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I just watched a YouTube video of some professor deriving the pseudo-inverse in terms of the SVD and there is something that I don't understand.

He represents our matrix $A\in \mathbb{R}^{m\times n}$ as $U\Sigma V^T$ where this should be the "economy SVD". Now going from $Ax=b$ he writes the following:

$$V\Sigma^{-1}U^TU\Sigma V^Tx = V\Sigma^{-1}U^Tb$$

I see that $\Sigma^{-1}U^TU\Sigma$ cancels away but what is with $VV^T$?

We are talking about the economy SVD and as far as I understand in general $V$ isn't an orthogonal matrix anymore(in general just a non quadratic matrix with orthonormal columns). Why is $VV^T$ still the identity?

Josh.K
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  • @Josh Indeed, $VV^T$ is not still the identity. It is not clear what was said in the video to make you think that $VV^T$ should be the identity. – Ben Grossmann Aug 17 '22 at 12:36
  • sorry for making a wrong statement – Vezen BU Aug 17 '22 at 13:03
  • @BenGrossmann I just thought that this was his intention... I don't know what the purpose of this equation should be then. Maybe he just made a mistake. If you are interested, here is the video: https://www.youtube.com/watch?v=PjeOmOz9jSY

    In minute 6:20 he starts with pseudo-inverse.

     – Josh.K Aug 20 '22 at 13:53
  • @Josh.K A bit after 8:20 he does say "this cancels to the identity". So yes, it seems as though he has made a mistake – Ben Grossmann Aug 20 '22 at 16:08
  • @Josh.K You might find my explanation of the relationship between the pseudoinverse and SVD that I give here to be helpful. Keep in mind that this explanation does not use the economy SVD, which means that $\Sigma$ doesn't have an inverse. – Ben Grossmann Aug 20 '22 at 16:08
  • @BenGrossmann: Unless I'm making a huge mistake here, if $m \geq n$ then the economy SVD should have $U \in O(m, n)$ and $V \in O(n)$, which means $VV^T = V^T V = I$, but it is $UU^T$ that is not necessarily equal to the identity. – VHarisop Aug 23 '22 at 18:40
  • @VHarisop That's correct; I meant to say that $VV^T$ cannot be guaranteed to be the identity, and I could have further specified that $VV^T$ will not be the identity in the case that $m<n$. Also, from the fact that $U^TU$ was "canceled" away and that $A$ was stated to be rectangular, I assumed we were considering the case that $m \leq n$, perhaps erroneously. – Ben Grossmann Aug 23 '22 at 20:02

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