# Left inverse question: if $R^+R = I$, then does this imply $R$ has linearly independent columns?

### 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!