Let M be a $r \times c$ matrix with entries in $\mathbb R$ that has a left inverse.

Does there exist $\epsilon > 0$ such that $|Mu| > \epsilon$ for every vector $u \in R^{c}$ of length $1$?

Or can one find a sequence of unit vectors ${u_{n}}$ such that $|Mu_{n}| \to 0$?

  • 2
    $\begingroup$ The unit sphere is compact and the map $u \mapsto \lVert M u \rVert$ is continuous. Therefore the image of the unit sphere is compact and either contains $0$ or avoids an open ball around $0$. $\endgroup$ – WimC Dec 3 '17 at 12:50
  • $\begingroup$ You presumably mean $n\times\n$? @WimC, you should've posted that as an answer. $\endgroup$ – user491874 Dec 3 '17 at 12:52
  • $\begingroup$ No, the matrix is not necessary quadratic. $\endgroup$ – mercury0114 Dec 3 '17 at 13:43
  • $\begingroup$ @mercury0114 “quadratic matrices” aren’t a thing. When $M$ is $n\time n$, that’s called being square. Only square matrices are invertable. $\endgroup$ – Stella Biderman Dec 3 '17 at 13:57
  • $\begingroup$ @user8734617 OK, my bad. I modified a question. $\endgroup$ – mercury0114 Dec 3 '17 at 15:22

If the sequence $Mu_{n}$ converges to $0$, then the sequence $u_{n} = M^{-1}Mu_{n}$ converges to $M^{-1}0 = 0$ (because the left inverse $M^{-1}$ exists and the function $x \to M^{-1}x$ is continuous). However, each $u_n$ is a unit vector, so $|u_{n}| \to 1$ and the sequence can not converge to $0$.

  • $\begingroup$ This answer isn’t complete without explaining why the sequence can’t converge. $\endgroup$ – Stella Biderman Dec 3 '17 at 13:59
  • $\begingroup$ I added a few extra words. Is that enough now? $\endgroup$ – mercury0114 Dec 3 '17 at 14:56

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