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If you have two matrices A and B such as through singular value decomposition:

AB = PSV

And any invertible matrix T such as

AB = $PS^{1/2}TT^{-1}S^{1/2}V$

Could you prove that B = $T^{-1}S^{1/2}V$

This question might be too out of context, so to situate it here is the page where I encontered the problem, it is from the book Biological Learning and Control by Sandro Mussa Ivaldi:

page 264

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  • $\begingroup$ You can't "prove" it- that is just one possible value of the many different A and B whose product will give that result. $\endgroup$ – user247327 Sep 10 '16 at 14:22
  • $\begingroup$ I am not sure to get your meaning. If A and (PS^1/2) are invertible... you could prove it by taking T to be T = (P * S^1/2)^-1 * A.. but then you would need to prove that T is also invertible.. no ? $\endgroup$ – Albert James Teddy Sep 10 '16 at 16:36
  • $\begingroup$ so you want to prove: there exists an invertible matrix $T$ such that $B=T^{-1}S^{1/2}V$? $\endgroup$ – Christiaan Hattingh Sep 10 '16 at 17:50
  • $\begingroup$ What i ultimately wanted to understand is Pr. Ivaldi reasoning in his book (see pic) ... but hmm I am probably missing something $\endgroup$ – Albert James Teddy Sep 10 '16 at 18:27
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The rank of $AB$ is less than or equal to the minimum of the rank of $A$ and $B$ (this is known as Sylvester's theorem I think). And the rank of $S^{1/2}$ is the same as the rank of $AB$ (SVD theory: rank of $AB$ equals the number of nonzero singular values). The equation $B=T^{-1}S^{1/2}V$ basically states that $B$ and $S^{1/2}$ are equivalent (matrix equivalence) and hence must have the same rank. But since the rank of $AB$ may very well be less than the rank of $B$ we see that this cannot be true in general. A trivial counter example: let $A=0$ and $B=I$.

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