my daughter was two when she learned the phrase 'lockdown drill' and enacted one. they can't do much at 2, but they try. presumably there is now a list of best practices that gets refined after every one of these.
you can develop base b log (b a positive integer) from the field axioms pretty quickly if you frame it in terms of "digit" counting. log_b(c) is a limit of a sequence counting base-b digits of c^n.
some of the properties are easier to prove than others.
$U,V$ are finite dimensional vector spaces. $W$ is a given vector space. Let $T\in L(U,V), S\in L(V,W)$ be linear transformations then it is to be shown that $\rm dimrange \,ST\le\min (dimrange\, T, dimrange \,S).$
Proof: For any $\rm x\in range \,ST, \exists y\in U$ such that $\rm STy=x$; and h...
yes. you could write it more cleanly. no backwards E's. the range of ST is a subset of the range of S, so its dimension is no larger than that of the range of S. and as you note, rank-nullity, applied to S as a map from range(T) to W, implies the last bit.
the elementwise stuff sorta justifies some of the hypotheses but if you were doing it an exam setting you might skip it.
"dimrange" also looks ugly, i prefer operatorname{rank} or \operatorname{dim} \operatorname{range}
rank nullity applied to maps whose domains and ranges are something other than all of some R^m and some R^n is a good reason all by itself to learn linear algebra.
Yes, query for 'favorite' votes, see the duplicate for why it's called that. Users that bookmarked a specific post will have voted with that type of vote on the post. — Martijn PietersOct 12, 2020 at 15:06
on some set of functions for which the operation makes sense.
the wikipedia page nods at this, "Clearly, the natural domain of definition of these partial differential operators is the space of {\displaystyle C^{1}}C^{1} functions on a domain {\displaystyle \Omega \subseteq \mathbb {R} ^{2},}{\displaystyle \Omega \subseteq \mathbb {R} ^{2},}"
you don't need to agree with the 'clearly' or have an independent understanding of the term 'natural domain of definition' to see that those things do have meanings as maps on those spaces.
sometimes the choice of domain is important. sometimes it is not.
@leslietownes does that mean $\frac{\partial}{\partial x}$ is the same as $\frac{\partial f}{\partial x}$ where $f$ is in the domain? (kind of starting to understand but not there yet)
kinda. it's a bit more like how sometimes people write f instead of f(x) as the function. you can sometimes omit what you are applying the function to, and don't always need to separately introduce notation for it.
you do a similar thing if you write ' instead of f'. by ' i mean the operation of taking the derivative; by f' i mean that operation applied to some f.
d/dx is to ' as f' is to df/dx. if that makes sense.
@AkivaWeinberger the proof I know is that $\Bbb Z^{\Bbb N}$ is too rigid to be free, its dual is the free group on a countable set, while it would have to be very uncountable if the group were free
@AkivaWeinberger here is a related fun fact. For a group $G$ define its dual to be $G^\ast=\mathrm{Hom}(G,\Bbb Z)$
Just like for vector spaces we have a canonical homomorphism $j\colon G\to G^{\ast\ast}$ given by $g\mapsto(f\mapsto f(g))$, that is $g$ is sent to the morphism "evaluation at $g$" on $G^\ast$. We say that $G$ is reflexive iff $j$ is an isomorphism
Now let $G$ be free Abelian. Must it be reflexive? The answer is yes if and only if $G$ has cardinality smaller than the least measurable cardinal!
It's easy to show that if $G=\bigoplus_{\alpha<\kappa}\Bbb Z$ with $\kappa$ measurable, then $j\colon G\to G^{\ast\ast}$ is not surjective, getting a measurable from a nonreflexive free abelian group is harder
The easiest (and the one I'm using in the linked answer) is that a cardinal $\kappa$ is measurable if there is a $\kappa$-complete nonprincipal ultrafilter (meaning that it is closed under intersections of less than $\kappa$ many elements) on $\kappa$
The definition actual set theorists like is the critical point (first ordinal that gets moved) of a nontrivial elementary embedding $j\colon V\to M$ for some class $M\subseteq V$
Wait so you're putting the universe in a subclass of the universe?
(In my mind a "class" is a "metaset" - like, whereas a set is just an object of the axiomatic theory, a class is a set in the metatheory. Is that a good perspective to have?)
