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12:53 AM
Let $T: \mathbb{R^n} \rightarrow \mathbb{R^n}$ and $h \in \mathbb{N}$ s.t $Im(T^h) = Ker(T^h).$
Show that every two chains in a chain basis for $T$ have the same length.
I know that if $(v_1, v_2, \dots, v_k)$ is a basis for $Ker(T^h)$ then we can choose $(u_1, \dots, u_k)$ such that $(T^h(u_1), \dots, T^h(u_k))$ forms a basis to $Im_T$ and for every $1 \leq i \leq n$, $(u_i, T(u_i), \dots, T^h(u_i))$ is a basis for $Z(T, u_i)$
I struggle to see how it connected if at all
What is chain length?
like every two chains in a chain basis for $T$ have the same number of vectors
Sorry, what is a chain basis?
@TedShifrin It seems that what I said may have nothing to do with the generators that you mentioned. Given a subgroup $H$ of cardinality 20, $S_5$ act transitively on $S_5/H$. On the other hand, since $|S_5/H|=6$, another subgroup that stabilizes one point is isomorphic to $S_5$ as well. MSE has someone work out the problem even better so that it is not at least two but exactly two to the answer.
reminds me a tad of the following
Mar 10 at 2:43, by leslie townes
also, FWIW, a lot of the terminology books introduce around jordan canonical form is not standardized to the point that it's 100% possible to know what a "cycle of generalized eigenvectors for T corresponding to lambda" is without being told in advance
1:04 AM
Sorry, @oscar, I don’t follow.
No problem.
It is just the above result that I wanted to say
@x4j what chain and chain basis means?
Are you referring to something like $\{v, Tv, \dots, T^{n-1}v\}$?
To me it looks like $T$ is a nilpotent that has order of $2h$
1:34 AM
A chain for a linear operator $T$ is a sequence of vectors $(v, T(v), \dots, T^k(v))$ and $T^{k+1}(v) = 0$. A chain basis for $T$ is a basis for $dom(T$ that is made by chains as above (.i.e if $C_1,\dots,C_k$ are chains for $T$ with $C_i \cap C_j = \emptyset$ for $i \neq j$ and $C_1\cup \dots \cup C_k$ forms a basis for $dom(T)$ then it is a chain basis).
1:50 AM
Ok, thanks, @X4J. Weird hypothesis … It implies $n$ even, for starters. I would suggest understanding $h=1$ and $h=2$ before dealing with the general case.
2:01 AM
I've just tried to make a "co-knot" theory, which deals with quotient maps instead of embeddings, but I cannot find a good demonstration. :(
In particular, I cannot prove that there is only one co-knot from $S^2$ to $S^1$.
I don’t think that’s a good word. There is no submersion (which is the reverse notion to an embedding) at all. What is your precise definition?
> Definition. For topological spaces $X$ and $Y$ and quotient maps $f, g: X \to Y$, a co-isotopy is a free homotopy $F: X \times I \to Y$ between $f$ and $g$ satisfying that $F(-,t)$ is a quotient map for every $t \in I$.
I'm trying to prove that there is only one co-isotopy class from $S^2$ to $S^1$.
Are the quotient maps surjective?
Yes, of course.
So what is an example of a quotient map $S^2\to S^1$?
2:10 AM
$f: S^2 \to I$, $f(x,y,z) = (x+1)/2$, let $g: I \to S^1$ be the quotient map identifying the endpoints of $I$, and give $g \circ f$.
I've just looked at the definition of a submersion; I suppose this particular $g \circ f$ is... wild?
No, not wild at all, but it drops rank at places. Embeddings can’t change rank.
Mapping to the interval $[-1,1]$ by going to $x$ is more intuitive to me.
Just project and then identify endpoints.
Quotient maps serve as the categorical dual notion to embeddings, so is my notion of "co-knots" to knots. I wanted to see whether this notion is effective/nontrivial.
Just shows I don’t like category theory.
what is this
im js curious because im new to the site
2:20 AM
This is a chatroom for general discussion of math.
oh ok
is it customary to use proper grammar or to chat like I'm texting someone?
