@MichaelGreinecker Interesting, you should write an answer based on this

@FedorPetrov I don't find this a convincing argument, personally. There are many people who would love their questions to be seen by a community of professional mathematicians, but this does not make their questions automatically on-topic here. But I know my opinion is not shared by everyone; this has been discussed on meta.mathoverflow.net/questions/4566 and meta.mathoverflow.net/questions/394 .

@MarkLewko True, but still I think that this question belongs on History of Science and Mathematics.

I’m voting to close this question because it has already been asked on hsm.se, where it is more on-topic

Great find! How did you find this counterexample? If it was with another Sage script, it may be useful for the answer to include its source as well, so you can "teach to fish".

Yes, that should work; thanks for fixing this hole in my proof.

Thanks for your answer! It is good to see even more applications. A remark: if I am not mistaken, inversion via PPTs is equivalent to Gauss--Jordan elimination; it is fine to use this algorithm with symbolic matrices, but it has worse backward stability properties if implemented in floating-point arithmetic; readers beware.

Does that function have an optimum at a smooth point or a discontinuity point? This seems like it could make a difference.

@Dirk Another interesting abstract way to think about it is the following: if you restrict the linear operator $M$ to the invariant subspace $\operatorname{Im} V$, then $\Lambda$ is the matrix that represents the restricted operator in the basis induced by the columns of $V$. In particular, this implies that any eigenvalue of $\Lambda$ is also an eigenvalue of $M$.

Are you sure about that necessary condition in the last line? Surely there are solutions in which the columns of $A^\top$ are not linearly independent, for instance you can take $A$ to be the matrix of all zeros.

Towards the end, did you mean to write that $\alpha$ has to be of the order of $\delta$? That bound is minimized when $C_1\frac{\delta}{\sqrt{\alpha}} = C_2 \sqrt{\alpha}$ by AM-GM, which means $\alpha = \frac{C_1}{C_2}\delta$.

@jlewk Seems a good suggestion, would you like to write an answer based on it? Comments shouldn't be used to answer questions.