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12:00 AM
Only meant to send that once, lol
Internet connection went spotty
If you wanted, you could write out the coordinates. Say, u=i in both cases, and v=i+0.2j (roughly) in the first case, and u=i and v=-i+0.2j in the second case
(i) x (i+0.2j) = ixi + 0.2ixj
(i) x (-i+0.2j) = -ixi + 0.2ixj
but i x i = 0 so the minus sign changes nothing
My phone died earlier, but to follow up
rip phone
@Xnero If you use the dot product, you get no such restriction
I don’t mind |a| as magnitude, if one is consistent about it
The cross product can't tell between theta and 180-theta, but the dot product can.
hm hold on
gif credit: Freya Holmér
But if you’re using double-bars for magnitude, then you’d better stick with it
12:10 AM
@AkivaWeinberger Because the dot product uses cosine and as theta is between 0 and 180, for each value of theta there is one unique cosine value?
Yes, precisely
In the image above, one of the vectors is j, and the other vector oscillates around the unit circle (let's call it ai+bj for some scalars a and b)
What's j.(ai+bj)?
@AkivaWeinberger b?
So the result is, in the GIF, the dot product is precisely the y coordinate of the second vector.
Looking at the numbers around the circle, this should make sense.
@AkivaWeinberger Yes. As the magnitude of i and j stay the same, the outside is basically cosine the angle between the two vectors?
It is the cosine of the angle, and it is also the y value of the second vector (these are equal - you can focus on the right triangle in the gif to check this)
well, it's easier if it stops moving maybe. Cosine is adjacent over hypoteneuse
12:17 AM
@AkivaWeinberger In this case, cosine is just adjacent as it is / 1?
Thanks for the help @AkivaWeinberger
Unfortunately, I don't think the textbook explains things very well.
That's a shame.
Have you done triple products?
For example, it explains about the cross product (formula and some pictures), then gives an example of finding a unit vector perpendicular to vector a and b. It never actually stated that the cross product of a and b is a vector perpendicular to a and b.
12:22 AM
i like the cross product definition via the innner product.
In the end I got that a x b = |a| |b| sin(theta) ncap and as |a| |b| sin(theta) is not a vector but a scalar, a x b = a scalar x ncap so a x b is a vector perpendicular to a and b multiplied by a scalar.
It should have spelled that out in words somewhere.
Is my understanding correct?
Yeah, a x b is a vector perpendicular to a and b
This means, by the way, that a.(a x b) = b.(a x b) = 0
@AkivaWeinberger That's the next section after using the cross product and vector properties to find the areas of shapes.
12:26 AM
(type check: axb is a vector, a is a vector, so a.(axb) makes sense and is a scalar)
Incidentally, the expression a.(bxc) is called the "triple product". It's very closely related to a construction called the "determinant", which you may learn about later
a.(axb) will always be 0?
@AkivaWeinberger I already know the determinant.
Yes. Because a and axb are perpendicular.
a.(bxc) is the determinant of the 3x3 matrix whose columns are a, b, and c
It also equals the volume of the parallelepiped determined by a, b, and c:
If you trust that that's true: a.(axb) is the determinant of a matrix with columns a, a, and b. Any matrix with repeated columns has zero determinant
One definition is $\langle x , a \times b \rangle = \det [ a \ b \ x ]$.
I like the determinant as that helps me to quickly figure out what i x j, i x k etc. are mentally.
perpendicularity is immediate from this. this is essentially the usual defn.
12:31 AM
Yeah - one way to think about the cross product is "b x c is defined to be the vector that makes the equation a.(bxc) = det(a,b,c) true"
this came up this morning in another question.
weird how that happens.
you go weeks without thinking about it, then lightning strikes in the same place twice.
sounds of twilight zone
Also, a weird way to think about the dot product a.b is "you write out b in terms of i, j, and k, and then replace i, j, and k with the three coordinates of a respectively"
it is exactly that definition, but formally.
