9.37 min read

Freedom in the SVD

Like the spectral decomposition, the SVD $A = U\Sigma V^\top$ is not unique. Sources of freedom include: reordering singular values (with corresponding reordering of columns of $U$ and $V$ ), simultaneous sign flips of corresponding columns of $U$ and $V$ , and when singular values repeat, choosing any orthonormal basis for the corresponding singular vector spaces.

The critical constraint: signs of $U$ and $V$ must be flipped in tandem. If you flip the sign of column $i$ of $U$ , you must also flip column $i$ of $V$ — otherwise the product $U\Sigma V^\top$ changes. The singular values themselves are uniquely determined by $A$ .

Formal View

Remark 9.2 — Non-Uniqueness of SVD

The singular values of

A

are unique. The singular vectors have freedom: 1. Reordering: permuting singular values with corresponding columns of

U

and

V

. 2. Sign flip: simultaneously replacing

\mathbf{u}_i \to -\mathbf{u}_i

and

\mathbf{v}_i \to -\mathbf{v}_i

(product unchanged). 3. Repeated singular values: any orthonormal basis for the corresponding spaces.

Why This Matters

Understanding SVD freedom prevents confusion when different implementations return different but equally valid factorizations.

MATLAB and NumPy may return different-sign singular vectors — both are correct.
In PCA, the sign of principal components is arbitrary (only direction matters).
When comparing SVD results across platforms, check that $U \Sigma V^\top$ matches $A$ , not that $U$ or $V$ individually match.

Learning Resources

SVD uniqueness and computation

Steve Brunton

Discussion of SVD uniqueness and practical computation.

20 min

Singular vectors and their signs

StatQuest

PCA and SVD: why sign conventions differ across implementations.

22 min

Quiz

Question 1

If you flip the sign of column $i$ of $U$ in the SVD, you must also flip column $i$ of $V$ to keep the factorization valid.

Question 2

Which part of the SVD is uniquely determined by $A$ ?

Common Mistakes

Flipping the sign of $\mathbf{u}_i$ without flipping $\mathbf{v}_i$ — this changes $A$ , invalidating the factorization.
Thinking all of $U$ and $V$ are arbitrary — only the freedoms listed are allowed.