Potentially a stupid question: $T$ cannot have rank $1$ and $0$ trace, can it?

Since $T$ has at least $n-1\ 0$ eigenvalues but then the condition on the trace forces that $T$ has all $0$ eigenvalues, whence it’s got rank $0$.

@JoeShmo It can’t?

Think nilpotent?

Oh dur

I said it was stupid

Potentially.

The potential has materialized

Potential converted to kinetic

Under promise over deliver, you know?

Hold on but I’m not following where the error is in my reasoning

yes, the eigenvalues are all zero

So isn’t it similar to the zero matrix ?

That’s probably the error

no, because it's not diagonalizable

except for the case where it's actually the zero matrix

Right

Ty

$\begin{pmatrix}0&1\\0&0\end{pmatrix}$ is an instructive example

Right yeah Ted said it

joe something you should think about is an operator like T(x,y) = (y,0) on R^2 or C^2

[sorry, i join any pile-on]

actually, it's fun to figure out that if you have such an operator, you can find a basis in which it looks like the matrix with zeros everywhere except a $1$ in the top right

so this example and its higher-dimensional generalizations are the only examples

How are Munchkin’s thistles doing tonight?

she's growing thistles?

She speaks with thistles in her mouth, like Eeyore.

she was yelling about the sun and planets on her t-shirt being in the wrong place

we went swimming and when she changed back into the shirt she had been wearing all day, she said the sun was on the wrong part of the shirt and kept trying to take the shirt off to 'fix' it. nothing was wrong

She sees it upside-down?

maybe something like that. although her main complaint was the sun was on the bottom of the shirt, when it 'should have' been on the top. i think maybe she just hadn't thought about that aspect of the shirt before.

there was also a disagreement about whether or not a planet with a prominent system of rings was mars

Saturnine …

@leslietownes Does her shirt have Pluto?

my wife has a hard time letting go of Pluto

it does.

they can't take pluto away any more than they can make zero a natural number.

They don't want to take Pluto away, just change its status.

sure. like the frenchies and zero.

They don't like saying "non-negative integers"; they consider it a double negative...

yeah, right

the double positive

is zero an unnatural number?

@robjohn here's a direct proof for Heine Borel theorem on $\mathbb R$:

To prove: I:=[a,b], a<b is compact. Let $\mathfrak G$ be any open cover for I. Define $H:=\{a\le x\le b: [a,x] \text{ is covered by finitely many members of $\mathfrak G$. }\}$

$H$ is non empty as $a\in H$. From here, we show that 1) H has supremum, 2) the supremum belongs to $H$, 3) the supremum is actually equal to $b$.

and we are done!

this proof is available in Manifolds book of M. Spivak.

that is a compact proof

From this you easily get the version for R^n

You just need to know that (finite) products of compact spaces are compact... and that R^n has product topology.

dang, just ruined my product pun

And that subspaces of products which are products themselves have product topology

I don't understand proposition 1.3 in Folland's :(.

Suppose that $\{X_\alpha\}_{\alpha \in A}$ be a family of non empty sets. Suppose that A is countable and define $X:=\Pi_{\alpha \in A} X_{\alpha}$. Let $\pi_{\alpha}: X\to X_{\alpha}$ be the coordinate map (i.e., it takes x in X and maps it to its $\alpha$th coordinate).

Suppose that $M_{\alpha}$ is a $\sigma-$ algebra on $X_{\alpha}$. I don't understand why the book says that if $E_{\alpha}\in M_{\alpha}$, then $\pi_{\alpha}^{-1}(E_{\alpha})=\Pi_{\beta\in A}E_{\beta}$, where $E_{\beta}=X$ for every $\beta\ne \alpha$.

@Koro Because $\pi_\alpha$ is a projection map

I think that is just wrong and the correct should be: $\pi_{\alpha}^{-1}(E_{\alpha})=\Pi_{\beta\in A} E_{\beta},$ where $E_{\beta}=X_{\beta}$ for every $\beta\ne \alpha$.

I think it's just a typo

yeah, I think so too.