Combine the inverse of that second thing with the first thing, and you end up with the usual way of computing the cross product
(the determinant of a matrix whose top row is [i,j,k])
@AkivaWeinberger Definitely. This book contains a lot of complicated chapters so I hope it's clearer in other chapters.
Fun fact: this is unknotted^
(meaning you can turn it into a circle without cutting it)
@Xnero At some point (maybe not now) you might want want to think about areas of arbitrary shapes
The area of a triangle (in the ij plane) with corners q1, q2, and 0 is 1/2(q1xq2) (times k)
@leslietownes That looks much harder than my question.
Ignore the "times k" for now. It turns out that the area of a polygon with corners q1,q2,q3,...,qn is 1/2(q1xq2+q2xq3+...+qnxq1)
Geometrically, it's the sum of a bunch of triangle areas.
@AkivaWeinberger Makes sense.
12:42 AM
In the picture on the left, you can see that the polygon is broken up into a bunch of triangles
and p (acting as the origin) is inside the polygon
The weird thing is, it works even when the origin is outside the polygon
(as shown in the second polygon)
because you end up with a bunch of negative area triangles canceling out a lot of the area
(recall that order is very important for cross products: axb=-bxa)
What is qi? A point? Why is it not qi -p?
qi is a point. The ∑ means "sum over all values of i". The neat thing is, it has the same value with and without the p!
If you expand it out, the ps all end up canceling
This means that the choice of p does not matter
@AkivaWeinberger I can't see that.
You have to use the fact that it's summed over all i
(q_i-p)x(q_(i+1)-p) = q_ixq_(i+1) - q_ixp - pxq_(i+1) + pxp
the pxp is zero, and -q_ixp is pxq_i
so, = q_ixq_(i+1) + pxq_i - pxq_(i+1)
Now ignore the q_i x q_(i+1) term. Think of what happens when we sum
p x q_i - p x q_(i+1)
over all the i.
That'll be (pxq1-pxq2)+(pxq2-pxq3)+...+(pxqn-pxq1)
Each term will be canceled out by one of the terms next to it
@AkivaWeinberger Why is p x p 0?
P is 0?
12:49 AM
Anything crossed by itself is zero
This is one of the very important properties of the cross product. pxp=0, and pxq=-qxp
Oh. It's cross product, sorry.
2 hours later…
2:23 AM
Hello....(Lionel Ritchie voice)
Howdy, dc3rd
Re hola
I never got the baguette + martini delivery.
I am stuck on a few of your questions sir....well stuck mid way through two and for the third it is just the idea of it all....
after i took it out of the toaster oven i realized it was going to cool down long before it got to san diego. i may have eaten it. and the martinis, i dunno, most of the afternoon is a blur.
2:29 AM
You’re getting a negative tip for service, leslie.
Well, for a fee, you may continue, dc3rd.
they're gonna ban me from the app if i don't get my ratings up. stupid doordash.
Maybe you need to do QAnon instead, leslie.
i honestly don't know how people did it in the mad men days. i had a two martini meeting with one of my attorney friends once and we were both nearly catatonic afterwards. tried again at a whisky bar, same result. we stopped meeting like that after that.
It’s all for incompetence and tax breaks. Typical US.
she did write off the whisky bar for some reason as a business expense. her firm gave her a budget for exactly that.
2:33 AM
The first question revloves around this one from the text:

Let $T: \mathbb{R}^n \to \mathbb{R}$ be a linear map. Prove that there is a vector $\mathbf{a} \in \mathbb{R}^n$ so that $T(\mathbf{x}) = \mathbf{a} \cdot \mathbf{x} $

Usually for an existence question I have to show the existence of the object. In this case that object is $\mathbf{a}$, which means I have to construct it. Or is it enough to show that a linear map in this case defined as $T(\mathbf{x}) = \mathbf{a} \cdot \mathbf{x}$ satisfies the "usual conditions of a linear map? i.e:
Yeah, omicron is scary.
i think ya wanna construct it. i think it's assumed that T is linear, and as you note it's easy to show that maps of that form are linear. the point is to show that T has that form.
