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 Dec 31 comment The Hadamard determinant problem: understanding a proof for the upper bound on matrix determinants Whoops, yea I meant alternating. Also, you can see that you can reach the maximum bound when $n$ is a power of $2$, since you can just take $A_1=\left[\begin{smallmatrix} 1 & -1 \\ 1 & 1 \end{smallmatrix}\right]$, and keep forming blocks, $A_n=\left[\begin{smallmatrix} A & -A \\ A & A \end{smallmatrix}\right]$, I think. Dec 31 awarded Editor Dec 31 revised The Hadamard determinant problem: understanding a proof for the upper bound on matrix determinants miswording Dec 31 answered The Hadamard determinant problem: understanding a proof for the upper bound on matrix determinants Aug 9 comment reference for linear algebra books that teach reverse Hermite method for symmetric matrices Sorry, what are you referring to in the link you gave? So, in looking for canonical form, we're looking for a change of basis matrix $P$, so that $P^tAP$ is diagonal. When you complete squares, what you have now is how the coordinates of vectors transform under $P$. But we know that if $e_i\mapsto Pe_i$, i.e. if $P$ is the change of basis matrix, then coordinate vectors change as $v\mapsto P^{-1}v$. So the matrix $Q$ you get from how the coordinates change is the inverse of $P$. That's why the above is true, and why you're going to need to take an inverse somewhere in the procedure. Aug 8 answered reference for linear algebra books that teach reverse Hermite method for symmetric matrices Aug 1 comment Trying to make sense of this proof in Hatcher Sorry to answer your question with a question, but why wouldn't there be two sums coming from the composition of two boundary maps? You apply $\partial_n$ to an $n$-simplex and you get a sum, say over $i$, of $(n-1)$-simplices. Then you apply $\partial_{n-1}$, move it into the sum via linearity, and so for each of the $(n-1)$-simplices, you get a sum, say over $j$, of $(n-2)$-simplices. Do you understand what kind of object $\Delta_n(X)$ is? Jul 26 comment Matrix is conjugate to its own transpose @MarcvanLeeuwen Sorry for the necromancy here, but I think I have a slightly more direct argument: Using rational canonical form as you suggested, it suffices to show that the companion matrix $C_p$ to $p(t)=t^n+a_{n-1}t^{n-1}+\ldots+a_0$ is similar to its transpose. I think we can prove this with just elementary linear algebra by noting: 1) The minimal polynomial of $C_p$ is $p(t)$ 2) In general, the minimal polynomials of $A$ and $A^t$ are equal 3) Any two matrices with cyclic bases (i.e. $v,Av,\ldots, A^{n-1}v$) and equal minimal polynomials are similar 4) $C_p^t$ has a cyclic basis Jul 26 awarded Critic Jul 20 comment Reconstruction of a function from its moments I'm not sure about this if you integrate on all of $(-\infty,\infty)$, but if you are integrating over some finite interval, $[a,b]$, and $f$ is continuous, then yes, $f$ is determined by its moments. This is a classic problem in undergraduate analysis: to use the Weierstrass approximation theorem to show that if $f$ is continuous on $[a,b]$ and $\int_a^b x^n f(x)\ dx=0$ for all natural numbers $n$, then $f=0$. (Can you see that this implies that the moments of a function are unique, by linearity of integration?) Jul 18 comment Galois Group of the splitting field of the polynomial of $x^{11} - 7$ over $\mathbb{Q}$ Can you explicitly write some generators of $\mathbb L$ over $\mathbb Q$? Hint: there are two natural generators which have degrees $11$ and $10$ over $\mathbb Q$. Jul 13 comment How to prove that the Schur algebra is isomorphic to a certain endomorphism ring? You should reproduce the theorem for those of us who don't have the book :P Jul 13 comment Exercise on a holomorphic $f$ on a strip satisfying $|f(z)|\leq A(1+|z|)^\eta$ A tighter bound would be to keep R a variable, include $(1+R)^\eta$ in your $A_n$, and then let $R\to 1$. Not that it matters for this problem in isolation Jul 12 comment find local inverse functions of $f(x,y)=(x\sin(y),x\cos(y)), (x,y)\in (0,\infty)\times (0,3\pi)$ Have you not noticed that $f$ is basically the change-of-coordinates function from Cartesian to polar? Jul 12 comment Questions about distributions on $l$-spaces. Another piece of intuition: for certain distributions (those of order 0), the Riesz representation applies, and we can write them as $\langle T,f\rangle=\int_{\mathbb{R}^n} f\ d\mu$, where $\mu$ is a measure and the integral is a genuine measure-theoretic integral. This is in Friedlander-Joshi somewhere early. Jul 12 comment Questions about distributions on $l$-spaces. In the Euclidean case, the Schwartz functions are defined as smooth functions $f: \mathbb{R}^n\to\mathbb{C}$ with compact support. The intuition for the integral sign you asked about also is comes from noting that any (locally Lebesgue) integrable function $f$ defines a distribution, usually also denoted $f$, by the map $g\mapsto\int_{\mathbb{R}^n} fg dV$. There are some technical conditions to do with analysis that we put on functionals here, in addition to having them in $S^*(X)$. A standard reference is Friedlander and Joshi's "Intro to the Theory of Distributions." Jul 12 comment Questions about distributions on $l$-spaces. It comes from the definition of distributions over, say, Euclidean spaces $\mathbb{R}^n$. The name distribution (I think) comes from charge distribution - the intuition is that on Euclidean space, they represent things you can integrate (test) functions against. I.e. the Dirac delta distribution, $\delta(f)=f(0)$, which would, in E&M, represent something like a point charge at the origin. Dec 23 awarded Organizer Dec 23 revised Banach Fixed Point Theorem problem contradiction This is not about Banach spaces, nor really about functional analysis Dec 23 suggested approved edit on Banach Fixed Point Theorem problem contradiction