# Does it make sense to use SVD to do a sort of “lossy compression”?

So - I know if you perform SVD to a matrix $$X$$, you can then use Echkart Young theorem to get the best rank $$r$$ approximation $$\overline{X}$$ to $$X$$ possible. Since the resultant $${\overline{X}}$$ will be of a lower rank, that means some of the rows will be linear combinations of the others, and so rather than store all of $${\overline{X}}$$ you could simply store some of the rows of $${\overline{X}}$$, with the combinations required to achieve the other rows. So it's a sort of "compression" technique.

Does this make sense/would it be sensible to do? I know there are clearly going to be better compression techniques out there, I'm just asking if in theory this would indeed provide a sort of lossy compression. Thanks!

• Are you asking if choosing $r$ equal to the rank of $X$ means that the SVD reconstruction is lossless? – Sycorax Jan 19 at 19:27
• @Sycorax sorry, I meant to say lossy*. If I choose the rank $r$ to be less than the rank of $X$* – Riemann'sPointyNose Jan 19 at 19:28
• Is $X^*$ the same as $\overline{X}$? – Sycorax Jan 19 at 19:29
• It's definitely possible to use the SVD for compression, and often with images, e.g. math.utah.edu/~goller/F15_M2270/BradyMathews_SVDImage.pdf – jld Jan 19 at 20:35
• – kjetil b halvorsen Jan 20 at 1:01