@psie if A and B are linear operators on a vector space and you can prove that for all v in some basis of that space that Av = Bv, then A = B
in context, and there is always context and you should rarely microscope down to the level of individual words because this eliminates relevant context, to say that "the behavior of R on the eigenvectors of T is uniquely determined" is to say that if R and S are positive operators and R^2 = S^2 = T, and if v is an eigenvector of T, the Rv = Sv
i don't think this is particularly good prose on axler's part, and underspecified use of the term "unique" is maybe often even a hallmark of bad mathematical prose and a source of errors, but in context it makes sense
if you know that "a positive operator that squares to T" has to do something specific on eigenvectors of T, and you know that the eigenvectors of T span the space, then you know "a positive operator that squares to T" has to do something specific on the entire space
to say that [some list of things] uniquely determines an object is to say that if A and B are objects [in the relevant universe, which is supplied by context exterior to this phrasing] that both satisfy the [some list of things] then A = B
here the relevant universe is positive operators that square to T and the list of things is the equations that specify what the operator does to eigenvectors of T
Let $\tilde{M}$ be the universal cover. You can use a transfer argument to show that $H^n(M;R)\cong H^n(\tilde{M};R)^{\mathbb{Z}/2\mathbb{Z}}$ for any ring $R$ in which $2$ is invertible. The superscript here denotes the invariants with respect to the induced action by the deck transformations. However, $H^n(\tilde{M};R)\cong R$ and $\mathbb{Z}/2\mathbb{Z}$ acts by multiplication with $-1$, so this group vanishes. Since $H^n(M)$ is f.g., this observation implies that it is $2$-torsion. The (rare) cohomological UCT implies that $H^n(M;\mathbb{Z})\otimes\mathbb{Z}/2\mathbb{Z}\cong H^n(M;\math…
admittedly, this probably involves more homological algebra tools than the standard argument, but I like it :)
I mean, geometric arguments are neat (and I wish I knew more), but they're typically suboptimal for algebraic ends (though can be enlightening in other ways)
My linear algebra is a bit rusty, so apologies if this is obvious. I'm reading this answer on the polar decomposition of a real matrix. I'm a tiny bit confused by the sentence where the author says $M^tM$ being positive definite and hence has the unique positive semi-definite square root $P=\sqrt{M^tM}$. Isn't it more precise to say that $P$ is also positive definite instead of positive semi-definite?
Let $A$ be a non-empty set. For all $n \in \mathbb N$ let $A_n \subseteq A$ and injective $f_n: A_n \to \mathbb N$ be given. Let $B = \bigcup_{n \in \mathbb N} A_n$. Show that $B$ is finite or countably infinite.
So, we need to find a $g: B \to \mathbb N$ that is injective. So, map $$b_n \mapsto (n, f_n(b_n)).$$ Then we have a $g': B \to \mathbb N \times \mathbb N$ and we know there is a bijection $\mathbb N \times \mathbb N \to \mathbb N$.
@ILikeMathematics can't you have an element that is contained in $A_n$ and $A_m$ for two distinct $m,n$? it's not clear what this maps to under this definition
@psie well, it would be accurate to say that, and it would provide more information to say that. "more precise" almost makes it sound like the person left out something needed to understand the argument, though, which they didn't. it reads like they are implicitly using a result that provides a PSD square root like a black box (i.e. not assuming it to be known a priori that "square rooting" preserves invertibility), and they immediately explain right after why the thing is invertible.
i would file this under, don't expect people to always provide the most amount of information possible at every sentence, or to infer from a phrasing that does not provide more information that more is not true.
as a side note there is an unfortunate amount of variation in terminology in this area. a lot of people i worked with would say just "positive" for "positive semi-definite" and have no special word for the invertible case.
> Consider taking the coefficients of the matrix $\bmod 3$ (treated as elements of $\Bbb Z[\sqrt2]$). This has the effect of simply reducing the coefficient of an element of the form $\Bbb Z\sqrt2$.
in https://math.stackexchange.com/a/1493249/745350). Could someone explain why can one drop $\sqrt2$ from the entries $\pm2\sqrt2$? Is this a ring homomorphism?