Nah, we know that that map is linear. The question is why every one must be of that form.
I’m too slow on my ipad.
it's stuff like this that gets mathematicians accused of plagiarism. what a coincidence that we would both write almost the same thing.
I saw that wrong "right".......
2:35 AM
Yeah, right.
i get phonetic after my martinis.
this is all speech to text, by the way. i'm on the floor right now.
Hint @dc3rd: What’s the matrix of $T$?
this is a one-liner if you play your cards right, dc3rd. this goes against my usual policy of acting as though space on paper or on a screen is free and unbounded.
Damn, have to concur again.
@leslie And you owe me a Pekin duck.
well using the whole standard basis vectors idea. Seems the matrix of $T$ is just $a_i$
2:41 AM
You mean $j$. Which is what in vector notation?
hmmm.... I want to say $a^t$ but I'm not completely clear why I'm saying it besides a vague idea of the dot product equivalence
The matrix of $T$ is $1\times n$, hence is the transpose of some vector. Done.
Again, it’s an issue of logic.
@dc3rd I think this is basically a corollary of Riesz representation theorem.
Ok...this is actually good because this idea kinda leads into a question I had about us finding $\|A\|$ when $A$ is a matrix of rank $1$. I did the exercise and treated it the way you did in the lecture. namely for example
$$ \begin{bmatrix}
1 \\
[1\ 1]$$
it's a corollary of T(x,y,z) = T(x e_1 + y e_2 + z e_3) = x T(e_1) + y T(e_2) + z T(e_3) = (T(e_1),T(e_2),T(e_3)) dot (x,y,z)
2:51 AM
Shut up, @Koro, seriously .
definition of vector addition, linearity, definition of dot product
whoo, matrix norms! i'm lovin it
I’m shut up 🤫. But I know I’m right.
You’re being an ass.
So you went from thinking of a "vector" to thinking of a "matrix" when we took it as the transpose.
Right, but vice-versa, dc3rd.
2:53 AM
Riesz Theorem is given much attention in Axler so that's where Koro is getting it from
I know.
I read that recently so I’m very tempted to say something when I see a related question professor Ted.
poor riesz must be spinning in his grave. he did other stuff too, you know.
so this "decomposition" idea....I clearly get the idea of it, but it doesn't "commute" to our set of linear algebra operations....
bad enough that he had a brother with the same last name and nobody can keep them straight.
2:54 AM
@Koro It’s abstract nonsense when we’re being very concrete. We are not sophisticated here.
What decomposition idea?
$$\begin{bmatrix} 1 \\ 1 \end{bmatrix} [1\ 1] = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$
We’re on something else now.
A rank $1$ matrix, perhaps?
Yes....in the original idea the rank $1$ matrices and finding the norm of them, so $\|A\|$
I'm just trying to reconcile the idea of it. Like you mentioned we've never looked at matrices like that before
I don’t remember a general problem, but show every rank 1 matrix is $vw^\top$.
Hmm...I don't think I've seen a problem like that in the text. But it seems to be just what I'm looking for in terms of the idea.
2:59 AM
Just think what rank 1 means.
rank $1$ being only one linearly independent row.
3:18 AM
So that row becomes what in my formula?
i think of rank in terms of subspaces
the row becomes $w^t$
my daughter learned about hanukkah today. seems age-appropriate, even if it is not a big holiday. yom kippur can wait until she is older.
i asked if she saw a menorah or a picture of one, and she said "a menorah is a thing you put candles in." she also identified a dreidel as "something that you spin." she also said a dreidel is blue, which i guess is a good color for one. A plus.
google has a spinnable dreidel that comes up if you google it.
ah...then the matrix will be a product of the entries of $w^t$ by $v$.....something like that.......visibly I get it, but just grounding it in the "logic"
are you going to sway her towards becoming a mathematician or will she be a statisticain like the wife?
one of our kid's teachers had them make dreidels. she was one of my favourite teachers.