@Koro sites.math.washington.edu/~folland/Homepage/oldreals.pdf

But then I don't understand why: 1) $\Pi_{\alpha \in A} E_{\alpha}=\cap_{\alpha\in A} \pi_{\alpha}^{-1}(E_{\alpha}). $

and 2) why is countability of A required here? Where was it used?

@onepotatotwopotato thanks :)

@Koro You can prove it directly by hand

hmm, 1) is okay.

$\sigma$-algebra. Should be countable

$\sigma$ algebra (if infinite) can't be countable.

there is an exercise in Folland, which I think asks to prove this.

I mean countable intersection.

Ahh, thanks a lot.

countable intersection must be in $\sigma$ algebra and that's how 1) will help me to conclude.

:-)

@Koro There are steps which require further proof. Do any of them rely on contradiction?

@Koro we want to equip the product with the smallest sigma-algebra which makes the projections measurable

funny thing: if we want products of measure spaces, we usually do it only for finite ones

but for probability spaces, we can extend it to infinite products, because the infinite product of one's is just one again

Fun fact: I actually used that statement in my point-set topology exam years ago without proof and TA deducted points because of that.

@robjohn if we use the fact that supremum of [x,y] is a limit point of [x,y] and hence contained in [x,y], then I think no?

preferably of the type of the first two questions i.e. find all integer solutions of $x^2-x=y^5-y$

or integer solutions of binom (x,2) = binom (y,5)

for the record i believe these two problems in particular are solved here

arxiv.org/pdf/0801.4459v4.pdf

@Koro You want to suppose that the supremum is $<b$ and derive a contradiction.

not required. sup<= b is true by definition of supremum and the completeness property. Isn't it?

Then, since sup lies in the interval I=[a,b], there exists a U in the open cover such that sup is in U.

This U prevents the sup to be < b.

@Koro doesn't it prevent it by contradiction?

...

(based on a discussion that happened a long time ago.)

Me explaining how I angered the witch into turning me into a snake: ♫ I said something wrong, now I long

("Yesterday" - Beatles)

You should be singing ONJ today.

what is the meaning of "canonical"?

"God"-given

Does it mean a form dependent on basis?

no, independent of choices

like I heard (or misheard) in class that $e_1,e_2,e_3$ form a canonical basis of $\mathbb R^3$.

That's misuse of the language.

That's the "standard" basis. That doesn't make it canonical. Indeed, it depends on a choice of coordinates.

Canonical things should automatically map over to canonical things if you take an isomorphism in whatever category. That is certainly not the case for linear isomorphisms here.

The isomorphism $V\to V^{**}$ for finite-dimensional vector spaces is "canonical" — i.e., independent of any basis.

I see- independent of basis.

thanks @Ted.

Yup

wow, Ted said "category" :P

I almost said "functorial."

What would your answer to Koro have been for the meaning of canonical?

it's complicated

for isomorphisms, I agree with your explanation

when people call $e_1,\dotsc,e_n$ the canonical basis of $\mathbb{R}^n$, however, it seems different

I never say that.

I think that is horrid mathematical language.

I'd still argue there's a truth in saying that, but it depends on how we formulate "naturality".

I say standard, as I said to Koro. So do all the textbooks I know.

Who's to say that the standard coordinate system is the only one?

Even with an inner product structure and a prescribed orientation, it's still far from canonical.

For example, $1$ is the canonical basis of $\mathbb{R}$ in the sense that the multiplicative unit is the only choice of basis simultaneously in all fields $K$ viewed as vector spaces over themselves that is compatible with base changes (of coefficients) coming from ring homomorphisms.

OK, I'm not going to debate the one-dimensional situation.

And if you have two vector spaces $V$ and $W$ with specified basis, there is a canonical choice of basis for $V\times W$ (defining the product in the category of vector spaces with chosen bases), so if $1$ is the canonical basis of $\mathbb{R}$, $e_1,\dotsc,e_n$ is the canonical basis of $\mathbb{R}^n$.

I do agree calling it the standard basis is a lot more convenient and avoids confusion

I will never agree that the word applies as you're using it.

