Mathematics - 2021-02-21 (page 1 of 3)

Mike Miller

12:01 AM

@user2103480 I'm sorry for being rude above. There was no reason to do that.

user2103480

12:16 AM

nah it's alright. I didn't see that BigSocks just asked "how many subobjects are there, up to isomorphism?"

which is, as you said, obviously a set

Mike Miller

Still, I could have just pointed to that. Thanks for your patience though.

user2103480

no worries, in exchange I'll just bug you again with some topology questions where I am getting confused in a flurry of definitions

Thorgott

got him

user2103480

tor, ext, hom and tensor feel like one just coin flipped whether some 5-ish properties hold for some functor

Thorgott

let me introduce you to the Grothendieck yoga

user2103480

12:26 AM

is Hom for abelian groups nice enough to turn arbitrary coproducts into products in the first coordinate and preserve arbitrary coproducts in the second coordinate

robjohn

@MadSpaces If it's a log scale, then $5$ divisions between $9$ and $10$ would be $9.000, 9.192, 9.387, 9.587, 9.791, 10.00$

. o O ( too late )

user2103480

and is there anything else that's useful for my purposes except $\mathrm{Hom}(\Bbb Z^n,G) \cong G^n$ and $\mathrm{Hom}(\Bbb Z_n, G) = $ n-torsion?

Mike Miller

@BigSocks OK, so I decoded what you are looking for.

You have the category of [some kind of] varieties, with morphisms the *rational* maps between them. (Maybe you are only considering surfaces? Unclear.) This actually is not a category: you cannot compose two rational maps in general (maybe the image of the first one lands where the second one is not defined). You have to restrict to *dominant rational maps*, those whose image is a dense subset of the codomain.

You want to understand whether there is a set of objects which you can test "equality of rational maps" with: for each pair of *…

Thorgott

@user2103480 that works in every category

user2103480

@Thorgott good

Mike Miller

12:33 AM

This is indeed not obvious. It is probably not true.

I can probably prove it's not. This all gets much easier to describe if you are very clear about what you are asking

Thorgott

though I guess there is a point to be made about internal hom vs external hom

works for both in this case

in any case, all these facts about Hom are just restatements of the respective universal properties

user2103480

Dude I will forget all of this after the exam, I'd rather take a numerics class than anything algebra

I find category stuff a la type theory nice enough, but man I couldn't care less about abelian groups and rings

Although I do find the whole machinery very impressive

Thorgott

bad take smh

user2103480

poincaré was on beast

BigSocks

@MikeMiller hmm well that’s cool- yeah maybe I have to think of dominant rational maps. I am not sure myself. Really I want to see what kind of “test-object” could help us rule out containing function fields of varieties of general type (other than the rational function field and maybe function fields of elliptic curves)

I think I might have to understand Faltings theorem so... that’ll be a bit

user2103480

12:48 AM

@Thorgott meet me solving ODEs in banach spaces

Thorgott

abstract linear algebra vs applied lienar algebra

user2103480

Theres so much semigroup theory that I have to learn properly ugh

https://math.stackexchange.com/questions/1697318/exponential-of-a-self-adjoint-operator
this lebesgue-stieltes integral against an operator-valued function is making my head hurt

Thorgott

semigroup theory is a meme

user2103480

I love it

Although I can't praise it too much since the physics prof makes it painful

He really just pretends it's all ODEs and the worst thing is that it works

Also, these kinds of identities

Lemma: For a pair of orthogonal projectors $P, Q$ with $P+Q=1$
the linear operator $R(t)=e^{-i \mathcal L t}$ fulfills
$$
\frac{d}{d t} R(t)=-i \mathcal{L} P R(t)-\int^{t} \mathcal{L} Q \widetilde{R}(t-s) Q \mathcal{L} P R(s) d s-i \mathcal{L} Q \tilde{R}(t)
$$

Thorgott

analogous to lienar algebra

user2103480

12:57 AM

where $\tilde{R}(t):=e^{-i Q \mathcal{L} Q t}$

Ted Shifrin

@robjohn This is what I said originally, but I think it's wrong. They're not spaced evenly. And when I used log-log paper in high school, the whole point was not to have to compute converted quantities.

user2103480

@Thorgott lienar algerba

Thorgott

lienar azumaya gerbe

BigSocks

$\mathfrak{lie}$nar

Astyx

1:12 AM

Am I missing the point if I claim affine varieties form a set of "test-objects" ?