@XanderHenderson i refereed a paper once where the author had made a mistake, "proving" a lemma for all positive operators (all "P >= 0") by observing it held for P = 0 and then proving it for "P > 0" (which in context assumes P to be invertible). i.e. forgetting about the case where P was not invertible and also not zero (which as it happens was the only interesting case). it was a relatively easy fix, "happens to the best of us" kind of thing. but very much in this vein
people who work with "operator inequalities" all the time are maybe a little too used to thinking in their own language. another one is, sometimes writing A >= B when A and B are individually not self adjoint but A - B is. there be dragons.
The original problem is to a continuous surjective mapping f:C^2⟶C such that the preimage of a point {0} is disconnected. I got an example $(z_1,z_2)\mapsto (z_1+z_2+1)^3-1$. But @SoumikMukherjee was asking me about another question, is there a continuous surjective mapping f:R⟶R such that the preimage of a point {0} is disconnected. I got an example $f(x)=x^2(x-1)$.
@XanderHenderson I think the function $x\mapsto x\sin(x)$ is surjective
I consider partitions of n=5, more precisely I consider the partition $\lambda=5$. I consider the young tableau 1,2,3 in the first row and 1,4,5 in the first column. Now I want to compute the row and column subgroup which are the subgroup of S_n with map every {1,...,n} into an element standing in the same row respectively the same column. I got that the row subgroup of this tableau is $P_\lambda:=\{(12),(13),(23),(123)\}$ and the column subgroup is $Q_\lambda=\{(14),(15),(45),(145)\}$
user: those aren't subgroups of S_5. you need the to add the identity element and (132) to your candidate P_lambda and the identity and (154) to your candidate Q_lambda
but the subgroups P_lambda and Q_lambda are certainly generated by the elements you list there
i guess the row subgroup would be a two-element subgroup of S_4 containing the identity and some transposition (corresponding to whatever entries you put in the row with two boxes), but you wouldn't know which one
I've been staring at this answer for 15 minutes or so, and I can't figure what the author means by $T_x$ in $\mathrm{Id}_{T_x M}$ and likewise $T_{f(x)}$ in $\mathrm{Id}_{T_{f(x)} N}$. First the subscript is used to denote a set, but I don't think $T_x M$ also means a set. This feels like an IQ test...
without clicking, T_x M would often denote the tangent space of a differentiable manifold M at the point x (which implicitly and without clicking on anything is an element M)
which is among other things a vector space
and Id_{T_x M} would be the identity operator on that vector space
I have a question about differential forms: Why $\omega=\frac{1}{x_{n+1}} \mathrm{~d} x_1 \wedge \mathrm{~d} x_2 \wedge \ldots \wedge \mathrm{~d} x_n$ is a non-vanishing n-form on $S^n$?
here is an example of an automorphism of a surface (what it means to me): take a plane, $P$ in R^3 equipped with a normal bundle $N$, and remove a line $l$ lying on $P$. This disconnects $P$ into 2 components $P_1,P_2$. Flip the plane $P_2$ s.t. $N$ reverses orientation (formally, $N$ is reflect by 180 deg.). Re-insert $l$. This transformation process is such that the identify map on $P_1$ doesn't match up with the reflected plane $P_2$ negating any sort of homeomorphism in the standard
sense, but the surface at the end still looks like a surface after these transformations
so it must be an automorphism (but not a homeomorphism)...?
leslie: there's this exercise in Spivak's Calculus on Manifolds. I'm wondering whether it is stating the same thing as in the link above. $f'(a)$ in Spivak's means the Jacobian of $f$ at $a$.
> Suppose $f:\mathbb R^n\to\mathbb R^n$ is differentiable and a has a differentiable inverse $f^{-1}:\mathbb R^n\to\mathbb R^n$. Show that $(f^{-1})'(a)=[f'(f^{-1}(a))]^{-1}$. (Hint: $f\circ f^{-1}(x)=x.)