And students have no idea what the word means.

I agree, for example, that if you have an inner product (say a Riemannian manifold) and an orientation, then there is a canonical volume element.

yeah, which is why I agree with you practically

there just is room for abstract nonsense which I'm exploting

But that's absurd on any level other than snobbish category players.

indeed

the volume element is canonical in every sense

I'm surprised leslie didn't pop in to interfere.

Maybe he's detained by his actual work.

There are definitions for canonical, but I prefer to use the word canonical informally. If something is natural, I call it natural and that has a precise meaning that is useful. Even if you can define formally what canonical means I still don't see how that is used for something

On the rare occasions that I use the term, I find it meaningful to do so. :)

It's worth understanding that a certain amount of structure gives you something with no arbitrary choices.

But I try to be careful with my use of mathematical language. As I tell my students, I never use the word "trivial" (much abused by math students and teachers) unless it's in the technical sense.

it's also bothersome that oftentimes people use "natural" when they mean "functorial"

Oh I am not saying that the word canonical is not useful, I'm just saying I don't see the point in formalising it

You don't mean a functorial transformation, Thor?

I also see usage of terms I never saw before, like: ets, tfae, wts etc. It's very nice to know these terms :).

Those aren't terms. They're acronyms.

And I never used such things when I taught.

I did sometimes abbreviate such that to s.t.

But not often.

Every mathematician has some working intuitive notion of canonical. Even if you can formalise it, what's point?

Because someone like Koro will come along and ask you what you mean when you say it.

You need to be able to answer.

Then you explain your intuition, no arbitrary choices involved

I don't think category theory would help Koro with that

natural transformations are what happens between functors, whereas functors happen between objects

I'd also say that the standard basis is not canonical

So you support Ted's position, for once.

Suppose that canonical is used to mean 'independent of basis', then there should be consistency in the usage of the term. It should not mean to refer to something contrary to earlier assumed meaning. Very informally speaking, linear maps are canonical but matrices of the linear maps are not.

I don't like saying linear maps are canonical.

I think I do tend to agree with you on occasion @Ted

No linear map — other than perhaps the identity — is a canonical thing.

what confused me was: $e_1,e_2,e_3$ is canonical basis of $\mathbb R^3$

I think your attempt at oversimplification is wrong, Koro.

I mean, given a linear map T from V to W, matrice of T depends upon bases of V and W. If I change the basis, the matrix will change.

However, even though T is determined completely by its values on basis of V, T will remain the same even if I take any other basis of V to determine T.

In this sense, T is independent of basis of V, but matrix of T is not.

Right, but that means that associating a matrix to a linear transformation is not canonical, but that doesn't mean linear transformation themselves are canonical

That is true, but that does not make an arbitrary linear transformation canonical in the least.

Don't go overusing your brand-new word.

0 and identity are canonical

Just throw it in the trash.

@LukasHeger yeah.

@TedShifrin right.

I'll restrict using the word for some time.

What is canonical depends on the context, say you have a subspace U of V. Then the inclusion of U into V and the projection of V onto V/U are canonical

But without the context of having a specific subspace of U of V, these are just some random linear transformations

@Koro I think you should just think of it as I said originally. Can you define/explain the notion without making any arbitrary choices? Back to the trash.

Examples from set theory: the inverse of a bijection is canonical, even unique, but there is no canonical right inverse for a surjection (although every surjection has a right inverse, but it involves arbitrary choices)

That's a good example to understand.

@Ted @Lukas so canonical is like "natural" i.e, independent of arbitrary choices. Like the zero map or the identity.

Except that "natural" has a technical meaning for fancy people. Yes.

@TedShifrin hence natural.

this makes sense.

I'm putting the words back in the trash now.

thanks a lot @TedShifrin , @LukasHeger et @Thorgott.

here's yet another meaning of "canonical": if I formulate the thing we're looking for as a problem (e.g. "Come up with an isomorphism $V\rightarrow V^{\ast\ast}$") and give it to a 100 different mathematicians, they will all return me the same answer :P

Some obstreperous mathematician might choose dual bases.

that always gives you the right isomorphism

but if you chose the dual basis of the dual basis, you do ge the canonical isomorphism into the double dual (but from this description it's not clear why that's independent of the choice of basis)

Hmm, why can't I scale the canonical one?