Since you're new, I strongly encourage to use MathJax.
Oh ok thank you as I said before I am new to this and this is the first stack exchange chat I have ever been in.
@DannyuNDos Ok thanks
I have a question
If you graph the gamma function on Desmos it looks nothing like the factorial function, is there a reason why or a error on Desmos's end
Because I know the gamma function is supposed to generalize the factorial
$n! = \Gamma(n+1)$
Oh that makes so much more sense I totally forgot about that when entering it into Desmos
Why is the Gamma function set up that way anyways?
2:30 AM
I constantly wonder that too.
It must have some use
Wasn't it Euler that made the function?
By the way I just plugged the gamma function in and it fits the x! function but is not defined for most negatives.
Bernoulli and Goldbach did.
And in fact, it's undefined for every negative integer.
Division by zero babe
Oh yea
By the way the gamma function works exactly like x! if $$f\left(z\right)=\int_{0}^{\infty}e^{-t}t^{z}dt$$
I also found this other formula online that I think is simpler and easier to understand where it came from:
$$f(x)=\lim_{N \to \infty} N^{x}\prod_{k=1}^{N}\frac{k}{x+k}$$
3:16 AM
you seriously think desmos is going to provide the gamma function and screw it up?
Curses, no, copper!
i found a few bugs in Desmos many years ago
my preoccupation with convex level sets
i'm still stuck on my little compact operator problem.
almost as bad as algebra
(my algebra, i mean. i wish i had the memory for algebra)
Early Mathematica had some truly embarrassing glitches.
It messed up some integrals every college calculus student could get correctly.
3:34 AM
i used to use Macsyma a lot (still do, from time to time). I view them with some degree of suspicion.
free, of course, had a lot to do with it
Blame MIT for that!
I did some stuff with Richard Fateman who had a lot to do with that. That was a long time ago.
One attraction for me was that it was Lisp based.
4:01 AM
@Jakobian Thank you very much @Jakobian Can you please explain what you are trying to mean bit more too? I still don't understand the point you try to say :) Thanks a lot and am very gateful
@Jakobian for your convenience I'm reposting my original question below
When we have two sets say A={1,2} and B={Ann, Jack, Janet}, the set operation - union, when applied on these two sets give A union B = {1,2,Ann, Jack, Janet} right?
I mean the set A union B can have elements of different types?
I have seen in some places they say a set can have elements of a same type only. But I feel doubtful about this
And if we can take union like above, we can have elements of different types too right?
As a set is just a collection of elements with no additional structure imposed on them
4:17 AM
there is no such thing as a "type", mathematically speaking
and in terms of CS, sets as a data type can hold objects of different types
4 hours later…
8:33 AM
I just took a seminar by a senior professor studying option pricing theory and he emphasized two things: don't invest and a radon-nikodym derivative is actually useful.
9:07 AM
Q: State a necessary and sufficient condition for a finite dimensional vector space $V$ over a field $F$ has precisely one basis.

Thomas FinleyState a necessary and sufficient condition for a finite dimensional vector space $V$ over a field $F$ has precisely one basis. I tried solving this problem as follows: Let $\dim V=n.$ So, we can assume, $B=\{v_1,v_2,...,v_n\}$ to be a basis for $V.$ Let $c\neq 0$ be in $F.$ Claim: $cB$ is a basis...

Can anyone please help me with this?