3:30 AM
or nowhere near academia and she is just going to be an "adventurous soul"
my wife is many things but she is not a statistician. she has a degree in statistics and knows her way around excel. we will keep her away from the dark arts.
we're hoping some kind of influencer career. whatever the next tiktok is.
just had a 2.5 in kensington
felt like someone kicked the house
Oh oh … I hope my youth house is ok.
Bike up there and check on it
If $A$ is an $m \times n$ matrix. Then $\|A\| \leq \sqrt{\sum_{i,j}a_{i,j}^2} \leq \sqrt{n} \|A\|$

so $Ax$ is going to be an $m \times 1$ vector of which we take the norm in this case it will be the unit vector that gives maximum norm. If we said $m = 2$ it will be $\sqrt{a_1^2 + a_2^2}$. Now assuming I understood the notation correctly $\sqrt{\sum_{i,j}a_{i,j}^2} = \sqrt{a_{1,1}^2 + a_{1,2}^2 + a_{2,1}^2 + a_{2,2}^2 }$. That part of the inequality makes sense.

But the second bit is throwing me off because if I take the same $\|A\|$ I still have the same $m \times 1$ vector I'm taking the
another small one a moment ago.
i might go for a ride tomorrow.
Is $\|\cdot\|$ the induced norm?
3:42 AM
$\|A\| = \max_{\|x\| = 1} \|Ax\|$
is that what induced norm means copper?
and the vector norm is Euclidean presumably
yes to induced
oh we going with the gov't here now?
as in gov't name.....that's how us young folks truncate things
@copper go to the top of Willamette and check on the north side
yes Euclidean norm...nothing fancy here at least not stated as such, but I'm sure Ted just used the easy names and later on we'll discover the norms have different names
@TedShifrin up behind the police department...
3:44 AM
@dc3rd You’re ignoring the columns here?
No, way up. At the corner of Purdue.
@dc3rd re this. you may have moved on but if the m rows of an mxn matrix A are all scalar multiples of the same single 1xn row vector w, say the rows are v_1 w, v_2 w, ..., v_m w, you should convince yourself that A is the product of the mx1 column vector v = (v_1, ..., v_m) with the 1xn vector w. i.e. A = v w^T if v and w are thought of as column vectors. it's just a degenerate case of matrix multiplication.
I'm trying not to....but I was just following the defn of the norm of $A$.
ted: the new owners need to maintain that yard a little better.
No, dc3rd, I don’t see that. Hint for the problem : you will need to use both definitions of $Ax$.
@leslietownes That is the thorough way to explain what I was thinking. Thanks.
both definitions you say?
hmmm....... going to eat and will attack it after that
4:01 AM
Here's something random i came up with after watching a Youtube video + comments
First, a fact: 2463199200/784060657 is a very good approximation to pi
(it's so good that it's what Casio calculators store it as internally, which is where this comes from)
To assess the quality of that, we can submit 2463199200/784060657/pi to WolframAlpha to get 1.0000000000000054...
to extract the deviation from 1, we subtract 1 from that. WA has no issues if you submit that and you get 5.40e-15
However: If you type that in without submitting, WA recognizes it's just an arithmetic calculation and shows you a "preview" of the answer
huh. why don't they just store the list of 15 or 16 digits internally?
which would be all well and good, except that the preview it gives you is 5.55e-15 :)
so somehow that input breaks its arithmetic preview
to see it for yourself, start here: link
wolfram alpha feels so hackish in some ways. i imagine it as having like twenty million hard-coded special cases for certain inputs.
i would imagine that casio stores pi as a fp number
always felt wa was hackish
my slide rule has pi to an infinite number of places
the conjecture is that Casio has that representation in storage like that b/c they know their main audience is high school students
so being able to recognize when your answer is some rational multiple of pi, is rather convenient
since 2463199200 is a multiple of 25200=3600*7, it's got the main factors you'd expect to ever see showing up as denominators for pi
that's the conjecture anyways
anyways, to see it in WA: go to what i linked, put -1 at the end of the input, and wait a moment
slide rule rules
especially when your mental arithmetic is as piss poor as mine
i'm imagining the litigation when someone uses casio's pi to compute the size of a fuel tank in a design and then the fuel runs out 0.000000000001 seconds too soon
9 hours ago, by leslie townes
every math book i have ever seen includes it in the definition. i think for some physics folks the quadratic form comes first and so they come at it differently.