BigSocks

Nah, I think you’re guessing pretty reasonably considering how ill-posed my question is

Also I realized I mucked it up a bit because elliptic curves are not varieties of general type I don’t think. But ignoring that implication what I said last would be the “nice thing” in mind, be it a “test-object” or whatever it really is

Astyx

I'm talking about Mike's translation of your question. I'm surprised because he seems to think it's probably not true

BigSocks

Oh ok. Yeah like I posted up there, it was kind of to rule out the weird nlab def of separator + depended on trusting what Zhen Lin was claiming in that one post, which I am reasonably comfortable with

Mike Miller

1:28 AM

@BigSocks If you are trying to make a category, yes, you need that. Rational maps cannot be composed in general.

@Astyx Seems good, I guess the point is there is a set of affine varieties because there is a set of multivariate polynomials over a fixed field.

I just got paranoid by thinking too much of the equivalence between dominant rational maps and maps between function fields.

BigSocks

Right Hartshorne section 4 or something

Astyx

I think that depends on how you define varieties

But I think in most cases you can get away with just dimension (hence you get a countable number of them up to iso, which forms a set?)

I'll think more about this tomorrow, gotta sleep, cya

robjohn

2:32 AM

. o O ( too late again! )

dc3rd

2:54 AM

Simple geometry question. I'm asked to determine the similar triangles. In the diagram it is clear which are the triangles, but I'm trying to solidify my reasoning.

So the criteria we're using to establish similarity is that all angles in each of the triangles are congruent to each other. I have the right angles, but how do I determine that the other angle in each triangle is congruent?....I know I'm missing something about the altitude segment, but can't put my finger on it

Ted Shifrin

3:23 AM

Just look at angles, nothing else.

Three triangles. All similar.

dc3rd

What angles?...Each of them has a right angle, fine. But the other two angles in each triangle what is the relation? I mean in the picture it is implied, but I'm trying to imagine if the picture wasn't so obvious.

Ted Shifrin

Do the calculation.

Call $\angle F = \alpha$. What are all the other angles?

dc3rd

Hmmm............very nice......I worked them out.

Not bad....not bad at all.............

I suppose the technique here is to designate an angle as $\alpha$ as you did here. Well to convince yourself at least.

robjohn

4:48 AM

@dc3rd the idea is to convince someone else.

dc3rd

Yes @robjohn, but if I can't sell myself on it, what chance do I have of selling the snake oil to somebody else?

robjohn

@dc3rd All the best snake oil salespeople convince others while staying immune to their own pitch.

robjohn

5:21 AM

@dc3rd the two acute angles in a right triangle are complementary (they sum to $\pi/2$).

thus, all it takes for two right triangles to be similar is that one of the other angles are the same.

dc3rd

@robjohn agreed. but my picture had no other "specified" angles. The suggestion from Ted did the trick in helping me see things.

Balarka Sen

6:49 AM

I have a function $f(x)$ on $(0, 1)$ such that $f'(x) \leq 1/x$, and $f(x_0) > c$.

Can I say something about the behavior of $f$ near $x_0$

Leaky Nun

8:12 AM

@BalarkaSen it doesn't tell you much right, I mean near x0 what we know is that f'(x) <= 1/x0

so it can point to any direction

love_sodam

11:35 AM

Is there anybody explain about induced retraction in example 0.15 of Hatcher's book?

Wolgwang

12:44 PM

@robjohn Can you add new links here?

Thorgott

2:32 PM

Say I glue two $4n$-dimensional manifolds along an orientation-reversing homeomorphism of their boundaries to obtain a manifold $M$ and the embedded copy of the boundaries I call $W$. How do I understand the orthogonal complement of the image of the restriction map $H^{2n}(M,W)\rightarrow H^{2n}(M)$ with respect to the intersection pairing on $H^{2n}(M)$?

robjohn

2:53 PM

@Wolgwang I found a new link for one of them (Google is your friend). I was unable to find a reference for the article by Perotti.

