As far as I understand, if I first generate 2 orthogonal matrices U and V, then generate a random diagonal matrix with random eigenvalues, I end up with a matrix, that might possibly have a different SVD decomposition. All of my matrices are 3×3.
Is there some kind of an extra condition, that I can enforce to assure, that the values used for the problem generation are the one and only possible solution?
Generate unique SVD decomposition problems
linear algebramatricessvd
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Best Answer
Uniqueness of SVD is not guaranteed in general. Take for example identity matrix.
We can construct a rotation matrix $\mathbf R$ for $\theta$ radians so that :
$${\mathbf I}_n = {\mathbf R} {\mathbf I}_n {\mathbf R}^T$$
for any value of $\theta$. In particular this happens if the eigenspace is non-simple.
If we have a projection instead of identity we will get even more degrees of freedom since we can rotate arbitrarily within the subspaces that have singular value 0.