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12:14 AM
Let V be a vector space of all m by n matrices over a field of 2 elements, then dimension of V?
I think that it's mn
Standard basis has mn elements $e_{ij}$
where $e_{ij}$ denotes m by n matrix whose every entry except ij-th entry is $0$, and ij-th entry is non identity element of the field.
is it correct?
 
yes. for any field F, dim M_{mxn}(F) as a vector space over F is mn, for the same reason.
 
It's something related to conic sections done by apollonius
 
12:46 AM
@leslietownes. Hello, Sorry if you don't like tagging, but I have a followup question about sigma distance please?
 
i'm sort of half here and half not, but ask away
 
@leslietownes Agreed.
 
i'm schroedingers cat
 
@leslietownes Or, at the very least, take the exam yourself three or four days after writing it.
@leslietownes My rule of thumb in precalculus: every minute it takes me to complete an exam is 7 minutes for my students. In calculus, the ratio is more like 1:5.
 
that sounds about right. i think you can go slightly lower in other fields, but still not that much lower. my wife (not in math) was getting negative evaluations a lot, and i kept telling her to make her exams shorter. she didn't believe me. i said, have a colleague take your exam. it took the colleague 30 minutes for an exam that the students had 60 minutes for.
she listened to her colleague but not me :)
 
12:49 AM
My ratio is way lower, although maybe for writing official solutions that’s close.
 
@leslietownes. **Lemma 2**: Suppose that during the execution of **BFS** on a graph $G=<V,E>$, the queue $Q$
contains the vertices $\left< v_1, v_2, \cdots, v_r \right>$, where $v_1$ is the head of $Q$ and $v_r$ is the tail. Then, $d[v_r] \le d[v_1] +1 $ and C 1 and $d[v_i] \le d[v_{i+1}]$ for $i=1,2, \cdots, r-1$.
I am not sure if you are familiar with BFS = breadth first search of a graph
 
i oppose capital punishment, so you are on your own with executing BFS.
i am not :)
 
$d[v]$ is the distance from source vertex $v$ to vertex $u$
That's fine :/
@leslietownes. Thank you
 
surely 25 to life would be enough.
 
1:10 AM
Munchkin is perverting what’s left of your mind.
 
1:25 AM
@leslietownes breadth-first-search?
BFS
 
@leslietownes thanks a lot :)
 
Ted: raspberry
 
Uh huh,
 
2:03 AM
Which came first, the binary operation on a single type or the cartesian product of two types?
If you say the latter, then you're referencing the binary operator on the colleciton of types called $\times$.
If you say the former, then you can't do the old $\star : S \times S \to S$ trick
because that references a product
 
2:28 AM
Is there a good algorithm for visualizing boundary conditions of a PDE as a 3D graph without actually finding the solution to the PDE? For example, I wanted to see the discontinuity mentioned in the comment here (math.stackexchange.com/questions/4249567/…). Is there some Mathematica or MATLAB code or something that will do this?
 
 
2 hours later…
4:29 AM
@user10478 are you into Python?
Teach a man to fish and so on
teach yourself python / Sympy and you've tought yourself how to come back and fish again
I could actually help get you started if you ping me
Will take donations when we're through
If you're satisfied
Sympy is not only a CAS but you write code in a real programming language, so that tying in other programming facilities (besides CAS) is easy later and you don't have to learn two things: a CAS language and a programming language.
 
I've done regular programming, very little Python but I went through a tutorial and made a small crypto price chart thing in Python a couple years ago. I'll scan what you linked tomorrow and if it has what I need I should be able to get it going. Thanks!
 
5:32 AM
Hi, I found this sentence in an equation "Augmenting variable x from Eqs 21 and 22 yields"
What does augmenting mean?
 
You need to provide more context.
 
yeah, 'augmenting' a variable 'from' an equation is nonstandard usage. never heard of it but might be able to guess with more context.
 
see the sentence just before eq. (23)
I can derive a wave equation from the eqns 21 and 22
which looks like d2S3/dt^2 = const times d2S3/dz^2
But no idea how to arrive at eq (23) and (24)
 
 
2 hours later…
7:34 AM
When people talk about matrix-vector multiplication in linear algebra where vectors are elements of vector space, do they imply that by vector they actually mean matrix representation of a vector ?
 
