 large primes, $a,b,c,d$ are large and their difference relatively small. The second does not always hold. Why do you think so? If you apply (multiply) the matrix on $\langle 1, -1 \rangle$, you get $\langle -16, -6 \rangle$, which is not good likeness to $\langle 1, -1 \rangle$ itself.

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  • $\begingroup$ What do you mean by substitution <-1,1> in the matrix? $\endgroup$
    – Widawensen
    Dec 9, 2016 at 14:23
  • $\begingroup$ sorry, I meant multiply the matrix by <-1,1>. Edited. $\endgroup$
    – Violapterin
    Dec 9, 2016 at 14:33
  • $\begingroup$ I see, but (-1,1) is only close approximation not exact value, so for such big numbers have decisive meaning... but your remark is very valuable.. $\endgroup$
    – Widawensen
    Dec 9, 2016 at 14:36
  • $\begingroup$ Well, basically @John Hughes is saying the same thing as me $\endgroup$
    – Violapterin
    Dec 9, 2016 at 14:38
  • $\begingroup$ Yes, it's true., but both of you don't clearly said that it is applicable for primes because they are sufficiently close to each other.. (John gave in comments some links to wikipedia when it is written so with this his answer can be accepted). $\endgroup$
    – Widawensen
    Dec 10, 2016 at 7:57

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