Wolgwang

3:11 PM

:-)

Akiva Weinberger

4:30 PM

Tricoloring knots, except one "color" is invisible and the other two are the same color

Vrouvrou

hello please someone tell me what is equal to

d/ds F(x,u(s))

satan 29

use the multivariable chain rule i guess

=$\partial F / \partial u$*$du/ds$

Eminem

Is $\mathbb Z_{25}$ a field?

Vrouvrou

and for x @satan29 do we obtain F(s,u(s))?

Akiva Weinberger

@Eminem $5\times5\equiv0\pmod{25}$

Vrouvrou

4:40 PM

the prof told us that we must find this @satan29

Akiva Weinberger

Therefore $5$ does not have a multiplicative inverse

(If it did, we would have $5=5\times5\times5^{-1}=0$)

Eminem

$\mathbb Z_n$ is a field $\iff n$ is prime?

Akiva Weinberger

$\mathbb Z_n$ means the integers mod $n$, right? Then yes

Eminem

But there is somehing about the powers of prime numbers

oh i know

order of a field must be a prime number or power of prime right?

finite fields ofcourse

Akiva Weinberger

Yes - there exists a field with 25 elements, but it is not $\mathbb Z_{25}$

BigSocks

4:45 PM

@AkivaWeinberger please say sike

Akiva Weinberger

Interesting - it looks like a cross between the Japanese kana characters め and の

More like の I guess

Mike Miller

5:04 PM

@Akiva The parallelogram with sides $(a,b)$ and $(c,d)$ and the parallelogram with sides $(a,c)$ and $(b,d)$ have the same area. There should be some geometric reason for this. What is it?

Vrouvrou

@AkivaWeinberger can you help me to calculate this d/ds F(x,u(s))

Akiva Weinberger

$\frac{\partial F}{\partial u}\frac{du}{ds}$

Where did you get this problem?

@MikeMiller Maybe first step is to rewrite in coordinates 45 degrees off

or not

Mike Miller

It's a mystery to me! I have been staring at pictures for a while. Perhaps this is wrong headed.

They two matrices are conjugate so some change of variables works. But that's not satisfying

Clarinetist

So, I haven't thought about polar coordinates in about 10 years...

In $\mathbb{R}^2$, suppose I have the region $\{x <0, y < kx\}$ where $k \neq 0$ is some given constant. How do I convert this to polar coordinates?

I get that $r \in (0, \infty)$, and the lower bound for $\theta$ should be $\pi$. The upper bound seems to differ depending on whether $k > 0$ or $k < 0$, but I'm not sure how to find the upper bound for $\theta$ explicitly. Because of the way $\tan^{-1}$ is defined, I think if $k < 0$, I can use $\tan^{-1}(k)$ given how $\tan^{-1}$ is defined, but not sure what to do for when…

Oh dang, I think I drew the region incorrectly when $k > 0$. Ugh.

Akiva Weinberger

5:19 PM

I think you get a piece of the lower-left quadrant

Clarinetist

Okay, when $k > 0$, we have that $\theta$ goes from some value to $2\pi$, but I'm not sure how to get that lower bound.

Akiva Weinberger

Shouldn't it just go to $\frac32\pi$? Since $x<0$ so it stops at the (bottom half of the) $y$-axis

Clarinetist

Lol I need to check this again... you're right

Akiva Weinberger

@MikeMiller So geometrically, you start with two points and a pair of perpendicular lines through a third point

Clarinetist

I shouldn't have tried to do this at 2AM last night

So in fact, if $k < 0$, we have the entire lower-left quadrant

Akiva Weinberger

5:23 PM

Yeah

Mike Miller

@AkivaWeinberger OK, I already buy that you're pointing me in the right direction

Vrouvrou

@AkivaWeinberger the prof todd me that we have to find F(s,u(s)) in the résultat

Clarinetist

If $k > 0$, we end up with that spot between the line and $\dfrac{3\pi}{2}$, agreed. Now, how do I figure out what that lower bound is?

Akiva Weinberger

I don't know the answer I'm thinking out loud @MikeMiller

Mike Miller

One of these is the image of the unit square under a matrix A and the other is A^T, so there's some dot product duality hiding here

I'm telling you that your thinking is the right way to think

Whether or not you have an answer

Akiva Weinberger

5:25 PM

We also can't stretch the two axes independently - we need to know that a length on one axis equals a length on the other

Clarinetist

From my digging through Stewart last night, my understanding is that $\tan(\theta) = k$ for that lower bound, which is not at all helpful.