 
1 hour later…
8:44 AM
0
Q: Question on tangent lines and the center of an ellipse

love_sodam Let $\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$ be an ellipse. Let $A = (x_0,y_0)$ be a point outside of an ellipse. Draw two lines that touch an ellipse, say $D_1 = (x_1,y_1),D_2 = (x_2,y_2)$. My question is that if I draw a line passing through $A$ and the middle point of $D_1$ and $D_2$ i.e., $(\fra...

Need a little help in here
 
@JamesGroon the representation is just the matrix-like array that facilitates speaking of the ordered tuples that vectors are
 
9:23 AM
hi everyone I had a small question on differential operators
actually, if I have a linear differential equation of order 2
let say $\frac{d^2y}{dx^2}-3\frac{dy}{dx}+2y=0$
 
9:35 AM
now the book converts this equation into a linear differential equation, using the differential operator, now my book convert this equation into linear differential equation
I know how to solve this using characteristic equation but IDK how to convert this into a separate differential equation can anyone help me in this?
and what is the actual difference when I solve the equation using differential operator
 
 
1 hour later…
10:41 AM
@shintuku Yeah, but I'm trying to be as rigorous as possible.
I've searched online what would be the formal definition of matrix eigenvalue and I've either stumbled upon transformation eigenvalue or things like Av = λv, where A is matrix and v is vector.
Now I still wonder what would be the correct way to define it, presumably using matrix representation of a vector.
 
10:54 AM
For the benefit of others who are visiting room (or reading the transcript), I'll add that the above question about ODEs is now posted on the main site: Question related to differential operator.
 
 
2 hours later…
12:31 PM
0
Q: Question related to differential operator

Jack RodI do not know the eg I am using can be on the site before so before . Actually, I was learning differential operators using a book given by our prof , and there was an example in it says like this> $\frac {d^2y}{dx^2}-3\frac{dy}{dx}+2y=e^{5x}$ now I have solved these kinds of question using the c...

 
1:23 PM
Notationally you’d write your ODE as $(D^2-3D+2)y(x)=e^{5x}$, with $D=d/dx$. @JackRod
Or w/e your prof uses for differentiation
One then can factor the left side as $(D-1)(D-2)y$, which makes evident that the characteristic polynomial has roots 1,2
The less-obvious part is how to get the particular solution
The simplest way is to guess it (method of undetermined coefficients)
Formally one does have $y=(D-1)^{-1}(D-2)^{-1}e^{5x}$ but figuring out how to apply that is the problem
 
1:47 PM
@Semiclassical excatly
I need to understand what happen after the last line u have written
 
2:09 PM
Right.
Continuing with formal nonsense, we do have the following:
\begin{align}
(D-2)^{-1} e^{5x}
&=-2(1-D/2)^{-1} e^{5x}\\
&=-2\left(1+\frac12 D+\frac14 D^2+\cdots\right )e^{5x}\\
&=-2(1+5/2+(5/2)^2+\cdots)e^{5x}\\
&=-2(1-5/2)^{-1} e^{5x}\\
&=\frac43 e^{5x}
\end{align}
hmm
the problem with formal nonsense arguments is that it's not easy to see when you've done something silly
 
how you expanded the second
 
geometric series.
and then i formally resummed the resulting series. however, neither series is actually convergent
hence why this is 'formal nonsense'
it can get the right answer, but it's very much not rigorous
 
@Semiclassical thanks
 
oh, i see my error
\begin{align} (D-2)^{-1} e^{5x} &=-\frac12 (1-D/2)^{-1} e^{5x}\\ &=-\frac12 \left(1+\frac12 D+\frac14 D^2+\cdots\right )e^{5x}\\ &=-\frac12(1+5/2+(5/2)^2+\cdots)e^{5x}\\ &=-\frac12 (1-5/2)^{-1} e^{5x}\\ &=\frac13 e^{5x} \end{align}
hmm, no, that's still not right
to see why it's not right, note that we definitely have $(D-2)e^{5x}=(5-2)e^{5x}=3e^{5x}$
oh
i guess that does work out then
multiply both sides by $\frac13 (D-2)^{-1}$
to get $\frac13 e^{5x}=(D-2)^{-1}e^{5x}$
huzzah for formal nonsense
that said
it was definitely easier to start out by anticipating that, if we want to apply $D-2$ and get $e^{5x}$, we should start with a multiple of $e^{5x}$ itself
 
