6

Is there a geometric interpretation of a non-symmetric matrix having only real eigenvalues?

It appears that multiplying random matrices with IID random entries eventually produces a matrix with only real eigenvalues, wondering if this can be turned into a statement about how random linear maps transform a vector.

For instance, code below multiples $10\ 2\times 2$ random matrices, eigenvalues are almost always real.

n = 2;
depth = 10;
dist = NormalDistribution[];
sample := RandomVariate[dist, {n, n}]/Sqrt[n];
sampled := Nest[sample . # &, sample, depth - 1];
Eigenvalues[sampled] (* {-0.859186, 0.613002} *)

Some trajectories of 1,1 vector evolving in accordance with random 2D linear transformation. It appears that power iteration converges to a line for all matrices with real eigenvalues but not for complex-valued ones

(code)

Real-valued eigenvalues

enter image description here

Complex valued eigenvalues

enter image description here

Yaroslav Bulatov
  • 5,327
  • It may be helpful to look at here and here. – kabenyuk Jul 30 '23 at 04:07
  • 1
    Somewhat loosely speaking, real eigenvalues correspond to an axis the matrix scales and nonreal eigenvalues correspond to a plane the matrix rotates. – Nicholas Todoroff Jul 31 '23 at 04:43
  • 1
    @BenGrossmann can the intuition of "non-real eigenvalues correspond to rotations" be made rigorous? – Yaroslav Bulatov Aug 02 '23 at 14:29
  • 2
    @YaroslavBulatov Not who you're at-ing, but if $v, \bar v$ are eigenvectors with conjugate nonreal eigenvalues and $v = v_1 + iv_2$ for real $v_1, v_2$, then you can compute that $av_1 + bv_2$ transforms to $$r[(a\cos\theta - b\sin\theta)v_1 + (b\cos\theta + a\sin\theta)v_2]$$ where $re^{\pm i\theta}$ are the eigenvalues. This is a rotation matrix applied to the $v_1, v_2$ basis followed by a scaling by $r$; it's not necessarily a rotation though because $v_1, v_2$ need not be orthogonal, but it's closely related. – Nicholas Todoroff Aug 02 '23 at 18:19

0 Answers0