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Consider a matrix A (containing elements that possess N as unit) that maps a vector b (containing elements that possess m/N as unit) to a vector c (containing elements that possess m as unit). What units do the singular values of A have? And why?

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When you do a singular value decomposition, $A = U\Sigma V^*$, you can put the unit whereever you like. You could even divide the unit between $U$, $\Sigma$ and $V^*$. So your question doesn't really make sense mathematically. The most natural thing would probably be to put it in $\Sigma$ and then the singular values would have the same units as your matrix entries. But that is really just a choice.

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  • $\begingroup$ It's pretty natural to put the units into the singular values so that $U$ and $V^*$ really are pure rotations (rather than a combination of a pure rotation and a "dimensional identity matrix"). $\endgroup$ – Ian Jun 2 at 16:27
  • $\begingroup$ @Ian I agree, as I said. $\endgroup$ – Klaus Jun 2 at 16:28
  • $\begingroup$ I think it's not just a choice to put units of A in Σ but necessary as U and V* are just rots. More details on this: n x n matrix V* rotates the n-dimensional vector b first. This rotated vector is then scaled and mapped onto an m-dim space by matrix Σ. This new vector in m-dim space is then rotated by m x m matrix U to give the m-dim vector c . Since Σ is responsible for physically altering the vector and for mapping the vector from an n-dim space to an m-dim space, units of elements of Σ and A should be same. Please correct me if I am wrong. $\endgroup$ – vik1212154 Jun 2 at 21:36
  • $\begingroup$ @vik1212154 You are not wrong and it is certainly the most natural thing to do. However, I don't think there is a consensus about units in mathematics other than to not use any at all. In what you are saying you choose $U$ and $V^*$ to be "just rots". You might as well choose $U$ to be the rotation times the unit and everything works out fine. Again, this does not seem natural to me either, but as there is no general consensus to my knowledge, this is also a fair choice. This makes this question mathematically ill-posed in my opinion. Of course, in physics your mileage may vary. $\endgroup$ – Klaus Jun 3 at 12:52

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