now why have u used D=5?
@Semiclassical
 
2:18 PM
we haven't used D=5 as such. what we've used is that $De^{5x}=5e^{5x}\implies D^2 e^{5x}=5^2 e^{5x}$ and so forth
so $D\neq 5$ when acting on arbitrary functions, but $D$ does behave like "multiplication by 5" when acting on $e^{5x}$
this is easy enough to justify at the level of $(D-2)e^{5x}=3e^{5x}$
the 'operator series' method is where things get dicey in terms of rigour
 
 
1 hour later…
3:45 PM
Hello,
Why can not we say here that $\pi[u] = t, r$ pleas?
Given that $pi$ is defined here as the predecessor
of u
 
huh?
 
4:04 PM
@copper.hat. :/
@copper.hat. What do you call the two parts $=>$ and $<=$ when proving if and only if statement?
I call $=>$ the forward statement and $<=$ the backward one
Does this has to do with sufficient and necessary conditions please?
In my notes, $=>$ is called necessary statement and the other one $<=$ the sufficient statement?
I am not sure if this is right!
 
Do $a^2$, $b^2$, $(ab)^2$, $(ba)^2$, $(aba)^2$, and $(bab)^2$ have any nontrivial relations?
 
@AkivaWeinberger. This is for me please?
 
No, sorry
unrelated
I think I've seen => called the forward direction and <= the backward direction
A => B means A is sufficient for B
and A <= B means A is necessary for B
Like, "water is necessary for life" is the same as "if there's life there's water"
 
@AkivaWeinberger. Thanks. Can you please reword and A <= B means A is necessary for B?
These two statements confuses me a little bit
 
"A <= B" means "A is necessary for B"
If B implies A, then you can't have B without A, so A is necessary
 
4:12 PM
"if there's life there's water" means water is necessary here
 
Yeah so "water <= life" means water is necessary
 
we have to prove both directions that necessity holds and sufficiency holds as well please?
is sufficiency stronger than necessity or we don't measure it in that way please?
 
Neither is stronger
If you have to prove "if and only if" you have to prove both directions
 
@AkivaWeinberger. Thank you. Got it!!
I will just ask one more question for anyone who have time. This is a puzzle, so suppose we have a punch of pennies and we entered a maze, we would like to find our way out of a maze with large supplies of pennies, what would be your approach please!
You have large supplies of pennies (given from problem)
I used pennies as an indicator we have been through a hallway, so if I travel, I leave a penny. Once I come back to the hallway and found that a penny is dropped there, then I don't enter it
The solution I found though suggests that we never travel down a hallway with two pennies? Any thought please?
 
4:28 PM
Can you pick the pennies back up?
I assume yeah
 
@Avra I use 'if' for the reverse direction and 'only if' for the forward direction :-).
 
Nope not given
In the statment
@AkivaWeinberger. What given is a large supply of pennies and nothing given about taking back some pennies. The solution I found suggested you never go down a hallway with 2 pennies! I have no idea why
Why not just dropping one and never go down a hallway...etc please?
@copper.hat. Thanks. Why 'if' for the reverse direction and 'only if' for the forward direction please?
 
@Avra: What does "IF P, then Q" mean?
 
@TedShifrin. $p$ implies $q$ meaning that the statement is true as long as both go along with truth table of the statement, T-T=T, T-F=F, F-F=T, F-T=T
I really understand it from truth table
That is how I understand the imply statement
 
4:44 PM
OK, but that's $P\implies Q$, right? So that's telling you if $P$, then $Q$. If you say "$Q$ if $P$" then that's this statement. The "only if" is the converse.
 
@TedShifrin. Thanks. Q if P is sufficent condition then please
If P then Q is necessary as Q might be false and in that case the whole statement will be false, so that means this is necessary statement
Q is necessary here for P in If P then Q
 
right, because once $P$ happens, $Q$ MUST happen.
 
@TedShifrin. Thanks. And the other direction Q if P please?
How you read it similar to how you did with If P then Q please?
Q must happen above though has nothing to do with Q being false or true I guess?
 
