# 1617-257/TUT-R-7

From Drorbn

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On 10/28/16, we discussed the following problem:

Suppose is with for all . Prove that is not invertible.

Let . At the end of the tutorial, a couple of students pointed out that it's also true that is not invertible for any . It's easy to check that when and that is in the kernel of .

It takes some more care to prove what the problem is asking because one can't immediately deduce that has non-trivial kernel with . A key observation here is that since is , invertibility of at implies invertibility of at a neighborhood of .