3 hours later…
12:09 PM
when people write $\frac{d}{dx}f(g(x))$ do they mean the derivative of $f$ evaluated at $g(x)$ or do they mean the derivative of the composition $f\circ g$? I think for composition one has to write $\frac{d}{dx}\Big(f(g(x))\Big)$
12:25 PM
@attack Interpret it as you'd like
Notation is for humans to read
as for what would be more correct, that's another issue
If anything, I'd say $\frac{d}{dx}(f\circ g)(x)$ would actually be more correct
The derivative is of the function $f\circ g$, and not of its value $(f\circ g)(x)$ at some point $x$
If someone were to write $\frac{d}{dx} f(g(x))$, I'd be more inclined to interpret it as the derivative of $f\circ g$ at $x$, since $x$ is used in the brackets and in $dx$, but you need to look at the context to really make sure this is what we're talking about
Also note that we often write functions, say $x\mapsto x^2$ simply as $x^2$
which is abuse of notation, but its a popular one, and someone writing the derivative might have thought of $f(g(x))$ with this abuse of notation in mind, as a function $x\mapsto f(g(x))$
1:19 PM
it's not abuse if you just call the identity function $x$
which is actually how you want to think when doing differential forms etc
Let $R$ be a commutative ring with unit. Show that an ideal $I$ is prime if and only if $I$ satisfies the following two conditions:
\item If $I = I_1 \cap I_2$ for two ideals $I_1, I_2$ in $R$, then $I = I_1$ or $I = I_2$.
\item If $a \in R$ and $a^n \in I$ for some positive integer $n$, then $a \in I$.
I am struggling with the if part, but I have solved the only if part. Any hints on this.
@attack: Both $\frac{d}{dx}f(g(x))$ and $\frac{d}{dx}\Big(f(g(x))\Big)$ mean the derivative of $f\circ g$, evaluated at $x$. However, it's worth noting that Leibnizian notation is not terribly precise from a modern point of view, and at times of doubt, it is better to use Lagrange's prime notation. There is some discussion of what $\frac{d}{dx}$ means in the first place here on MathOverflow.
@attack: For what's its worth, I would interpret the notation $\frac{d}{dx}f(x)$ as follows: the $x$ on the "denominator" of $\frac{d}{dx}$ is serving two purposes: it tells us that the function we are differentiating is $x\mapsto f(x)$, i.e. $f$, and it tells us to evaluate the derivative $f'$ at the point $x$.
This "double duty" – where $x$ is both interpreted as both a free and a bound variable – is part of the reason why Leibnizian notation is considered to be imprecise.
1:36 PM
superb, thanks
I will stick to $\frac{d}{dx}(f \circ g)(x)$ for composition and $(\frac{d}{dx}f)(g(x))$ for non-composition
@attack: No problem! Actually, I think it is better to write $f'(g(x))$ in the latter case, or if you insist upon Leibnizian notation, $\frac{df(u)}{du}\big\vert_{u=g(x)}$. The reason I say that is that the notation $\frac{df}{dx}$ (with $f$ a function in the modern sense), while somewhat common, does not really make sense from the point of view I am advocating.
I am reading the proof of the following theorem:
let G H be connected Lie groups , and f a lie group homomorphism, then f is a covering iff the differential an isomorphism:
proof: if f is a covering, it is a local diffeomorphism, then follows theat the differential and isomorphism
my problem is, we need to have it such, that f is a global diffeomorphism, for the differential to be a vector space isomorphis, right?
and do we mean a normal vectorspace isomorphism ,or a lie algebra isomorphism (respecting the bracket) in this case, it is not clear in the theorem
Q: If $f :X \to Y$ is a diffeomorphism, then $df_x$ is an isomorphism of vector spaces.

Johnver${\bf Proposition}$: If $f: X \to Y$ is a diffeomorphism, then $df_x$ is an isomorphism of tangent spaces. ${\bf Proof}$: Let $f: X \to Y$ be a diffeomorphism. Then $f$ is a smooth bijection and so is $f^{-1} : Y \to X$. Since $f$ is a bijection, $\dim X = \dim Y$. Therefore, $\dim T_x(X) = \dim...

@attack: After all, it's kind of weird to use both $\frac{df(x)}{dx}$ and $\frac{df}{dx}$.