@AkivaWeinberger a function $j:V\to M$ that preserves truth of formulas, meaning that for every formula $\phi(x_1,\ldots,x_n)$ and every $v_1,\ldots,v_n\in V$ we have $V\vDash\phi(v_1,\ldots,v_n)\iff M\vDash\phi(j(v_1),\ldots,j(v_n))$
(this makes sense for maps of structures in any language, there's nothing special about ZFC)
@NeelRayal If there were an element of order 5, then the subgroup generated by that element would have order 5, and 5 would therefore divide the order of G. Does 5 divide 32?
aghh Im so confused. Im actually studying calc 2 but I am stuck on a basic concept, I derived a function and now have a rational function with bunch of xs on the numerator and denominator, and I can cancel out but the thing is in the denominator there is abs(x), but I will only ever put in 0 for the x
so I have to cancel out so I dont get an undefined expression
Hello! in this problem, is the $\Delta$ here the same as $\nabla ^2 $ ? https://math.stackexchange.com/questions/345878/proving-that-f-is-analytic-if-f-and-z-fz-are-harmonic
Hello everyone! I'll repeat my question 😅 in this problem, is the $\Delta$ referring to the aplacian operator $\nabla ^2 $ ? https://math.stackexchange.com/questions/345878/proving-that-f-is-analytic-if-f-and-z-fz-are-harmonic
@Silidrone That's a bit hard to read. What does w'/o/ mean? Is that power of v supposed to be 3? Please see math.meta.stackexchange.com/q/33075/207316 Unfortunately, Stack Exchange Chat doesn't directly support MathJax, but there's a link at the top of the page to ChatJax, which uses bookmarks to render MathJax here.
I feel a bit sorry for the prime sieve guy math.stackexchange.com/q/4455158 He's putting in a lot of effort to convert his notation to MathJax, but that by itself won't make the question worthy of reopening. He's still using vaguely defined terminology, or using terminology that already has a well-defined meaning, and expecting readers to just know what he means by those terms.
Maybe it's just me, but this is virtually buzz word salad:
> It's a prime numbers sieve that unravels as harmonic patterns of co-primes of an ever growing sequence of prime generators numbers (G) that depicts the harmonic distribution of prime numbers.
if D could vary for a given f, one thing i would worry about would be using language that suggested D was the set on which the thing was harmonic, that the set of points at which f was harmonic was D and not a larger set. but neither 'on' or 'in' carry that connotation, in my mind. and maybe that isn't even a relevant consideration in your setting.
Hello, I dont find any way to specify an expert prior for that (I am noob). It has to be limited 0,1. It has to have a 95% CI. The most probable value of the diagnostic sensitivity and specificity of the test being used was determined as 0.85 and 0.95, with a 95% probability interval of 0.82−0.90 and 0.90−0.97, respectively. Translate this as good as possible in a prior distribution for sensitivity Se and specificity Sp.
I knew that limits are defined for limit points because, otherwise, for example a function from Z to R could have any possible limit L because the condition $0<|x-x_0|<\delta \implies |f(x)-L|<\epsilon$ is vacuously true. However, I always thought that the use of limit points is also to modellize the concept of "go near $x_0$ as much as I want", because that is the intuitive meaning of $x \to x_0$.
Is this latter intuitive reason correct too, or the reason why we need limit points is just to have the implication not always vacuously true?
@AlekMurt that's not difficult to prove for yourself
Let $M$ be a Noetherian module and let $N$ be a finitely generated module over a ring $A$. Then there's a surjection $A^n\to N$ for some $n$. Now tensor that surjection with $M$, to get a surjection $M^n \to M \otimes_A N$. Can you conclude from there?
that's a general property of tensor products, people call that "right exact"
and I do mean $M^n$. we have $A^n\otimes_A M \cong M^n$
the general statement is that if $f:X\to Y$ is an $A$-linear surjection, then for any $A$-module $Z$, the induced $A$-linear map $Z \otimes_A X \to Z\otimes_A Y$ is also surjective
ok so you can find in many sources on commutative algebra e.g. Atiyah-Macdonald the statement that if $0 \to M' \to M \to M'' \to 0$ is exact, then $ N\otimes M' \to N \otimes M \to N \otimes M'' \to 0$ is exact
this implies the statement about surjections above
If $M$ is finitely generated, we have an epimorphism $A^{n} \rightarrow M$ for some $n$. This induces a monomorphism $hom _{A}(M, N) \hookrightarrow \operatorname{hom}_{A}\left(A^{n}, N\right)=N^{n}$.
Hey, I'm trying to see why is this true, how come we can induce such injective function and how do we have the last equality.
you can also do this directly. If $M$ is generated by $m_1, \dots, m_n$, then you can define an injection $\mathrm{Hom}(M,N)\hookrightarrow N^{n}$ via $f \mapsto (f(m_1), \dots, f(m_n))$
Hello everyone. I have the following short question. Is $\Bbb{Z}/6\Bbb{Z}$ not a UFD since for example [4] can not be factorized with irreducible elements?
