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Question

Does $R^+ R = I$ imply that $R$ has linearly independent columns, and why? ($R^+$ is the psuedoinverse of $R$).

Sources

I can readily find sources that say if $R$ has linearly independent columns, then $R^+ R$. [1][2]. I don't see the converse as often, but I can find it: for example, Property 2 under "Properties of generalized inverse of matrix" from [3]. However, I can't find a reason why that is true.

Attempt

Let $R$ be an $m \times n$ matrix. $$\begin{aligned} I &= R^+ R \\ &= (V \Sigma^+ U^T)(U \Sigma V^T) \\ &= V \Sigma^+ \Sigma V^T \end{aligned}$$ I believe this implies that $\Sigma$ needs to have nonzero entries all along the diagonal, and this in implies $R$ has to be rank $n$. I'm not positive about this, though. My linear algebra is not very good...if anyone could confirm this and flesh out the explanation, I would really appreciate it!

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Yes: a function has a left inverse if and only if it is injective. A linear function is injective iff the columns of its matrix are linearly independent.

This is very basic, there is no need to use SVD or anything of the sort. It also works for arbitrary matrices, not just real/complex ones.

Note that this does not immediately imply the converse, as it is not a priori obvious that the inverse is linear (indeed, when the function is not an isomorphism, then there will be non-linear left inverses).

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  • $\begingroup$ Thank you for your response! I'm not used to thinking of matrices as functions (as I've said, my linear algebra knowledge is very rudimentary). The statements you mentioned, about a function having a left inverse iff injective, and a linear function being injective iff the columns of it matrix being linearly independent, are not familiar to me. If possible, could you tell me more about why these properties are true, or point me to some reading? $\endgroup$ – Mr Bear Jun 11 at 4:20
  • $\begingroup$ The first one is quite basic: in one direction, if you glue two things together (map them to the same thing), there is no way to unglue them. In the other direction, strictly speaking, you need to assume that the domain is nonempty. Then the left inverse is given by simply reversing the function (which is then a bijection onto its image) and mapping all the other elements of the codomain to an arbitrary point of the domain (which is possible if the domain is nonempty). $\endgroup$ – tomasz Jun 12 at 1:30
  • $\begingroup$ The second one is also not so difficult: a linear map is injective iff its kernel is trivial. Elements of the kernel of a map given by a matrix correspond exactly to zero linear combinations of its columns. $\endgroup$ – tomasz Jun 12 at 1:31
  • $\begingroup$ Got it. I appreciate the response and clarification. Thank you for your answer. $\endgroup$ – Mr Bear Jun 12 at 2:40

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