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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
.