The point is that usually, $\frac{d}{dx}$ operates on "expressions" (or "functions of $x$"), but $f$ is just a function (not a "function of $x$").
apparently this is the theorem
For manifolds, however, does this apply in both directions? because the theorem apparently uses the same premise in both directions, and then we speak merly of a vector space isomorphism (not a bracket respecting one)
1:59 PM
@userunkown your TeX is not rendering properly
nvm found my answer
@Mad no, the differential at a point only depends on what the function does in an arbitrarily small neighborhood of that point. think back to analysis
@Mad the differential of a Lie group homomorphism at the identity is automatically a homomorphism of Lie algebras, so it is a Lie algebra isomorphism iff it is a vector space isomorphism iff it is bijective
@Thorgott I don't know how to make it work. But, here is the question without TeX: Let R be a commutative ring with unit. Show that an ideal I is prime if and only if I satisfies the following two conditions
(a) IfI=I1∩I2 fortwoidealsI1,I2 inR,thenI=I1 orI=I2. (b) Ifa∈Randan ∈I forsomepositiveintegern,thena∈I.
@Thorgott Okay, so we have "f local diffeo imply df iso"?
Let me try. Is this good?: Let $R$ be a commutative ring with unit. Show that an ideal $I$ is prime if and only if $I$ satisfies the following two conditions:
(a) If $I = I_1 \cap I_2$ for two ideals $I_1, I_2$ in $R$, then $I = I_1$ or $I = I_2$.
(b) If $a \in R$ and $a^n \in I$ for some positive integer $n$, then $a \in I$.
2:04 PM
And its even two way implication right?
yes, Mad, and you can also formulate this at any specific point rather than just at all points at once
@userunkown that's readable!
@Thorgott by point you mean an element in the tangent space ) IE df_x)?
no I mean a point on the manifold
@Thorgott Good to know. Do you have any hints for the only if part. In particular, I am having a hard time using the hypothesis conditions (a) and (b) to get that $I$ is a prime ideal. I keep reaching a dead end.
@userunkown that's "if", not "only if". are you familiar with the notion of a radical?
2:09 PM
which radical
@Thorgott Sorry, i got confused. Yes, I am familiar with the notion of a radical. But, I don't see how that is helpful.
@Jakobian If you have to ask... :P
have you done the sister exercise to this one before where you instead have the condition $I=I_1I_2$ for two ideals $I_1,I_2$ implies $I=I_1$ or $I=I_2$?
@Thorgott No, I haven't. But, suppose I have done that, how could it be used to solve this problem?
your condition (b) says that $I$ is a radical ideal, so you can take radicals on both sides of such an equality and leverage that there is a relationship between intersections, products and radicals
2:18 PM
@Thorgott Thank you. Although in my course and book there is no mention of radical so far. Is there a way to do this without using radicals?
Oh so we're talking about radicals of ideals?
I mean, you can always just expand the definition at every point to remove an explicit mention of radicals. That just makes it a lot less streamlined.
is there a theorem that states that kernels of local diffeomorphisms are discrete?
apparently so:
2:39 PM
yes, the fibers of local diffeomorphisms are discrete
A question that I saw recently on the AoPS community: Given 2 tangents to a circle and a point on the circumference of the circle (not at the contact point), can we construct the circle? [2 circles can satisfy the condition, can we construct any or both] [I was thinking of first drawing the bisector of the tangents at their intersection, the center of the circle must be on the bisector… but what then?]
@Thorgott Is there an elementary way of doing this? Also, how would one use the condition (a): If $I = I_1 \cap I_2$ for two ideals $I_1, I_2$ in $R$, then $I = I_1$ or $I = I_2$.
3:46 PM
@Mad yes, regular value theorem
1 hour later…
5:04 PM
@Mad Preimage, not kernel
In fact, it has nothing to do with inverse function theorem or even differentiability. It's true for local homeomorphisms, too. If there were a limit point, you couldn't have a neighborhood of that point on which the function is a homeomorphism to its image.
@userunkown using radicals is entirely elementary. it's just a definition that helps us organize arguments. there's no "deep" element that I'm sweeping under the rug or anything.
@Thor How radical of you!
lenin taught us that you need to organize radicals to sweep the old ways under the rug
you need "deep" elements, professional cadres, to arm the proletariat materially and theoretically
5:26 PM
Will that involve dictatorship on the first day?