in terms of history i do wonder where that divergence occurs
like, physicists had von Neumann laying down the axioms for QM
i had copies of a bunch of von neumann stuff and schroedinger stuff, i think at least some of it was quadratic formish
4:14 AM
all lost by the neerdowells at the USPS
it's a matter of how bra-ket notation works, up to a point
whats the big deal of | over , ffs
you start with kets as vectors in Hilbert space, define bra's as dual vectors
bras already have a handful of meanings
sorry, i couldn't help it.
4:15 AM
at which point <f|g> being antilinear in f is inevitable
yes, the way it should be
you gotta admit that it looks sillier to write ,a> or <b, all by themselves than it does to use |
only if you go all obsesquilinear over it
i feel like the main reason we use bra-ket notation tbh is so we can put longer stuff inside the ket
ohh, the possibilities are opening up
4:17 AM
e.g. writing the Schrodinger cat state as |alive> +|dead>
poor cat. if ever there was such a thing
or writing spin states as |Sx=up> which is notation I do like
looks like a cat sat on the kb
the other reason i think the physicist convention makes more sense to me
if you think of your vectors as complex column vectors, then the way you get duals is to take the conjugate transpose
in which case they act from the left
so once you've signed on to having vectors be columns by default, then having the first vector in the inner product be the antilinear one just sorta fits
which, in turn, is consistent with the default of acting on vectors from the left rather than the right
in that sense the typical math convention is sorta weird to me
books specific to operator algebras waver back and forth between the 'math' convention and the 'physics' convention. i took a few pure math classes where we used the wrong way.
this is why i quit math, i couldn't take it anymore.
4:23 AM
lol, conventions being "wrong"
except the convention that electrons have negative charges, that one is genuinely awful. curse you Ben Franklin
i'm ANTI-"linear in the second argument." you get me?
i'm pro linear in the correct second parameter. lawyers always call them arguments
life would be so much easier if the electron was positive, but Franklin made a guess one way and guessed wrong
i made a lot of sign errors in my E&M class. i think i was using the left hand rule a lot of the time, but let's blame ben franklin, who was left handed.
4:25 AM
i used the middle finger orientation rule
if we used the opposite convention for charge then we'd just write the force as $\vec{F}=q\vec{B}\times \vec{v}$ instead
no need for a left-hand rule, just swap the order of the cross product
the right hand rule actually favors lefties because you can use it without dropping your pen or pencil.
screw that
you can also do the right-hand rule using your left hand
just a different assignment of fingers
4:27 AM
its so arbitrary
not really
it must be
it's just the fact that coordinate axes are ordered x-y-z
there's the arbitrariness of "positive rotations = CCW" i guess
which, come to think of it, you really can't do with your left hand
it's good to be a leftie.
i am unassigned
left, right, its all the same.
4:29 AM
but the left-hand version of the right-hand rule is just: pointer finger in direction of first vector, thumb in direction of second, palm in direction of cross product
or, translating for physickists: left| right| its all the same.
i was disappointed when my daughter turned out right handed. someone other than me is gonna have to teach her how to tie her shoes.
only difference for right-hand is swapping the thumb/pointer
i think we have two lefties on my side
and ultimately you always can think of it as "first vector is in the +x direction, second vector is in the xy plane with y>0 => output is in the +z direction"
right-hand rule is just a mnemonic for that at the end of the day
4:31 AM
i want to eliminate the m in mnemonic
along with bras
and kets
sorry, hit cr accidentally
the m is there to remind you of the value of antiquity
one thing i want to go back to learning about in terms of history is the history of Hilbert space methods in probability/statistics
they did make inroads eventually, but surprisingly late
the way i remember it is that you can't spell "remember" without two m's, so you can't spell mnemonic without two m's either.
