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 17h awarded Popular Question Nov23 comment Minimal polynomials and cyclic subspaces @JeremyDaniel: I almost said no. The eigenvectors are $v_1$ and $v_2$, and we can't get from one to the other. Then I had a look at $v_1 + v_2$; its image is linearly independent. And so if by "cyclic" we mean the whole space is generated by one vector, then I think this proves it is :). The problem is, I'm not sure why I chose that linear combination besides the fact that I've seen similar choices made before. As for why one does that, no idea. In particular, the connection between what I just did and the eigenvectors eludes me. Can you recommend a source, online or otherwise? Nov23 comment Minimal polynomials and cyclic subspaces @Timbuc: Thanks for the hint. Are you saying, for example, that $V$ might be written as a direct sum of, say, two cyclic subspaces, but that one of those subspaces might be contained in the other? Nov23 revised Minimal polynomials and cyclic subspaces Made headline more precise Nov23 asked Minimal polynomials and cyclic subspaces Oct21 comment If $f$ is increasing, then for all $n\in\mathbb{N}$ there exists $P_n$ : $U(f,P)-L(f,P) \leq (b-a)/n$ @Did: Second response: As someone who almost went for a linguistics degree before he went for a math degree, there's something kinda annoying about that :). (See this.) Granted, I have a constant times $n$, but it's not $n$. I assume, in general, it's impossible to do with $1\cdot n$? Oct21 comment If $f$ is increasing, then for all $n\in\mathbb{N}$ there exists $P_n$ : $U(f,P)-L(f,P) \leq (b-a)/n$ @Did: Thought about this on the ride home from work. First of two responses: I think I see where you're heading. Let $D = \lfloor f(b) - f(a) \rfloor + 1$. Define $P_n = \{ a + i((b-a)/Dn) : 0 \leq i \leq Dn \}$. Then everything seems to work out. Oct21 asked If $f$ is increasing, then for all $n\in\mathbb{N}$ there exists $P_n$ : $U(f,P)-L(f,P) \leq (b-a)/n$ Oct19 comment For groups and orders show $\forall a,b\in G: |ab|=|ba|$ and $\forall G$ with $|G| = p^n$ for $p$ prime has a subgroup of order $p$ @user149868, Apologies, apparently I got some wires crossed when I was answering the second portion. I've gone ahead and deleted that part, since it wasn't helpful. But to answer your question: If $|G|=p$, it is indeed cyclic. If $|G|=p^n$, though, it need not be cyclic; for example, $|V_4|=2^2$ and $|D_8|=2^3$, but neither is cyclic. Oct19 revised For groups and orders show $\forall a,b\in G: |ab|=|ba|$ and $\forall G$ with $|G| = p^n$ for $p$ prime has a subgroup of order $p$ deleted the nonsense Oct16 answered For groups and orders show $\forall a,b\in G: |ab|=|ba|$ and $\forall G$ with $|G| = p^n$ for $p$ prime has a subgroup of order $p$ Sep30 awarded Explainer Sep24 awarded Autobiographer Sep16 comment How to find system of equations from solution space @almagest, I'm working on a similar problem, and I have a few questions. First, why should we expect just one equation? Second, I follow up until $8w + 2x - y = 0$; where does that come from? Sep3 asked Prove that $p \mapsto \inf\{d(p,s) : s \in S\}$ is uniformly continuous Aug24 revised Positive integer solutions to $p^2 + q^2 \leq 4^k$ added 159 characters in body Aug24 comment Positive integer solutions to $p^2 + q^2 \leq 4^k$ @flawr: Thank you. I'll edit my post: What you write is certainly true, and it "systematic," to use my original word; what I meant, though -- at four in the morning :) -- was: Is there a closed form for the answer? Aug24 comment Positive integer solutions to $p^2 + q^2 \leq 4^k$ @EwanDelanoy: Thanks for your response. I know what Riemann sums are; I've never done them, like, rigorously, though. (The problem is from Chap. 1 of Pugh's Real Mathematical Analysis.) I'm not sure it'll fly with my professor if I tie up my proof by referring to some "well-known" result :). Especially since, to be frank, I'm a little skeptical: I wouldn't know how to avoid writing $\lim_{n\rightarrow\infty} S(n) = n\cdot\infty^2$. Suggestions? Aug24 asked Positive integer solutions to $p^2 + q^2 \leq 4^k$ Aug24 accepted Unit disc contains finitely many dyadic squares whose total area exceeds $\pi - \epsilon$