It's just the converse. If Q, then P is saying that Q is sufficient for P, because whenever Q happens, P must happen.
 
@TedShifrin. I got it! Thanks...
 
4:53 PM
@Avra Read it as $P$ if $Q$ which means that when $Q$ is true then $P$ is true.
 
@copper.hat. You mean the backward <= statement is better read as P if Q
 
@TedShifrin Question I thought of, that I don't know the answer to
Let $F_2$ be the free group on $\{a,b\}$, and $F_2^+$ be the subset where all the exponents are positive
For a set $S$, let $S^2=\{g^2:g\in S\}$
Is $\langle (F_2^+)^2\rangle$ (the group generated by $(F_2^+)^2$ the same as $\langle (F_2)^2\rangle$?
For example, can $(a^{-1}b)^2$ be written as a product of squares, each with all-positive or all-negative exponents?
 
DogAteMy, you know I'm not the right person for this :P
 
if i had to bet, i'd bet no. but i don't know much about the theory of free groups and have a serious gambling problem
 
5:08 PM
@BalarkaSen Thoughts? ^
 
Bizarre question
 
Heya @Balarka
 
Hi @Ted
and @leslie, @Akiva, @copper
 
ted, california seems to have dodged a bullet.
good morning or evening, balarka
 
Well, I wouldn't phrase it that way. It was a decisive trouncing.
 
5:12 PM
at least it was cheap to administrate.
 
Hi Balarka!
 
The idiot Trompies should have to pay for all that.
 
it shows a dangerous crack in the system that gives huge leverage to those with a few spare million. and there seems to be many of those.
 
How can $(a^{-1}b)^2$ be written with only positive squares? That's ridiculous.
@copper That was something like the fourth time they'd tried to recall him. The idiot court gave them 4 extra months to get signatures cuz COVID.
 
not a real fan of gavin. but let him do the job he was elected for. (not that there were any alternatives that were acceptable to me at the time.)
i did like arnie, but think davis should have been left to finish term. many of his issues were much bigger than him.
 
5:15 PM
0
Q: Can all squares in a free group be made from squares in the free monoid?

Akiva WeinbergerHere's a question I thought of, that I don't know the answer to. Let $F_2$ be the free group on $\{a,b\}$, and $F_2^+$ be the subset where all the exponents are positive. For a set $S$, let $S^2=\{g^2:g\in S\}$. Is $\langle (F_2^+)^2\rangle$ (the group generated by $(F_2^+)^2$ the same as $\langl...

@TedShifrin You can invert the positive squares
 
they began trying to recall newsom like two weeks after he was elected. it was never about him.
 
so $(b^2)^{-1}$ can happen
 
when did the republicans become snowflakes?
 
Oh, OK, so one level of difficulty removed.
DogAteMy, you might page Thor and ask him. He's our most algebraic person these days.
 
@Thorgott
I have a question
 
5:20 PM
@leslie How's the lame munchkin doing? Maneuvering like a pro?
 
If you can write $(a^{-1} b)^2$ as a product of squares you can write $[a, b]$ as a product of squares as well.
That shouldn't be possible by topology, but I have to run now
 
I wondered if Balarka would have a topology proof. Seems reasonable.
 
@leslietownes does she have crutches?
 
is this a functional $f_{.3}(x) \mapsto \int_0^1 f_{.3}(x)~dx $
 
the crutches we had growing up (forearm) were much different than the ww2 version i had here (armpit), esp for kids.
What does th $.3$ mean?
 
5:24 PM
$t=.3$
for parameter $t$
like a pdf but NO because no normalising constant
 
no crutches, but she can crawl around pretty well.
 
the map $g \to \int_0^1 g dx$ defines a functional on whatever space $g$ is in. the $.3$ is just noise.
 
Linear functional, indeed. Sometimes the language is more general.
 
okay
linear functional
 
she hasn't complained of any pain since getting the cast, which is great, although it means she has more energy to devote to bad behavior
 
5:26 PM
functional usually means scalar values, as Ted pointed out your's is liear.
 
@copper.hat should it be $\mapsto$ ?
 
Calculus of variations deals with all sorts of very nonlinear functionals, for example.
Yes, $\mapsto$ is correct.
 