the word that always gives me trouble with spelling is "separate" for some reason. i always want to write it as "seperate"
4:35 AM
semi i wonder if they use the methods or independently kind of do the same thing without realizing it's also this other thing.
probably because how i pronounce it is closer to the latter than the former
to remember that i have to mentally link up "separate" with "parity"
conditional expectoration has a nice interpretation in hilbert spaces
@copper.hat yeah, exactly
its parity time
i have in mind this paper specifically: economics.soton.ac.uk/staff/aldrich/yule%20gauss.pdf
4:38 AM
the dismal science
specifically section 10 (pp. 72-74)
basically statisticians found Hilbert space interesting b/c it gave them an orthogonalization procedure, not b/c of the geometry
that is geometry
yeah, but for them the procedure was more about separating signals into uncorrelated components
the fact that that was equivalent to being orthgonal was secondary, at least at first
there is something in that paper that makes me wonder, but i have no easy way to explain it
least squares as a minimization problem was simple enough to understand. least squares as being equivalent to a geometry problem was not. but once you do view it that way, the connection to minimization is easy enough (shortest distance between two points)
in particular, there's the following sentence in that article:
"Halmos’s influential Finite Dimensional Vector Spaces starts from the observation that the “seemingly separate” Hilbert space theory and elementary matrix theory are “intimately associated”, are actually different ways of doing the same thing. The implications of this situation for regression theory were only worked out on any scale in the 1960s, when the matrix formalism was dominant."
whereas in physics i dont' think Halmos/Stone show up in the history very much. we already had von Neumann founding QM axioms in the Hilbert space context
yeah, i think it's weird to attribute that to halmos, although he probably crystallized it well in something quotable in his textbook.
halmos was working with von neumann when he wrote that book of course.
things is bad when i switch between mse looking for convex psqs and youtube looking for shuffle dance videos
i returned my son to ucsc today
4:51 AM
have the doldrums returned?
@leslietownes yeah. the paper seems to point to Stone's work as being significant in this regard as well
so Halmos/Stone as vehicles for convincing stats to take it seriously
the doldrums are beating loud at the moment...
whereas physicists already had von Neumann
that's a pretty narrow part of the paper, though, so i'm not sure how exhaustive that part of it is
This limit is to be calculated: $\lim _{n\rightarrow \infty }\frac{\pi }{n}\left(\sin\frac{\pi }{n} +\sin\frac{2\pi }{n} +\cdots +\sin\frac{n-1}{2} \pi \right) $.
i feel bad for people who are knee deep in the mathematical side of stats, or more broadly the 'developing/refining methods' side of stats. so many users of their work just ignore whatever may have inspired the methods and run hacked together code cribbed from whatever other people or doing or what they see at a conference. and that's kind of the public face of it that leaks out into other disciplines.
4:55 AM
$\displaystyle =\lim \frac{\pi }{n}\frac{\sin( \pi /2)}{\sin\left(\frac{\pi }{2n}\right)}\sin( n-1)\frac{\pi }{2n} \sim \lim 2\ ( n-1) \pi /2n\sim \pi $
I think this is correct.
However, I have one confusion.
that seems plausible. you're basically sampling $\sin\theta$ at the points pi/n, 2pi/n, ...
koro, looks a bit like a riemann sum, no?
if you have your exact formula i guess there's that too. it isn't where i would have gone first.
4:57 AM
is that limit same as $\displaystyle =\pi \int _{0}^{1}\sin( \pi x) dx$?
i don't understand the last term
that last term confuses me too
all the rest have $n$ below, the last on top
$\sin(\frac {(n-1)\pi}n)$ is the last term. I made a typo earlier.
ahh, that makes more sense
4:58 AM
leslie, I thought so too but then I thought: shouldn't f be monotonic on $[1,\infty)$ to write the limit as integral?
can you use the geometric series here
no? riemann sums don't assume monotonicity

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