I find this is the best site to post this question, even though I considered cs.

It is a Monte Carlo experiment over the set of 10.000 n×n matrices.
If a single matrix eigenvalue is complex then python numpy package will return all the eigenvalues as `numpy.complex128`

type, else it will return all eigenvalues as `numpy.float64`

type.

Here is the algorithm:

```
# Monte carlo experiment in numpy
import numpy as np
from numpy import linalg as LA
e=10000 # examples
# n - matrix rank
for n in range(1,10):
cc=0 # complex counter
for i in range (e):
m=np.random.randn(n,n)# m=np.random.random((n,n))
w, _ = LA.eig(m)
if (type(w[0]).__name__ == "complex128"):
cc+=1
print(f'matrix:{n}x{n}, total cases: {e}, complex cases: {cc}, ratio: {cc/e}')
```

The output of this algorithm is as follows:

```
matrix:1x1, total cases: 10000, complex cases: 0, ratio: 0.0
matrix:2x2, total cases: 10000, complex cases: 2851, ratio: 0.2851
matrix:3x3, total cases: 10000, complex cases: 6481, ratio: 0.6481
matrix:4x4, total cases: 10000, complex cases: 8782, ratio: 0.8782
matrix:5x5, total cases: 10000, complex cases: 9674, ratio: 0.9674
matrix:6x6, total cases: 10000, complex cases: 9944, ratio: 0.9944
matrix:7x7, total cases: 10000, complex cases: 9998, ratio: 0.9998
matrix:8x8, total cases: 10000, complex cases: 9999, ratio: 0.9999
matrix:9x9, total cases: 10000, complex cases: 10000, ratio: 1.0
```

It is very easy to understand the 1x1 case, since we take values from floating point numbers (set $\mathbb Q$) the eigenvalue will be the number itself (also in $\mathbb Q$).

For the case 2x2 and later, we may get complex eigenvalues ($\mathbb Z$).

I am counting the complex cases (where at least one eigenvalue is complex) and providing the ratio.

I would like to get some thoughts about this ratio possible the formula how to calculate one analytically.

My thought is that the ratio is not dependent on floating point arithmetic. I used the normal distribution $\sim \mathcal N(0,1)$, but one may test also the uniform distribution I commented.