@geocalc33 I like $\mapsto$ but there are a few conventions.
 
okay
 
I never use it, personally. I use $\rightsquigarrow$ :P
 
5:28 PM
i think ambiguity in function notation/definition and same with derivatives leads to much downstream educational confusion.
 
Not to mention equalities where functions magically turn into derivatives on the next line.
I almost never used the $\mapsto$ or $\rightsquigarrow$ notation in teaching.
 
my grannie had parkinsons and i loved her letters to me and tried to imitate the wiggles. when i se $\rightsquigarrow$ it reminds me of that.
 
I went to lecture and didn't understand a single thing after 30 min for information bombardment.
 
so $g \mapsto \int_0^1 g ~dx$ defines a linear functional. And I want to make sure I understand - say $g$ is a distribution (not a probability distribution). When one takes an integral transform of the distribution function to get $G(s)$ what is really going on here? The domain is converted from say $x$ to $s,$ from what I understand
 
Number theory combined with computer science gives heart attack.
 
5:35 PM
a distribution is a functional on a particular sort of test space
 
when I say its a distribution I mean in the sense that $\lim_{s \to \infty} G(s)=0$ and $\lim_{s \to 0} G(s)=1$
 
you need to clarify what you are talking about
the word distribution means different things to different people.
 
Indeed. At least three meanings for me. Generalized functions, probability, and subbundles of the tangent bundle.
 
I mean that $g_s(x)$ is a probability distribution but without the normaliseign constant thing
but it obeys the limits^^^
so I can't say its a probability distribution really
 
It's a nonnegative $L^1$ function.
 
5:38 PM
is it monotonic?
 
I know that the functional is monotonic because the area increases smoothly as the parameter $s$ decreases
and the area decreases smoothly as the parameter $s$ increases
I want to try to see what a functional with those properties could be applied to
so in probability distribution you integrate over the support of the fun action to get $1$
but I just wanna integrate over the support and get a scalar $k\in (0,1)$
and then the limits dictate the upper and lower values the functional takes
so the functional takes on 1 at s=0 and the functional takes on 0 at s=infinity
 
6:06 PM
$g_s(x) \mapsto \int_0^1 g_s(x) ~dx = G(s)$
I think I need to understand where this breaks down notationally
start with the distribution function $g$ with parameter $s$ and real var. $x.$ Then this looks like an integral transform mapping from the variables $x$ to $s$ to transform $g_s(x)$ from a function of $x$ to a function $G(s).$
if $x \in (0,1)$ $g \in (0,1)$ and $s \in (0, \infty)$ - then $G(s)$ should be defined for $s \in (0,\infty)$
okay that makes more sense. So I think I should just write $\int_0^1 g_s(x) ~dx = G(s)$ which is an integral transform with the kernel being $g_s(x)$
which you can separate algebraically
$\int_0^1 g(s)g(x) ~dx = G(s)$
 
6:22 PM
say what
 
which letting $g(s)=e^s$ then it's basically a Fourier transform of $g(x)$
 
until you said "which you can separate algebraically," it looked like g was either a two variable function, or a set of one-variable functions indexed by a parameter (there's no formal distinction between these two, but notationally you might write $g(s,x)$ for one and $g_s(x)$ for the other)
now g is a function of one variable? what is the background here?
 
oh okay
yeah I guess I am conflating the two things
@leslietownes that makes sense to me now ty
@leslietownes what do you mean the background?
the background meaning the motivation ?
 
what are you assuming about g, what do you want to know, is there context that would flesh out the story
yeah
 
well I wanted to convert each 1-var. function indexed by a parameter $s$
and then somehow capture the space of the converted family using a function
or a family of functions
and then OTOH I guess I also wanted to look at converting the two var. function $g(s,x)$ from the x-domain to the s-domain
now I realize these are entirely different things
bothers me a little that I can't express a mapping of all the 1-var. functions (with parameter s) at the same time... because stitching each of the 1 var. functions together using $s$ as the parameter is something of a an analytic foliation of some space, and then I could use some intro manifold theory to literally transform a real manifold from one space to another - like a more general version of integral transform
I really doubt that the integral transform of each 1-var. function (initially in x-space) with parameter s can be collected together in s-space an result in a foliation of s-space
I don't know - I give up
81
A: Does a Fourier transformation on a (pseudo-)Riemannian manifold make sense?

