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Jan
12
answered Vector subspaces of $\mathbb{R}[x]_n$ result
Jan
9
awarded  Announcer
Jan
4
comment If every vector is an eigenvector, the operator must be a scalar multiple of the identity operator?
Related
Dec
29
comment Prove $A=\{x\in \mathbb{R}|f(x)=x\}$ is closed subset of $\mathbb{R}$
Sequentially closedness.
Dec
20
awarded  Constituent
Dec
12
comment On the problem of polynomial bijection from $\mathbb Q\times\mathbb Q$ to $\mathbb Q$
What this seems like is material worth a blogpost.
Dec
11
revised $\lim_{p\to \infty}\Vert f\Vert_{p}=\Vert f\Vert_{\infty}$?
added 5 characters in body; edited title
Dec
11
comment
Relevant
Dec
10
comment If $\sigma$ is a cycle of length $r$, then it has order $r$?
@darijgrinberg Would you like to put your comment as an answer. I'll upvote it.
Dec
10
comment
@PedroTamaroff I don't chat as often as I used to, the few times I've been there recently, I found some things that made me form some opinion. For what is worth, I remember I used to appreciate being in chat when you were around. More, when the talk was about math. Just to make clear that I'm not trying to screw you on this elections.
Dec
9
comment
The good thing about being active in chat is that many people gets to know you. This kind of behavior seems pretty much in opposite direction to lead by example and being patient and fair, not to talk about respect.
Dec
8
awarded  Caucus
Dec
6
awarded  Nice Question
Dec
6
revised Prove that $f $ is constant
edited title
Nov
28
revised How do I find the characteristic polynomial and eigenvalues?
deleted 4 characters in body
Nov
26
revised Verifying an equivalence relation
Formatting
Nov
26
revised Verifying an equivalence relation
added 7 characters in body
Nov
16
comment what is the set $\mathbb R[X]$ defined as?
Nice answer. Multiplication can also be defined the same way for the formal power series, so multiplication of polynomials can be realized as the restriction of that one.
Nov
15
answered Eigenvalues of operator $p(T)$ in terms of the eigenvalues of $T$, where $p$ is a polynomial
Nov
13
revised $A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$.How to prove this?
added 39 characters in body