Compute the **rank of a matrix**, which is the number of linearly independent rows (or columns) in a matrix. This helps you understand how much “unique information” the matrix contains in a linear algebra sense.

The rank can be defined as the number of non-zero singular values from an SVD:
$$
\text{rank}(A) = \left|\{\, i \;:\; \sigma_i > \varepsilon \,\}\right|
$$
where $\sigma_i$ are the singular values of $A$ and $\varepsilon$ is a small tolerance.

## Requirements

Implement the function
```python
import numpy as np

def matrix_rank(A, eps=1e-10):
    \"""
    Computes the numerical rank of a matrix using singular values.

    Args:
        A (list[list[float]]): 2D matrix of shape (m, n).
        eps (float): Tolerance for treating a singular value as non-zero.

    Returns:
        int: The numerical rank of A.
    \"""
    # Your solution below
```

**Rules:**
- Treat `A` as a 2D list (not a NumPy array input), but you may convert it to NumPy internally.
- Compute rank by counting singular values strictly greater than `eps`.
- Do not use `np.linalg.matrix_rank` (or any other built-in rank helper).
- Return a single integer.
- Keep it in one function.

## Example

```python
A = [
    [1.0, 2.0, 3.0],
    [2.0, 4.0, 6.0],
    [1.0, 0.0, 1.0]
]

matrix_rank(A, eps=1e-10)
```

Output:
```python
2
```

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Matrix rank

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1577. Matrix rank

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Similar Questions

Argument	Type
A	list
eps	float