Akiva Weinberger

Hm. In general, for arbitrary vectors $u$ and $v$, do we have $\det(au+bv,cu+dv)=\det(au+cv,bu+dv)$?

Yes - they're both $(ad-bv)\det(u,v)$

Mike Miller

Yeah, let B = (u, v), then you've written AB and A^T B; now det(AB) = det(A) det(B) = det(A^T) det(B) = det(A^T B)

When B = I as here, two of these equalities are boring and I am asking you why det(A) = det(A^T) :)

user2103480

@Clarinetist it's the angle of $\left(\frac{-1}{\sqrt{1+k^2}},\frac{-k}{\sqrt{1+k^2}}\right)$ if that is of any help

Akiva Weinberger

OK so now we have a version that doesn't depend on perpendicularity

Clarinetist

5:29 PM

I should add: for context, this is for a double-integration problem

user2103480

maybe you can find a nice sine-cosine expression of this

or you can say that $\tan(\theta) = k$ so $arctan(k) = \theta$

Clarinetist

No, because $\arctan$ does not have range in that lower-left quadrant.

That's why I've been struggling with this.

user2103480

shift by $\pi$

or something like that

Clarinetist

Making pancakes at the moment, but I'll look into what you mean in a sec

Akiva Weinberger

@MikeMiller Hm actually the $A^\top$ thing makes me think we want to think about covectors

Astyx

5:35 PM

@MikeMiller You got the bounds wrong right?

You mean (a,b) and (c,d) = (c,b) and (a,d)

Akiva Weinberger

$ad-bc\ne cd-ab$

Astyx

Wait I'm confused, what is (a,b) here? a point, an interval, two lengths ?

Akiva Weinberger

A point

Coordinates

user2103480

@Clarinetist the angle of $\left(\frac{-1}{\sqrt{1+k^2}},\frac{-k}{\sqrt{1+k^2}}\right)$ in the left plane is the same as the angle of $\left(\frac{1}{\sqrt{1+k^2}},\frac{k}{\sqrt{1+k^2}}\right)$ plus $\pi$. Since multiplication by minus one is just going around a half circle and this yields an angle in $(-\pi/2, \pi/2)$.

But valid point, I missed the fact that the arctan is not defined there

So $arctan(k) + \pi$ is your least angle

@robjohn the educated formulation of Biggie Smalls' 4th Crack Commandment

geocalc33

are rotations different in different coordinate systems?

anakhro

5:49 PM

They certainly look different.

Like just think about Cartesian vs. polar in the plane. Rotating around the unit circle looks very different coordinate-wise. In polar coordinates, one coordinate is fixed, while in Cartesian, neither are fixed.

geocalc33

yeah

that makes sense now, thanks

Akiva Weinberger

@MikeMiller Maybe leverage the similarity between $A^\top$ and $A^{-1}$

$A^{-1}A=I$ means the columns of $A^{-1}$ and $A^\top$ are perpendicular

(Not just in 2x2 but in general)

Er that's not what I mean

The $k$th column of $A^\top$ is perpendicular to the span of all columns but the $k$th of $A^{-1}$

robjohn

@Vrouvrou Is $x$ independent of $s$?

geocalc33

@anakhro so a standard rotation matrix in x-y coordinates will look different than a rotation matrix in u-v coordinates right?

Mike Miller

@AkivaWeinberger Sure, if I understand 2x2 I will understand n x n

BigSocks

6:07 PM

or backwards- reminds me of the volume for a parallelepiped. Apparently for an n-parallelotope you get the "volume" as a similar looking determinant

Parallelepiped

In geometry, a parallelepiped is a three-dimensional figure formed by six parallelograms (the term rhomboid is also sometimes used with this meaning). By analogy, it relates to a parallelogram just as a cube relates to a square. In Euclidean geometry, the four concepts—parallelepiped and cube in three dimensions, parallelogram and square in two dimensions—are defined, but in the context of a more general affine geometry, in which angles are not differentiated, only parallelograms and parallelepipeds exist. Three equivalent definitions of parallelepiped are a polyhedron with six faces (hexahedron...

anakhro

@geocalc33 you can compute a change of basis matrix.

