I would like to calculate the singular value decomposition of a matrix and the order of the singular values is important. By default, it seems `numpy.linalg.svd`

(and `scipy.linalg.svd`

) sort the singular values, which makes it impossible for me to tell which column corresponds to each singular value.

Example:

```
import numpy as np
X = np.array([[-74, 80, 18, -56, -112],
[14, -69, 21, 52, 104],
[66, -72, -5, 764, 1528],
[-12, 66, -30, 4096, 8192],
[3, 8, -7, -13276, -26552],
[4, -12, 4, 8421, 16842]])
U, D, V = np.linalg.svd(X)
print(D)
```

Returns:

```
array([3.63684045e+04, 1.70701331e+02, 6.05331879e+01, 7.60190176e+00,
1.17158094e-12])
```

When I need:

```
array([1.70701331e+02, 6.05331879e+01, 7.60190176e+00, 3.63684045e+04,
1.17158094e-12])
```

Is there a way to get the singular values (D) such that they are unsorted? The relation X = UDV^T must also be preserved.

**Edit:** Some context was needed here to elucidate my misunderstanding. I was trying to reproduce Section 2.3, Variance Decomposition Method in this paper.

When you say:

I think you're making a mistake, there are no unique order in the singular values in the "Singular value decomposition", all that matters is that order of column vectors of U, D & V are such that:

`U * D * V == X`

That's why, by convention, they are put in descending order, but obviously the vertical vectors of the unitary basis U and the conjugate transpose V are also set in an order such that the formula above holds.

If you want a proof, to compute X back from U, D & V, you have to do: