### Compute dimension of vector subspace

I have a reasonably looking number field K of degree 20 (by reasonably looking I mean that the defining polynomial has coefficients < 1000). I have a stream that produces me reasonably looking vectors in K^30, that I call v1, v2, .... For each n I want to compute the dimension of V_n := span (v1, ..., vn).

If I go through (K^30).subspace([...]) then Sage runs the echelon_form and produces me matrices with coefficients of size 10^(100000).

Here is an example showing the issue

```
sage: x = polygen(QQ, 'x')
sage: K = NumberField(x^20 - 3, 'a')
sage: V = K**20
sage: U = V.subspace([])
....: for _ in range(20):
....: U += V.subspace([V.random_element()])
....: s = max(max(abs(coeff.numerator()) + abs(coeff.denominator()) for coeff in z.polynomial()) for z in U.matrix().list() if z)
....: print("coeff size: %d" % len(str(s)))
....: print("space dim: %d" % U.dimension())
....:
coeff size: 131
space dim: 1
coeff size: 301
space dim: 2
coeff size: 487
space dim: 3
coeff size: 640
space dim: 4
coeff size: 877
space dim: 5
coeff size: 1141
space dim: 6
coeff size: 1501
space dim: 7
coeff size: 1959
space dim: 8
```

Alternatively, you can consider the non streamed version

```
sage: K = NumberField(x^20 - 3, 'a')
sage: M = MatrixSpace(K, 20)
sage: m = M.random_element()
sage: m.rank()
```

This is not practical at all and basically stops around dim 15. Is there any alternative?

Note that for problems of the same size over rational numbers, Sage does succeed

```
sage: M = MatrixSpace(QQ, 2000)
sage: m = M.random_element()
sage: m[1400] = m[0] - 30 * m[18] + 25 * m[1500]
sage: %time m.rank()
CPU times: user 22.6 s, sys: 149 ms, total: 22.8 s
Wall time: 22.8 s
1999
```

(if the matrix is full rank the problem is way easier)

See also: this thread on sage-devel.