But I think something more interesting is happening there. Since changing basis is keeping your same coordinate system, while changing between polar and cartesian is changing your geometric interpretation.

geocalc33

yeah I need this to solve my problem I posed

anakhro

Which problem?

Mike Miller

This is how you ought to understand determinant, @BigSocks

This is why the change of variables theorem in multivariable calculus has a determinant in it

det A = 0 <-> A is invertible is because:

if A is invertible then Im(A) = R^n and this paralleliped (the image of the standard unit cube under A) bounds some convex body, which has nonzero volume
If A is NOT invertible then Im(A) has dimension <= n-1. The n-dimensional volume of A(box) is zero, because it is constrained to lie in a smaller-dimensional space!

Thorgott

that tells you why det A=0 <=> A is not surjective

BigSocks

6:19 PM

right cool so does this clear up your confusion?

Astyx

This is so weird

Mike Miller

@Thorgott i assume one has already done the basic work in understanding invertibility = surjectivity = injectivity for square matrices...

@BigSocks No of course not the question is why, geometrically, $$\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \det \begin{pmatrix} a & c \\ b & d \end{pmatrix}.$$ Yes, you can argue "The first is the area of the first guy, the second is the area of the second guy, and then computing these the formulas match up" but that gives me no geometric insight, it's a magic trick, and I hate magicians

Astyx

geogebra.org/geometry/dkq5jvqh

anakhro

I am not sure, the defining rule for the transpose seems to suggest that most geometric features are preserved by the transpose.

Since the determinant, being the n-dim signed volume, is expressible in these features, then is it really surprising that the determinant is blind to the transpose?

BigSocks

"det is preserved under transpose"?

& the det in question gives you the area of the parallelogram ~ geometry

I am unsure as to why you are confused

Astyx

6:32 PM

It's not about confusion, it's about interpretation

Akiva Weinberger

This is closely related to "What is the geometric meaning of the transpose"

geocalc33

@anakhro The problem about projecting the 3d vector field in (0,1)^3 onto x-y plane, x-z plane and y-z planes

Mike Miller

"Determinant is preserved under transpose" is an algebraic statement. I am asking you to explain why that is true using geometry, not taking it for granted.

A good answer to this will lead one to further insight into the transpose.

Akiva Weinberger

(which Ted likes to explain as $x\cdot Ay=A^\top x\cdot y$)

Mike Miller

If you think I don't understand what a determinant is or what a transpose is or why those are equal from the algebraic POV, then I think we are talking past each other.

BigSocks

6:33 PM

@MikeMiller @Astyx I guess I interpreted "mystery" here as "confusion"?

@MikeMiller i don't think that

anakhro

Surely there is a good "geoemtric interpretation of det(A) = det(A^\top)" answer on stackexchange somewhere.

@geocalc33 Why are you wanting to change coordinates in this question?

Mike Miller

I did not see a useful one, @anakhro. I saw one that basically said "Row operations get turned into column operations and determinant does the same thing to both", which is true and does give a proof, but it is algebra.

Thorgott

idk if the transpose is well-suited to be interpreted in this geometric context

anakhro

@MikeMiller the "The Transpose Inverts the Rotation but Keeps the Scaling" one didn't satisfy you either?

Akiva Weinberger

Applying $A$ and asking for the $x$-coordinate is equivalent to dotting with $(a,c)$. Applying $A$ and asking for the $y$-coordinate is equivalent to dotting with $(b,d)$.

Mike Miller

6:36 PM

Algebra.

Akiva Weinberger

That is how one gets $(a,c)$ and $(b,d)$ from $(a,b)$ and $(c,d)$ geometrically.

anakhro

92

A: Geometric interpretation of $\det(A^T) = \det(A)$

A geometric interpretation in four intuitive steps.... The Determinant is the Volume Change Factor Think of the matrix as a geometric transformation, mapping points (column vectors) to points: $x \mapsto Mx$. The determinant $\mbox{det}(M)$ gives the factor by which volumes change under this ma...

Hopefully it links to the one I am referring to.

Alessandro Codenotti

The transpose is a wonky adjoint, so what's a geometric interpretation of the adjoint?