LOL, @Thor.
Almost everywhere version of chain rule: If $f, g$ are differentiable a.e. and $f$ maps null sets to null sets, then $(f\circ g)' = (f'\circ g)\cdot g'$ a.e.
For example, $f$ maps null sets to null sets and is a.e. differentiable if $f$ is an indefinite integral of a gauge integrable function. In particular, it holds when $f$ is absolutely continuous, or when $f$ is continuous and $f'$ exists except for a countable set.
in 1d
does thsi make sense : Consider $f(x,y) = \vec{r}$ where $\vec{r}=r_1\hat{i}+r_2\hat{j}$ where the $\vec{r}$ is a vector in N^2
im on my phone so cant read latex rn
does that map N^2 to N^2 depending on r
@Obliv how is $f$ a function of $x, y$?
if not how would I write down a function to map N^2 to a subset of N^2
what is N^2
5:40 PM
so you're considering a constant map from $\mathbb{N}^2$ to $\mathbb{N}^2$?
the vector notation confuses me here
you can just write ordered pairs as ordered pairs
@leslietownes it's NxN just too lazy to type it in my phone
what is N, please
the vector notation is an unusual choice for analysis of functions on integers or positive integers or nonnegative integers
hence my question
ordinarily i would not be doing this
its supposed to map NxN to a subset NxN where first pair element is smaller than or eq to second element in the pair
natural numbers sorry
hmm... sure then
5:42 PM
I just thought to map a pair to a pair we have to use vector notation or something
so the constant function f(x,y) = (1,2) would map N^2 to a one-element subset of N^2 satisfying that condition
It seems to me like you're doing something from elementary set theory
but your notation suggests some analytic geometry or physics
there seem to be a couple of things going on here, one being how do you write down a function "generally" and then maybe also how do you construct one having specified properties
so its kind of confusing
sorry ill post the question to clear things up
number 14
5:45 PM
If you take a constant function then it won't be a bijection then
i was trying to write it in vec notation so i could arbitrarily define x,y like r_1 would be strictly smaller than r_2 somehow
You can just write $f(x, y) = (r_1, r_2)$
its fine
do your natural numbers contain 0
My suggestion is to try going for $f(x, y) = (x, r)$ and then trying to figure out what $r$ would have to be for this to work
so something like $f(x,y) = (x,x+1)?
5:48 PM
thats not a bijection
That doesn't have (x, x) in the image
but you're close
why does that need to be in the image exactly?
Because $x\leq x$
Oh right
5:52 PM
yep. Good
cool thank u :)
Consider the lattice of N×N, take any horizontal strip of it of and map it to the vertical strip starting at a diagonal point
Of same y coordinate
6:07 PM
what's a lattice of NxN? i think i remember seeing that as a section in an abstract alg book
im reading that Q is countably infinite, and I wonder is the list of rational numbers impossible to order?
Give an example of a linear operator $T$ on $\Bbb R^3$ such that $T\neq 0,T^2\neq 0$ but $T^3= 0.$ Can anyone think about an example of this one?
like a nilpotent matrix?
can you find a linear operator $S$ on $\mathbb{R}^2$ such that $S\neq0$, $S^2=0$
if you can do that, you can extrapolate
@Thorgott No, that's the problem. Can't think of an example like that right now...
@SineoftheTime maybe..but what should that be if I ask, explicitly ?
The defining of countability by being able to list elements seems weak. Like couldn't u just include all the irrational and transcendental numbers in the list?