If I have a basis of eigenvectors and I apply Gram Schmidt to them to orthogonllize them and then proceed to normalize them as well, would this orthonormal basis still be eignevectors in general? I know in the case of self adjoint opertors it will and also for normal operators in a complex inner product space. But I was curious if it happens in general. I tried a quick rough proof but it didn't yield much
@D.C.theIII clearly not, or every diagonalizable linear operator would be normal!
the version of the spectral theorem that I learned says that on an inner product space over the complex numbers, a linear operator is normal if and only if there is an orthonormal basis of eigenvectors
there is a generalization of the spectral theorem, called the singular value decomposition that also works for non-normal operators. But it won't give you a basis of eigenvectors in general
@D.C.theIII The usual argument for the spectral theorem is to apply GS only to eigenspaces of dimension >1. And the eigenspaces for distinct eigenvalues are already orthogonal.
yeah. and that little piece of the argument works anywhere, you can GS a basis for an eigenspace (of any operator) and get an ONB for the eigenspace the same way you can GS a basis for any vector space W and get an ONB for W. there's just less of any reason to do that if the operator isn't normal.
just as some people wish they could see a movie or play or something again for the first time, i wish i could re-learn the spectral theorem with fresh eyes.
Is a linear operator $T$ that is normal, normal regardless of the ordered basis while the matrix representation of that operator must be on an orthonormal basis?
I was attempting an exercise that asked to produce a counter example to the false claim that the linear operator $T$ is normal iff the ...
Here's an easier version for you. You know what a symmetric matrix is. A symmetric linear map satisfies $(Tx,y) = (x,Ty)$ for all $x,y$. When is the matrix of a symmetric linear map a symmetric matrix?
This is something many students fail to understand.
the question in the post, as phrased, comes awfully close to inviting a type mismatch. it is using two definitions of "normal." one for operators, one (implicitly) for boxes of numbers. the adjoint operation that you get from the operator definition only coincides with the box-of-numbers definition under certain hypotheses. in general it's just some random matrix operation.
i understand your confusion but some of it is in the setup of the problem. "T is normal but [T]_b is not." well, that's implicitly two different adjoints there. not surprising that they don't give the same thing.
maybe our pal sheldon was on to something when he nearly banished matrices from linear algebra. at least, you wouldn't run into this.
@TedShifrin only thing that comes to mind is when the matrix is diagonal, but that's because I can't picture anything else.
@leslietownes I did not consider this.....but now that you point it out, it makes sense. Since matrices are their own linear transformations and us associating a linear transformation $T$ to a matrix representation is going to be some sort of isomorphism...
it also explains throughout the text why Insel and co have been stressing $T$ and the matrix of $T$ separately...
@leslietownes I'm not sure I see what I should "expect" but the only notion that comes to mind is that the coefficients of each column are the scalars for the linear combination of basis vectors used to write out each adjoint operator being applied to a basis vector......
i guess i've signaled one way of sorting it out. in my world it's unnatural to think of matrix conjugate transpose as an "adjoint" operation. because we use matrices for so many purposes. the matrix op is a box of numbers thing that might or might not be the adjoint.
of course if you open a book on C* algebras people will, indeed, introduce tons of examples in matrix algebras where it's the involution, and maybe not even say that they are implicitly defining that as their involution. i can't fix america.
the new mountain lions at the OC zoo are really cute.
Even though I had read it, it has not fully taken, but the matrix of a symmetric linear map will be symmetric if the basis I use to write the matrix is orthonormal.
glad to know I'm not losing my mind when I ask myself these questions
I really shouldn't, but I think I'm going to get some Momos (Tibeten delicacy). Be back in 30 mins. to rack my mind. Then later on to work on showing the cross product of compact sets is compact... :)
what would be the motivation behind choosing that?
when I attempted it. I made two arbitrary matrices that I tried to get to not be normal, then extracted their linear maps. Of course this is wrong now that I have talked abut these ideas but that was my first attempt
rotation matrices are part of the toolkit. diagonal matrices (not super helpful here :)) are another.
of course it's not the only example. your approach could have worked but it probably just began looking so ugly that it looked wrong even if it wasn't.
I don’t see the issue. What you posted is crap. But understanding a symmetric map even in an orthogonal basis (with different lengths) destroys symmetry immediately.
I see what you're saying Ted....I have to get out of the tendency of thinking I could produce everything without writing it down.....a simple application of the definition of the adjoint had me find $T^*$. Just have to improve on having a stock of examples.