Willie WongThere are three main ways of interpreting the Fourier transform. Decomposition relative to eigenfunctions of the Laplacian On $\mathbb{R}^n$, the plane waves $E_\xi(x) = \exp( i \xi\cdot x)$ can be interpreted as generalized eigenfunctions of the Laplacian. That is, let $$ \triangle = \sum_{i=1}^...

 
6:44 PM
Hey, the average of a holomorphic or harmonic function on a perfect circle is the value at the circle's center
Does that work for squares instead of circles?
 
7:34 PM
Let $\Omega \subseteq \Bbb{C}$ be a domain. If each $f_n : \Omega \to \Bbb{C}$ is holomorphic and $f_n \to f : \Omega \to \Bbb{C}$ uniformly on compact subsets of $\Omega$, does it follow that $f$ is continuous? If it does, I know Morera's theorem will imply $f$ is in fact holomorphic also. But I'm just wondering about continuity.
 
yeah, because continuity is a local property. given any p, you can find a little closed ball contained in Omega, with center p, on which f_n converges to f uniformly
and a uniform limit of continuous functions something something
 
Something indeed.
Surely the answer to Akiva's question is NO, because the usual proof certainly uses circles. But I need to think about it.
 
7:52 PM
@TedShifrin Wolfram Alpha suggests the average for a square around the origin for $z^n$, $n=1,2,3,4$ is $0$, but it fails for $n=5$
 
So, you're talking about an integral $ds$, which in the circle case turns into $dz/z$, of course.
So one approach is to convert $ds$ into $dz$ integrals in the case of a square. But I just want to verify that this is what you intend.
 
Is $ds=|dz|$?
I think so, so you get an average value
 
Yes, $ds = |dz|$.
When $n$ is odd, you get $0$ automatically.
But by symmetry, you also get $0$ when $n$ is even. Hmm.
 
Oh. So I must have mistyped somehow.
It probably fails for $n=4$
 
Nah, I don't think so.
This is bothersome.
Oh, duh. No, my symmetry for $n$ even is wrong. $ds$ doesn't care which direction we go on the line segment, so things double instead of canceling.
Right, it should fail for $n$ even.
 
8:30 PM
howdy @robjohn ... I assume you are much relieved today.
 
@TedShifrin Not for $n=2\mod4$ I think
'cause then $(ix)^n=-x^n$ and so the top and bottom sides cancel the left and right sides
 
8:46 PM
OK, that seems right.
 
@AkivaWeinberger hmm, my intuition says no, but I don't have an argument
 
"I've come for an argument." "No, you haven't." ...
 
This isn't an argument! It's just a contradiction!
 
You're no fun — ruining Monty Python.
 
I thought I was quoting Monty Python. :(
 
9:09 PM
Oh, I had forgotten that one. My apologies.
This is what happens when I haven't watched the movie in 95 years.
 
@BalarkaSen My StackExchange post (on $\langle(F_2^+)^2\rangle$) got an answer. Would you like to hear what it is, or to keep on thinking about it?
 
Oh, already?
 
Yup
@Thorgott
 
That's pretty crazy. But I never have had any intuition for this sort of symbolic stuff.
 
Was it Monty Python who had the exchange - "Who are you and how did you get in here!?" "I'm a locksmith, and I'm a locksmith." -?
 
9:24 PM
Never heard of it. Google says Police Squad.
 
9:35 PM
Police Squad was Leslie Nielsen in magnificent form
The Naked Gun movies were nominally sequels to it
But Nielsen was the only character played by the same actor as in the TV series
 
I don't think I watched none of those.
 
Oh, there’s actually two other actors who appeared in both. But only Nielsen in a lead role
One of the things which made the film age poorly, unfortunately, is OJ Simpson in a supporting role
 
9:54 PM
who?
just kidding
 
The munchkin keeping you hopping?
 
she's refusing to nap. she demanded to have her nap on the couch in my office. instead of sleeping she sat there asking me what i was doing over and over, and kept phbthtbtbht'ing at the cat when the cat tried to share the couch with her.
she's playing by herself with some legos on the floor of the office now. who knows how long this will last.
 
