Lepidopterist

Does anyone have access to academic articles here? Looking for this paper: jpm.iijournals.com/content/45/1/141

isn't the kernel never trivial? since x=0 makes the image zero

the claim is that $k[[x,y,z]]$ can inject into $k[[u,v]]$ via :"Take a map from $k[[x,y,z]]$ to $k[[u,v]]$ that sends $x$ to $u$, $y$ to $uv$ and $z$ to $uf(v)$ for some $f\in k[[v]]$ and think about what the kernel is. It isn't hard to see that only for countably many choices of ff can it possibly be zero."

Can someone explain this interesting MO answer: mathoverflow.net/a/25231/13542

i figured it out. gershgorin circle theorem gives you this

why is the maximum-column-sum norm larger than the spectral norm of a matrix?

or u for unit

i was thinking about e

What's a good choice of name for a variable that takes values -1 or 1. I would like not to use 's'.

does anyone here know about martingales? i have an extremely elementary question

@Semiclassical actually it's easy, in the end you just have to show (by your method) that the trace of $(AB^{-1}A+A)^{-1}$ is positive. which is easy since $AB^{-1}A$ is positive definite since it is symmetric

what i said is true, it just won't be enough here

actually, nevermind. that only gives an upper bound. we need it lower bounded away from zero

actually @Semiclassical, can't you use the fact that $\operatorname{trace}(AB)\le\operatorname{trace}(A)\operatorname{trace}(B)$?

i guess if they commute they are simultaneously diagonalizable, right @Semiclassical?

i understand. the question was for @Semiclassical

thanks @PedroTamaroff. @Semiclassical how does this help?

I can show it's true using the Weyl inequalities, but I feel that this must be trivial

If $A$ and $B$ are positive definite, is it obvious that $\text{tr}(A^{-1})> \text{tr}((A+B)^{-1})$?

are there any useful bounds on the Trace$(A^2)$ in terms of Trace($A)$

For positive definite matrices $A$ and $B$, it is a fact that det$(A+B)\ge \text{det}(A)+\text{det}(B)$. Does anyone know if it is possible to have the reverse inequality, if one introduces coefficients (perhaps involving the dimensionality) in front of $\det(B)$ or $\det(A)$?

If one takes an outer product of two vectors, we can think of it as creating a set of all the products of an entry from the first vector times an entry of the second vector (with 1 redundant entry per product). Is there something which generalizes this to all triple products of two vectors?

try Z as a monoid under multiplication

@Lucas no

Ok, Razieh. I will upvote your answer because I think you are probably right. I have to read it carefully but I trust you. However, if you want it to be generally understood you should use the standard notation. It's an interesting theorem, but I'm not sure I get the proof

unless you have spent tons of time with it. he's not using any standard notation

it would be nice if you made an argument instead of citing a text that is impossible to read

this text is unreadable. you are citing D49 to prove the inequality?

@RaziehNoori what is $E_i$?

i understand that

@RaziehNoori the spectral norm is defined on page 397

it even says these are the $l_p$ norms. that is not the spectral norm

isn't that just the $l_\infty$ norm?

earlier he defines it as $\|\cdot\|_{sp}$

i don't see it

@RaziehNoori on what page does he define this?

yes @RaziehNoori look at wikipedia's definition of the infinity norm. it's the maximum column 1-norm

the maximum column 1-norm, that is

@RaziehNoori i don't think the infinity norm is the spectral norm. it's the maximum column norm as i recall

thanks for thinking through it with me. i would have prefered the spectral norm, but the frobenius norm is probably enough for my purposes

that made sense. but you implied you could relate it to the spectral norm because the trace and inner product were linear. i don't get what you mean

@Huy i don't quite understand what your argument is

yes, @Huy. As @RaziehNoori says the spectral norm is the largest singular value

@Huy What you did works because the H-S norm IS the trace of a certain matrix

@Huy I guess I'm not sure what you mean unless you want to use some other inequality between the trace and spectral norm

@Huy maybe i'm missing something, let me see

@RaziehNoori how does that relate to the spectral norm and the trace?

To that end, this looks related: cs224d.stanford.edu/reports/BartleAric.pdf

@Aksakal This does not address the point, which is that even if that is true this analysis needs to account for the likely fact that even controlling for treatment-by-IT female preferences may be different from male preferences, on average.

All of this discussion/analysis should control for the possible/probable fact that women may not value a STEM career as much as men do, on average.