@TedShifrin always wrote out "without loss of generality"?

I very rarely used that, actually, but I did succumb to WLOG ... but not the ones Koro listed.

I think the notion showed up a lot in differential topology/manifolds, but I often wrote "we may just assume that ..." :P

I have used it, but not much any more.

For example, assuming the first $k\times k$ submatrix is nonsingular in a $k\times n$ matrix of maximal rank.

if $e_1, \dots e_n$ is a basis and $e_i^*$ is the dual basis and $e_i^{}$ is the double dual basis, then for all $i,j$, we have $e_i^{}(e_j)=\delta_{ij}=e_i^*(e_j)$, thus the linear map extending $e_i \mapsto e_i^{**}$ is the canonical isomorphism

@robjohn Since you're here, can you make heads or tails (or some part thereof) of this?

@Lukas, I know that, of course. But why can't I multiply by $5$ in the final construction?

we had a professor who in his script wrote WLOG once (well the German equivalent oBdA) and in the lecture it took him 30 minutes to explain exactly how to reduce the general case to the one with that assumption

how, not who

If it takes 30 minutes, then you should not be so blithe so as to assert it nonchalantly.

@TedShifrin damn permutations

Maybe $S_{50}$ suffices.

eys htta si yevr lusfeu

I read that it takes certain things in the brain to look at something like that and immediately know what the words are.

If you permuted among the words, I think it might be very challenging.

@TedShifrin "surface of the unit-sphere" is adding unnecessary words. Perhaps they are getting mixed up between a geodesic and a geodesic circle (which on the sphere is the intersection of a plane with the sphere).

@TedShifrin I think you still get a natural transformation, but it feels a lot less canonical (maybe cause it doesn't work in char 5 :P)

@Ted: I know you know that a geodesic circle is this intersection, just a possible way of describing it to them.

The OP actually figured out what I said, just didn't know that he/she had figured it out. But I have no idea what Fritz John — as transcribed — was babbling about.

@TedShifrin which sane mathematician multiplies things by $5$ for no reason

You didn't say the isomorphism had to be boring.

(I admit most mathematicians are boring.)

well, I also wouldn't say a multiplication by $5$ makes something more interesting

Fine. Make it $e^{5\pi}$.

it also doesn't work in characteristic $5$

OK, back to the trash bin.

follow up on last night. my solution disagrees with the posted solution. $T$ is rank $1$ and trace$T = 0$. then find determinant and inverse of $aI + T$. So since all eigenvalues are $0$, we have $\det(aI + T) = p_T(-a) = (-a)^n$. Now $T = uv^T$, and $0 = \text{trace}T = v^Tu$, so $T^2 = uv^Tuv^T = (v^Tu) uv^T = 0$. So $T = PJP^{-1}$ where $J_{1n} = 1$ and $0$ everywhere else. So $aI + T = P(aI + J)P^{-1}$; where the inverse of $aI + J$ should be $a^{-1}I - a^{-2}J$,

so that the (inverse of $aI + T) = P(a^{-1}I - a^{-2}J)P^{-1} = a^{-1}I - a^{-2}T$. Agreed?

the posted solution is different, and is a humongous overkill (although the techniques are interesting), and disagrees with what I have here ^

$T$ has all zero eigenvalues since it cannot have two distinct nonzero eigenvalues. they would correspond to linearly independent eigenvectors, making the rank of $T \ge 2$.

this is linear algebra quals

the posted solution is that the inverse is $a[I−(−a^{−1})T]$. They're off by a sign and a factor. and they claim the determinant is $a^n$, but I think it should be $(-a)^n$

oh no, they're obviously right on the determinant

its an upper triangular matrix under change of coordinates with all $a$'s on the diagonal

also what is $- (-a^{-1})T$? Why not just write $a^{-1}T$

@TedShifrin Not again, I just got out of the trash bin...