Mike Miller

OK, that's actually quite a bit better than I anticipated. I assumed they were using the decomposition of GL_2(R) into rotations, shears, and scalings, which is a reduction to algebra.

They're instead viewing determinant as a product of eigenvalues. Unsuitable for an introduction, but geometric.

Akiva Weinberger

@anakhro Oh I like this

anakhro

6:38 PM

@AlessandroCodenotti I personally think the definition of the trace with the inner product seems adequately geometric.

Akiva Weinberger

I view trace like this. Visualize $A$ as a vector field, assigning the vector $Av$ to every point $v$. Flow along $A$ for an infinitesimal amount of time. How much did the volume change?

geocalc33

@anakhro I am wanting to change coordinates in this question because the boundary vector fields (on the cubes faces) are set in a specific coordinate system, so I think the interior 3d vector field should reflect the same coordinate system as the boundary vector fields I guess

Akiva Weinberger

In symbols, $\operatorname{tr}A=\frac d{dt}\det(I+At)$

Oh I misread a comment

I thought someone asked about the trace, sorry

anakhro

Oh that was me who said trace when I meant transpose.

I AM THE SORRY ONE, Akiva.

BigSocks

what about "the column space of $A$ is the row space of $A^T$" here? spaces sound pretty geometric

Thorgott

6:47 PM

trace counts fixed points, kinda, sometimes, loosely

anakhro

@BigSocks column and row isn't very geometric. That's just referencing the orientation of the notation.

geocalc33

The change of basis of the past is not the change of basis of the future!

The enemy cannot find you if you keep changing your basis

BigSocks

@anakhro unsure if you missed the "space" part. that isn't really just about the orientation of notation

Mike Miller

Column space is R^2, row space is R^2, neither of those see the determinant, only that it is nonzero.

This will lead you to the Gauss-Jordan elimination proof that $\det(A) = \det(A^\top)$ which is fine I suppose

@Astyx I'd like to use this applet for illustration later in the semester, did you make it and if so, do you mind if I do, and then do you want to be attributed?

anakhro

7:06 PM

They will probably prefer you use their full name, Eel Astyx.

user2103480

@Thorgott no for positive definite operators it's a variance man

pretending there's one absolute intuition

Mike Miller

I don't pretend to know what a trace is.

user2103480

But it's actually kinda neat. Positive definite trace class operators are covariance matrices of normal distributions on hilbert spaces, and for such a random variable $X$ with mean vector $m$ and covariance operator $Q$, $\int \| X - m\|^2 \,\mathrm{d}\Bbb P = \mathrm{tr}\,Q$

Astyx

@MikeMiller I did make it, feel free to use it! No need to attribute me. I'm glad to know it'll be somewhat useful :)

anakhro

@MikeMiller Hint: It's something students do and often results in you marking a dozen of identical assignments.

Mike Miller

7:18 PM

@AkivaWeinberger This joke doesn't work, as lambda is a reflection of y, not a rotation. Of course it can be fixed.

Akiva Weinberger

Sorry I meant $ʎ$ not $\lambda$

Mike Miller

:D

robjohn

7:33 PM

@MikeMiller That's why I like the $90^{\large\circ}$ rotation of $\infty$ equals $8$. Pretty standard characters.

Mike Miller

Yeah it's a good one.

@anakhro I missed this, but it's amusing.

Bad choice on their part, but I'm no stranger to bad choices. Don't understand the homework and you won't pass the tests. It's how it be.

anakhro

Is there something extra we can do to help students avoid copying answers out of books and online answer websites? It seems offering more office hours doesn't seem to help./

Ted Shifrin

@anakhro This was a challenge long before COVID. Now, probably hopeless.

Mike Miller

Personally, my approach is to remove the grading scheme that punishes students that don't copy from the internet, and to make tests they cannot pass if they do not understand what they're doing.

Past that IMO incentives are not my problem.

anakhro

I guess that's a good point about not/punishing students who don't copy from the internet.

Mike Miller

7:40 PM

@TedShifrin I was asking earlier for a geometric reason that det(A) = det(A^T); I phrased this as asking why two parallelograms have the same area. I know multiple proofs of the first fact but I would characterize most as algebraic. (Decomposition into elementary column operations and using that column operations and row operations do the same thing to determinant is algebraic.) Do you have insight?