What I'm doubting is that Spivak does not require that $f$ be a $C^1$ diffeomorphism, i.e. not continuously differentiable, only differentiable.
there is an enormous amount of variation from reference to reference in what exactly they assume about their manifolds. i don't see the question posted on main linked above as even being express about that. it kind of doesn't matter, the question is just the chain rule. if you have the chain rule then you have that.
the answerer assumes the smooth context. that is more than necessary for the chain rule to hold, but it is certainly enough, and the asker seems to be basically asking a question that is answered by "the chain rule" whatever the hypotheses may be.
i don't mean to imply that definitions and hypotheses don't matter, but at the level of generality that people ask questions on MSE it is often the case that it sort of doesn't matter. i.e. if you aren't asking some question that depends on the details of the setup, then someone can assume whatever setup they like that is compatible with the question, and answer it in that setup, and everybody goes away happy
but again, a running theme in a lot of these questions is, if someone states that X is true in Y context, they usually aren't also stating "and by the way X is never true outside of Y context." that is overwhelmingly not how people write mathematics
"if Y, then X" is not usually not a transmission of secret signals about what happens if not Y
it says nothing about what happens if not Y, and should generally be interpreted as authorial silence without any secret implication whatsoever about what may happen if not Y
Can someone help me find a group $G$ and a subgroup $H$ of G s.t there exists $g \in G$ with $gcd(o(g), [G:H])=1$ whereas $g \notin H$. I know that for a normal subgroup $H$, one cannot find such an element
After completing the system of linearly independent vectors {(1, −2, 1), (1, −1, 1)} into a basis of R3, determine the matrix associated with the endomorphism with respect to it
Does it follow by Sylow theorem that any finite group G can be decomposed into a multiplication of disjoint sylow subgroups such that each subgroup corresponds to a disitinct prime number dividing the order of G?
I'm reading about the change of variables formula. There's a lot going on in the proof and the author has this auxiliary lemma where they use a so-called "mean value inequality". I would have to type a lot if I'd provide all details, so I try to keep it short by risking missing some details.
We have $C^1$-diffeomorphism $\varphi:U\to D$ on a compact set $K$, where $K,U,D\subset \mathbb R^d$. The author simply says.
> [Fix $\epsilon>0$ small.] Since $\varphi'$ is continuous, the mean value inequality allows us to choose $\delta>0$ small enough so that $\delta<\frac{1}{d} \mathrm{dist}(K,U^c)$ and, for every $u_0\in K$ and every $u\in \mathbb R^d$ such that $0<|u-u_0|<d\delta$ we have $$|\varphi(u)-\varphi(u_0)-\varphi'(u_0)(u-u_0)|<\epsilon|u-u_0|.\tag1$$
Note, $\varphi'(u_0)$ is a matrix. I think $(1)$ is just a restatement of the definition of what it means for $\varphi$ to be differentiable, but what is meant by the "mean value inequality"? I've tried to look this up, but haven't found anything sensible (nor do I really understand how the author is using it).
it is not helpful to divorce things from the references they are in, but if i just heard in a vacuum the terms 'mean value inequality' i'd assume it was something like, |f(x) - f(y)| <= [something involving a derivative of f] |x - y|.
the one variable MVT is that if f is nice enough then f(x) - f(y) is literally equal to f'(c) (x - y) for some c between x and y. this stops being true very quickly in the multidimensional context but once you start putting on norms and relaxing literal equality to inequality you still get useful stuff.
yes, but there is no mean value inequality for vector valued functions of a vector variable, as far as my understanding on my search on the internet has taken me
well your internet search should take you further, or perhaps you should read further into the reference that said "mean value inequality" without comment. i don't know.
rudin's PMA gives an example of something worthy of being called a mean value inequality although i forget if it is in the body of the text or the exercises.
there is no literal mean value theorem for vector valued functions of a real variable, generalizing the 1d MVT in the literal sense i mentioned above. you should perhaps reassess your views as to the rest of what you just said.
i dunno. i don't have rudin's PMA nearby or on this device.
there absolutely are multivariable statements that generalize things that would be consequences of f(x) - f(y) = f'(c) (x - y) if something like that were known to be true as it is often known to be true in the 1d context.
@BenSteffan I mean, taking coefficients in Z/2 already gives you the 2-torsion component is Z/2, but by which argument are you excluding non-torsion and p-torsion for odd primes p?
I'm using a theorem of hatcher which says that if $M$ is not $R$-or. then $H_n(M; R) \to H_n(M, M \setminus \{x\}; R) \cong R$ is injective with image $\{r \in R \mid 2r = 0\}$
(especially the part where I temporarily broke my TeX installation cause I hadn't updated it in two years, but needed to so that I can get the package that draws the triple arrow)