6:14 PM
well, try a bunch of examples
the diagonalization argument makes more sense to me
@Obliv how
@ThomasFinley the space of linear trasformations is isomorphic to the space of matrices, if the dimension is finite. So you can find a nilpotent matrix and then consider the operator associated to it. I don't know if I misinterpreted your question
@Thorgott maybe instead of the two headers for columns&rows being natural numbers for which to take a division of to represent members of Q, you can define the members of the table as all possible relations on $\mathbb{Z}\times \mathbb{Z}$ like the square root operation, division, stuff like that and if there's a finite amount of such operations to get irrational numbers then you'd have a countable list?
but stuff like $\pi$ idk how you'd get. $\sqrt{2}$ is easy to define
at best, the numbers you would get using such methods are the algebraic numbers (roots of integer polynomials) and the set of those is indeed countable
6:21 PM
Ah ok, is there no way to generate non algebraic numbers then?
idk what "generate" means
there certainly is no way to make a countable list of transcendental numbers, because there's uncountably many of them
oh i guess not, by definition. Like by defining a non algebraic number by some formula lol but thats literally the definition of non algebraic
@Thorgott sounds tautological
{pi/n: n = 1, 2, 3, ...} is a countable list of transcendental numbers
@Obliv of course, all true statements are tautologies
@Thorgott also this one?
6:26 PM
this is all getting pretty far away from soumik's excellent suggestion. i think all that was meant by 'lattice' was to think of ordered pairs as points in the cartesian plane. if you draw pictures of {(n,m): n <= m} and NxN it is not too hard to see how to biject one into the other by moving vertical "strips" of these "regions" up or down
scare quotes indicating references to the cartesian plane picture of those sets
then there's the challenge of coming up with a "formula" for such a map, which soumik helpfully left for the reader
@Obliv lattices are certain subgroups of $\mathbb{R}^n$
from what I know
but there is also another meaning as a partial order of special kind (or algebraic structure, if you want to think about it this way)
that's irrelevant to the discussion, but I think lattices are pretty neat because they're partial orders which can be thought of algebraically
can they be thought of transcendentally
so $f(x,y)=(a,b)$ where a,b are related by a line with slope 1 and y intercepts at least greater than 0?
Apparently Collatz conjecture reduces to a question of "Is $n$ a modular power of 2?" as I'm gonna call it.
Not hard to see (even for me) that for the rules given, let $a_0$ be the first iteration, then $a_1$ is always even.
I did some other work, however. $3n + 1 = 4^n$ when $n =\lfloor\frac{4^n}{3}\rfloor$ (identity of $n = \frac{4^n - 1}{3}$).
@leslietownes There is a concept of a polynomial for a given universal algebra
6:40 PM
Happens to be directly related to $xy = 2^p (2^q - 1)$ which is how I get this answer for $n$.
But I think there'd be a problem of what a "root" means in such setting
obliv: in your exercise you are asked to provide a 'formula' for the map. so this notation is a start, but implicitly a and b are both functions of x and y here, so you can be more explicit (and the exercise, in asking for a "formula" and not simply a proof that the sets are in bijection with one another, presumably expects that)
i.e. there are a specific a(x,y) = [something involving x and y only] and b(x,y) = [something involving x and y only] that will work
at which point you might just write f(x,y) = (the thing involving x and y, the other thing involving x and y)
using vector notation with is and js, or calling the first coordinate a and the second coordinate b, is not really doing any work here
For a lattice such "polynomial" would consist of some combination of variables, meets and joins
Like $(x_1\vee x_2)\wedge (x_3\vee x_4)$
@AMDG Sorry, I always forget to make sure the variable names are distinct sometimes. Say $3x + 1 = 4^n$.
@Jakobian I guess we should also add constants
6:48 PM
and then just have $x$ on the LHS of the two latter equations I gave.
A: Prove that the eigenvalues of a real symmetric matrix are real

naturerLet $Ax=\lambda x$ with $x\ne 0$, with $\lambda\in\mathbb{C}$, then \begin{align} \lambda \bar x^T x &= \bar x^T(\lambda x)\\ &=\bar x^T A x \\ &=(A^T \bar{x})^T x \\ &=(A \bar x)^T x \\ &=(\bar A \bar x)^T x \\ &=(\bar\lambda\bar x)^T x\\ &=\bar \lambda \bar x^T x.\\ \end{align} Because $x\ne 0$...

another problem would be to see what a non-trivial polynomial is I suppose
Can anyone exllain what is $\overline x$ mean ?
@ThomasFinley Probably conjugation.