The best argument I saw abandoned the parallelograms and thought eigenvalues by drawing ellipses.

Ted Shifrin

Yes, @MikeM, even in my honors multivariable class I had a few students who cheated on homework and had good scores but repeatedly did failing work on exams. Nevertheless, I continued to count homework for something like 35% of the grade, since all the challenging proofs were on homework only.

@MikeM: That's a great question.

Mike Miller

Astyx made a program to illustrate this: geogebra.org/geometry/dkq5jvqh

Which makes it obvious that it is true, but not obvious why it is true!

Ted Shifrin

Is it obvious that the dual map should have the same eigenvalues? I don't think so.

Mike Miller

I also don't think so. The best answer I saw used what amounts to SVD.

Which does give good geometric insight if you already believe in SVD.

Ted Shifrin

That's pretty fancy.

Mike Miller

7:44 PM

Right, I liked the picture, but not the pedagogy.

Ted Shifrin

Shame on me for never having asked this question.

So if we take a basis of eigenvectors, what is the dual basis? Nothing interesting?

Mike Miller

I am not sure.

Let me try an example.

Ted Shifrin

My computation seems to say the dual basis should be eigenvectors of the dual map.

BigSocks

@TedShifrin eigenvectors of the dual space?

neat

Ted Shifrin

dual space???

BigSocks

7:49 PM

subspace spanned by eigenvectors is a space yea

copper.hat

Reducing $A$ to Schur form maybe?

Ted Shifrin

One talks about eigenvectors of a linear map, @Socks, not of a space.

What is Schur form?

Mike Miller

I suppose I could have done the computation before writing down my example. :) I agree with your computation.

Ted Shifrin

So that probably answers the question, although this too is a bit less geometric than I'd like.

Mike Miller

It seems one has no choice but to abandon parallelograms here.

Ted Shifrin

7:51 PM

On the other hand, the only way to "understand" the transpose is as the dual map.

Yeah, I always observed the miracle with parallelograms as a consequence of the algebra.

BigSocks

@TedShifrin yeah shoulda said that span

copper.hat

Can write $A=QTQ^*$ where $T$ is upper triangular.

Ted Shifrin

Still not to the point, Socks.

Not clear how the Schur form of the transpose is related, I guess @copper.

Transposing the equation gives nothing good.

copper.hat

Just if it is upper triangular the transpose is lower triangular so the $\det$ is trivially the same.

I should say Hermitian...

BigSocks

@copper.hat 2algebraic

copper.hat

7:54 PM

@BigSocks Not really, it has a nice geometric interpretation.

Mike Miller

I suppose I will accept determinants of upper triangular matrices as geometric, because that's about shear invariance of volume. But I am skeptical of factorizations.

copper.hat

Depending on your definition of nice.

Its a Schur thing...

BigSocks

lol ok

copper.hat

Why, it is easy to demonstrate in the plane?

Ted Shifrin

Oh, sure, @copper. But what's the Schur form of the transpose?

copper.hat

7:56 PM

Well, it just needs to be triangular.

Ted Shifrin

Plus, you need to know that $\det Q = \pm 1$. That begs the question, I think.

copper.hat

True.

Ted Shifrin

tosses this attempt in the trash

Mike Miller

Now that I have you hooked on the problem I need to stop thinking about determinants and start thinking about obstructed gluing problems.

copper.hat

But it is a useful factoriSation from a numerical perspective as well.

Ted Shifrin

7:57 PM

LOL

Yes, it's one of the many Lie group decompositions that I always forget.

I think my dual/eigenvector thing is solid, @MikeM, but not elementary.

Mike Miller

I only ever remember Iwasawa as a topologist.

Yes, I think that's the clearest so far. But you do have to give up on parallelograms. This could be considered a positive: "the fastest route to understanding this forces us to get a good grasp on the theory..."

And I'm not committed to that being the only geometric interpretation of determinant by any means. Far from it.

copper.hat

The fact is you need a basis to define $\det$, so that probably says something.

Even though it is basis independent.

Mike Miller

You need less than a basis, but yes, you need a notion of signed volume.

Ted Shifrin

Well, ultimately, $\det = \Lambda^n$. :)

Mike Miller

(A basis gives a notion of signed volume.)

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$Mathematics$ Mathematics