Reputation
6,303
Next privilege 10,000 Rep.
Access moderator tools
Badges
3 18 52
Impact
~139k people reached

Apr
28
awarded  Popular Question
Apr
17
revised Instructive proofs in functional analysis
clarified what looked like a confusing sentence
Mar
30
comment Constructive proof of the existence of an algebraic closure
@Rubertos You can show that there exists a set that contains all algebraic extensions up to isomorphism
Jan
11
comment What properties do you lose when you extend your number set?
@fvel If they don't fully answer your questions, you should explain why. Then maybe someone can answer whatever is left that you want to know.
Jan
9
revised What properties do you lose when you extend your number set?
edited tags
Jan
9
answered What properties do you lose when you extend your number set?
Dec
23
revised function such that $f(x\cdot t)=f(x)g(t)$
the title was confusing, and different from the statement of the question
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
@JaVaPG Yes, exactly. But as MPW (and Kaladin) mentioned, the two problems are basically equivalent (since $AB-I=-(I-BA)$).
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
@JaVaPG Correct.
Dec
23
revised Proof If $AB-I$ Invertible then $BA-I$ invertible.
improved formatting
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
@JaVaPG You're close, but that's not quite the answer. Recall that you are trying to prove $BA-I$ is invertible, not $BA$.
Dec
23
revised Proof If $AB-I$ Invertible then $BA-I$ invertible.
added 87 characters in body
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
I don't think this is a counter-example. I may have made a computation mistake, but I have $$AB-I = \begin{pmatrix}1&1\\-2&-2\end{pmatrix}$$ and also $$BA-I = \begin{pmatrix}0&1\\0&-1\end{pmatrix},$$ and neither are invertible. Moreover, you give a proof for when $A$ is invertible, which is the case in your "counter-example".
Dec
23
revised A counter example
it is more specific than general topology
Dec
23
revised Proof If $AB-I$ Invertible then $BA-I$ invertible.
the OPs question is not formulated as an "iff" statement
Dec
23
answered Proof If $AB-I$ Invertible then $BA-I$ invertible.
Dec
19
awarded  Constituent
Dec
10
awarded  Caucus
Nov
13
awarded  Yearling
Oct
16
revised How to get to the formula for the sum of squares of first n numbers?
edited body