6:51 PM
@Jakobian But I am familiar with conjugation of a complex number , but never did I hear about conjugation of an $n\times 1$ vector like $x$ .
Am I missing something?
Elementwise conjugation?
If $A$ is a matrix, the conjugate $\overline{A}$ is a matrix obtained by conjugating each entry of $A$
$\overline{A}^T = \overline{A^T}$ is sometimes called the Hermitian transpose of $A$
But then again, if $t\in \Bbb R$ then, what is meant by $\overline t$? Is it the same as $t$ ? @XanderHenderson and @Jakobian
6:56 PM
For $\overline{a+ib}=a-ib$ and all real nos. are complex nos. as well, so $t-i0=t=\overline{t+i0}=\bar t$
@Thorgott I am still stuck. Could you please help me get started, or give me a more explicit hint.
For the inner product of complex matrices, rather than just transposing, you need to transpose and conjugate.
This is so the map $(A, B)\mapsto \overline{A^T}B$ is sesquilinear
In mathematics, a sesquilinear form is a generalization of a bilinear form that, in turn, is a generalization of the concept of the dot product of Euclidean space. A bilinear form is linear in each of its arguments, but a sesquilinear form allows one of the arguments to be "twisted" in a semilinear manner, thus the name; which originates from the Latin numerical prefix sesqui- meaning "one and a half". The basic concept of the dot product – producing a scalar from a pair of vectors – can be generalized by allowing a broader range of scalar values and, perhaps simultaneously, by widening the...
vector spaces over complex numbers are a bit different than the real ones
@Jakobian Sorry, but I did not get that. Actually, I never heard about anything called, "sequilinear" till now.
It means that it needs to be linear in one coordinate, but linear with conjugation in the other coordinate (when it comes to scalars)
instead of bilinear like for real matrices
$f(ax, y) = a\cdot f(x, y)$ but $f(x, ay) = \overline{a}\cdot f(x, y)$
7:19 PM
@userunkown which step are you having issues with
some perverts like conjugation on the second variable.
i gave someone a hint on a compact operator problem but after scratching around a while i am unable to solve it myself. i keep checking, hoping that someone has a hint or a solution, but to no avail.
@copper.hat Hey, baby.
You called my name?
compact operator?
"pervert who likes conjugation on the second variable". :P
or a 2nd variable deviant?
7:26 PM
Love that conjugation on the second coordinate.
tiny things like that throw me. one of my fave real analysis books is Kolmogorov & Fomim, but they (like most of the era) assume that all Hilbert spaces are separable.
@copper.hat Wait... what?
it was an assumption of the era for many
basically, for them, there's just $l^2$.
Yeah... exactly.
What else could you possibly do?
I don't understand...
All Hilbert spaces (worth studying) are separable.
7:29 PM
i made a fairly assertive statement on the basis of that once, to be quickly and appropriately shot down
Getting thrown isn't failure...not getting back up is.
i'm an expert at getting back up.
@user85795 It is if you break your neck in the process...
(Horses are dangerous.)
true dat
7:31 PM
generally when i ride an animal they try to scrape me off be going under branches
@copper.hat They'll do that.
elephants too, i discovered.
Some animals are not meant to be ridden, I guess.
@copper.hat not me
I'm all for $\mathbb{R}$-vector spaces
the only problem with reals is that you can't go negative while circumventing zero.
7:45 PM
You customarily ride elephants? Tigers too?
I found a picture of copper.hat.
copper is the green tiger?
Might be a good likeness.
He man
Macho man
7:56 PM
@TedShifrin Maybe he was singing?
Or writing the Rime of the Ancient Mariner?
I was trying to be poetic.
i think i may have ridden an ostrich once. certainly was bruised by one. never a tiger
i wonder what the response would be locally if i were to ride my bike on speedos like the chap above?
@copper.hat Dude! What are you even doing with your life?!
nothing, negative even
i peaked too soon :-)
7:59 PM
Well, copper, not to be outdone by you ... I found out today that I am due for cataract surgery in the next month.
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