She's bored.
 
day care said they would be willing to try having her back for half days next week, but they don't really have staff to move her around.
 
Yeah, that seems beyond their pay grade.
 
10:39 PM
@AkivaWeinberger I haven't really spent any thought on it if I'm honest
Looked up answer. Surprising
 
@TedShifrin I am relieved that Elder is not governor.
 
Precisely.
 
Sorry, I've been out of it for a while. Not feeling well (separate from the election)
 
10:58 PM
What's the intuition for the dual of a vector space?
 
@Derivative It is like a vector space, only it likes to fight.
:P
 
@robjohn i’m sorry. I hope you’ll be better soon. Did they find out what was wrong with your pup?
smacks Xander
 
In the particularly nice case of square integrable functions (e.g. $L^2(\mathbb{R})$), the space is self-dual. This means that every element of the dual space is in correspondence with some element of the original space. The "usual" way in which this is represented is via integration: if $u \in (L^2)^*$, then there exists some $g \in L^2$ such that $(f,u) = \int f \,\mathrm{d}g$.
 
Columns versus rows, Derivative
You need to say context.
 
derivative, i don't know that there is one single intuition for it. depends on the application. something you meet early on is that a solution to a system of linear equations is the intersection of level sets of functions of the form f(x_1, x_2, ..., x_n) = a_1 x_1 + ... + a_n x_n. those functions happen to be linear functionals.
 
11:10 PM
right, so basically we haven't uncovered any deep facts about the nature of god's mind and the universe, we've just changed notation?
 
So a dual object is, essentially, a measure (indeed, my recollection is that this is how Dieudonne presents things).
But, as others have suggested, context matters.
 
I'm just a little scared because the professor dumped the definition on us today and I don't see what's the point of it
 
@Derivative In what context are you being presented with the definition?
 
if all you're given is the definition, it may be healthy to delay the deep questions until later.
 
linear algebra/ class. We're talking mostly about finite dimensional vector spaces over the reals or the complex numbers
 
11:12 PM
Oh, shoot. My example is useless. :P
@TedShifrin's "columns vs rows" comment is much more relevant. :D
 
okay, so by the syllabus we're in week 4 and in week 6 we get to isomorphisms and eigenthings, so hopefully things will start to wrap up. I think I'm just going to calm down
 
In any event, I'm exhausted. Imma go home now.
 
sounds like a good idea
goodnight
 
"Night" is four hours away (more or less). But it is time for the work day to be over. Later.
 
the dual of a vector space consists of the linear functionals on it. you can think of these as things you use to "probe" elements of the vector space. e.g. an element of a vector space is zero iff all functionals evaluate to zero on it.
more precisely, the dual space induces a certain geometric duality (a special case of so-called Grassmann duality) in that lines, i.e. 1-dimensional subspaces, in the vector space get naturally identified with hyperplanes, i.e. codimension 1 subspaces, in its dual.
most of this isn't immediately useful, but they're useful things to keep in mind and understand over time
 
11:19 PM
Always a good idea to calm down @Derivative
You're not expected to understand the mind of god in undergrad :P
 
Dot product with a vector is an element of the dual, Derivative. You can think of this as a linear map whose matrix is the transpose of the vector.
 
right, and you get a basis of the dual for free if you have a basis of the original vector space, which means they look like each other, which is why it's called the dual. Duh. It sounds trivial when you put it like that. Thanks, regardless of how difficult or easy it might be
 
Well, dual because it acts on the original. I never covered this in regular linear algebra but did in my multivariable math course, because these ideas build differential forms.
 
what book do you use for that? I'm reading Axler on the side for the linear algebra class and Vector Calculus by Susan Jane Colley for multivariable calculus, but that last one was probably a bad choice because it doesn't prove a lot of things
 
11:35 PM
My own :) The YouTube lectures are free, though ;)
 
nice
 
When Colley was a graduate student, she took my grad course on manifolds and geometry.
 
wow what
how do you know all the math authors?
 
Not all.
 
but you did write the exercises for Spivak
 
11:39 PM
Some, yes.
He taught me manifolds and geometry in grad school :)
 
ted is everywhere.
 
Nowhere dense.
Oh wait …
 

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