Rudy the Reindeer
Reputation
21,620
77/100 score
 Aug 19 comment Slick proof that if an open set contains $\mathbb Q$ it has all irrational numbers, except a countable amount. @dREaM Oh I see, the "not countable" referred to the missing numbers. Aug 19 comment Showing $\sup \{ \sin n \mid n\in \mathbb N \} =1$ Nice answer. Where you write $n\alpha$ is in that interval didn't you mean $\operatorname{fractional part}(n\alpha)$ is in that interval? Aug 19 comment Slick proof that if an open set contains $\mathbb Q$ it has all irrational numbers, except a countable amount. @dREaM What is wrong with taking $\mathbb R$ minus $\{i_1, \dots, i_n\}$ where $i_k$ are irrationals? Aug 18 comment Basis for a linear space @DavidP Tru dat. I will not delete my comment though as I think it might be useful to others who read this thread. Aug 18 comment Basis for a linear space Showing that the set spans the space is not enough: one has to show that it's linearly independent, too. Aug 9 comment Are all matrices linear operators? But I would like to be convinced otherwise, of course. Aug 9 comment Are all matrices linear operators? In particular, if $A$ is such a matrix and $x$ some vector and we let $Ax$ be the usual multiplication then this seems to me is not a map $\{1,\dots,m\}\times\{1,\dots,n\} \to \mathbb{K}$ but a map $\mathbb R^m \to \mathbb R^n$. Aug 9 comment Are all matrices linear operators? I will think about this some time. For now I do understand what you meant in your previous comment but I still don't see why a matrix does not define a linear map. Aug 9 comment Are all matrices linear operators? Honestly, I really really don't understand how a matrix is a map $\{1,\dots,m\}\times\{1,\dots,n\} \to \mathbb{K}$. It seems to me that it takes vectors of arbitrary real numbers as argument not just integer pairs in these finite sets...? Aug 9 comment Are all matrices linear operators? An $n \times m$ matrix with real entries is a map $\mathbb R^m \to \mathbb R^n$. So please do continue. Aug 9 comment Are all matrices linear operators? Alright. Can you humour me and point out to me which of the parts of the definition of a linear map a matrix does not satisfy? Aug 9 comment How do I find a root of $A^2$? Yes! Exactly! : ) Aug 9 comment Are all matrices linear operators? Sorry, does your first sentence really say that matrices do not satisfy the definition of a linear map? Aug 9 comment Are all matrices linear operators? Yes. ${}{}{}{}{}$ Aug 9 comment How do I find a root of $A^2$? The first thing that came to my mind when I read your question was "Hm... can I diagonalise $A$ and then take the root of $D$?" Aug 9 comment Find matrix representing Linear Transformation As far as I can tell you are not "given the correspondence between the two (0,0)" either. Hence I would assume that the vertices are mapped in order. May 8 comment Express unit sphere as countable union of great circles? @Misakov The version I state is equivalent to yours if you replace union with intersection (see e.g. here. In your other comment, if you replace dense with open dense in the conclusion then yes, it seems to me that you can argue like that. May 8 comment Express unit sphere as countable union of great circles? @Misakov I elaborated on this hint. Maybe it helps you answer your comment here. Apr 27 comment $p$-adic completion of integers @user135520 Which sequence are you talking about? It's quite some time since I wrote this post... Mar 7 comment Intersection of all neighborhoods of zero is a subgroup @TobiasKildetoft Thank you for your comment!