1

If we have a SVD of $A$ such as $U \Sigma V^*$ then $U$ and $V$ are transformation matrices with columns made from eigenvectors of $AA^*$ and $A^*A$ respectively. Now if I multiply a column of $U$ by $-1$, and call the new matrix $W$, I will still have a set of orthonormal eigenvectors of $AA^*$. Does this mean that my SVD relation will still hold ? i.e. $A = W \Sigma V^*$ ?

I was writing a code and multiplied a column by $-1$ and I couldn't reproduce $A$ from $W\Sigma V^*$, so I was wondering if I am missing something mathematically?

Rodrigo de Azevedo
  • 1
Ramin
  • 11
  • Related: https://math.stackexchange.com/q/1805191/339790 – Rodrigo de Azevedo Nov 29 '19 at 12:12
  • Probably then $\Sigma$ also has to change. – Berci Nov 29 '19 at 12:18
  • You need to multiply the corresponding row of $V^\ast$ by $-1$ too. Note that $U$ and $V$ are not any matrix with eigen vectors, you need to chose them jointly. – P. Quinton Nov 29 '19 at 12:24
  • @P.Quinton In the definition which I had seen on SVD, they left singular and right singular are known to be "a" set of normalized eigenvectors of MM* and M*M, so that is confusing me, because we should be able to calculate them independently – Ramin Nov 29 '19 at 12:35
  • The definition states that it is a set of [...], not any set of [...] right ? – P. Quinton Nov 29 '19 at 12:39
  • Well, obviously my definition was not complete. It did not refer to the ambiguity which exists while choosing a set of orthonormal eigenvectors. So my problem is resolved. The first link that Rodrigo posted cleared this up. Thank you – Ramin Nov 29 '19 at 12:44

0 Answers0