1

The famous Mandelbrot fractal can be calculated by starting $z_0=0$ and then iterating

$$z_{n+1} = {z_n}^2+c$$

and seeing which $c$ makes $z$ run off to complex infinity ( and if so, how fast ).

Can we represent this with a matrix? If we can, would that let us analyze it's behavior with linear algebra?

mathreadler
  • 25,824
  • 1
    Complex numbers can be represented as matrices (see: https://math.stackexchange.com/questions/180849/why-is-the-complex-number-z-abi-equivalent-to-the-matrix-form-left-begins ) so thanks to this isomorphism probably it is possible to generate the fractal from convergence of matrices. – Widawensen Apr 25 '17 at 13:38
  • @Widawensen I am hoping for something like that, but I am not sure I could show/prove it. – mathreadler Apr 25 '17 at 13:41
  • For this I have no idea, but there is also one additional alternative way of generating fractals in sheer geometrical way and this famous Mandelbrot operation can be represented as rotating a vector, dilating it and adding to it other vector. So we have three isomorphic ways of generating these very interesting objects. Moreover quaternions can be also used for generating the fractals ( multidimensional) and quaternions have also a matrix representation. – Widawensen Apr 25 '17 at 13:48

2 Answers2

2

Have you heard about IFS (https://en.wikipedia.org/wiki/Iterated_function_system) (http://www.hiddendimension.com/fractalmath/ifs_fractals_main.html)? They use linear algebra and generate fractals in the same "spirit" as Mandelbrot fractals (https://mathoverflow.net/q/199286).

Jean Marie
  • 81,803
  • Yes, they are interesting! That is one reason I thought maybe mandelbrots were possible to analyze the same way! – mathreadler Apr 22 '17 at 09:40
1

That's depend on what you mean by linear algebra. The equation is certainly not linear and it does not become linear because you use matrices (from linear algebra). After all the matrices are not more "linear" than the real numbers (as the real numbers and $1\times 1$ matrices are basically the same).

If you extend the recursive formula to matrices then one can easily see that it's required that the matrices are all square and of the same dimension. I don't think there's much understanding to be gained by this either, for example if you (can) diagonalize $c$ you see that the recursion formula only decomposes to a set of recursion formulas of the original kind (you'll only get back to square one again).

skyking
  • 16,654
  • I mean are there some methods from linear algebra, matrix decompositions, factorizations, canonical forms, linear transformations et.c. we could use to help understand it's behaviour. – mathreadler Apr 22 '17 at 09:14