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visits member for 3 years, 2 months
seen Jan 26 at 18:04

I'm a second year PhD student in Biostatistics. Previously, I wrote a Masters thesis in number theory. In my thesis, I investigated one of the many connections between representation theory and arithmetic geometry. Specifically, I worked on the connection between Langlands functoriality and algebraic cycles on Shimura varieties.


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
Sep
30
awarded  Explainer
Sep
9
awarded  Popular Question
Aug
12
reviewed Leave Closed Why is the covariance matrix of i.i.d. Gaussian variables the identity matrix?