I guess I can have one more post for the road. Most voters in every election anywhere ever only care about domestic policy. Well ok most voters don't seem to care about specifically policy so to speak, but they only care about things that affect them. Acting like being self-centered is in some way uniquely American is nonsense, the only difference here is that the US foreign policy has an outsized influence on the rest of the world.
ehh fuck it this post is my commitment to not return here until I finish the brothers karamazov (or give up on it entirely)
eating more meat and cheese recently, which means i need to shower more often. human bodies are disgusting i wish i was one of those ethereal waif creatures from racially insensitive scifi/fantasy tv shows
it is maddening how bloodthirsty and punitive the average person seems to be
While it probably won't ever be as efficient as just walking upright, I'm confident that it would only take the right method and enough practice to learn how to walk on all fours at a reasonable pace and relative ease. I'm also confident that I could learn to do this, and that it could prove useful one day.
Imagine walking home late at night, trying to not be spooked while still being alert of your surroundings just in case someone will try to fuck with you, and then you spot some guy just scampering down the street on all fours, barefoot and wearing shoes on his hands. Doing so like that's just his regular form of locomotion. And you're now certain of two things and nothing else: you want nothing to do with whatever that is, and that absolutely nobody is wanting to fuck with that guy.
reminds me of my old technique which is to do it facing upwards. not actually difficult and i could get pretty fast with it. mightve beaten my walking speed actually
...facing upwards?
every sufficiently top-down approach is indistinguishable from magic and every sufficiently bottom-up approach is indistinguishable from formal logic
one more day and then i can stop seeing reminders to vote on my dash for at least several months
growing up I never quite understood higher category theory, categorical logic, and homotopy type theory. as a kid in elementary school I just couldn't master the n-categorical perspective. so I always felt like I wasn't cut out for math.
To be fair the n-categorical perspective is kinda dogshit. Not sure why they're still teaching it in elementary schools, classic U.S. school system moment I guess.
Love and support your trans math colleagues
why does the amount of linear algebra any graduate math course expects you to know greatly exceed the amount of linear algebra any math course will ever teach you
This is true. If there's one thing you need to know it's linear algebra.
Here's a fun linear algebra problem to test your skills!
For which n can you find a set of n nxn matrices, A_1, ..., A_n, such that for all n dimensional vectors x, the set A_1 x, ..., A_n x is linearly independent?
I’d really like to see the answer to this
don't you have a trivial counterexample where x is the zero vector?
how about for all nonzero n dimensional vectors x
yeah that's what i'm working through now - i think it only works for n = 1. otherwise you always can find some vector x that forms a linearly dependent set. having trouble formalizing it tho, several years since i've written a linalg proof.
all A_n have to be invertible because ker{A_n} needs to be trivial. In the n=2 case that means we need to find
A x = k B x
(B^-1)A x = k x
Which reduces to whether (B^-1)A has at least one eigenvalue. Taking [[0, 1], [-1, 0]], the answer is yes in C but no in R.
so e.g. a solution for real x for n=2 is {identity, [[0, -1], [1, 0]]}, right.
(conceptually, because your resulting system of eqns is: x1 = k x2, x2 = -k x1, right.)
In general, if some matrix isn't invertible then we also get a trivial solution.
If we're working over any algebraically closed field that works for all n.
If A_1 and A_2 are invertible, we can find an eigenvalue k of (A_1^-1)A_2, which gives us a solution.
Consider now the case R^n for odd n>1. The characteristic polynomial of (A_1^-1)A_2 is of degree n, so it has a real root and a real eigenvalue. The case for even n>2 uhh idk.
What about A_i[i,i] = 1, and all other entries are 0? This picks out the ith element of x, and returns a vector that is 0 for all other entries. Therefore A_i x and A_j x have only one non-zero entry each (the ith and jth element respectively). If i != j then (A_i x)^T (A_j x) must be zero. So the set {A_1 x, …, A_n x} must be linearly independent for all x.
Set v = [1, 0, 0,...
Then A_2(v) + A_3(v) = 0 is a linear dependence. Or am I thinking about this wrong?
why does the amount of linear algebra any graduate math course expects you to know greatly exceed the amount of linear algebra any math course will ever teach you
This is true. If there's one thing you need to know it's linear algebra.
Here's a fun linear algebra problem to test your skills!
For which n can you find a set of n nxn matrices, A_1, ..., A_n, such that for all n dimensional vectors x, the set A_1 x, ..., A_n x is linearly independent?
I’d really like to see the answer to this
don't you have a trivial counterexample where x is the zero vector?
how about for all nonzero n dimensional vectors x
yeah that's what i'm working through now - i think it only works for n = 1. otherwise you always can find some vector x that forms a linearly dependent set. having trouble formalizing it tho, several years since i've written a linalg proof.
all A_n have to be invertible because ker{A_n} needs to be trivial. In the n=2 case that means we need to find
A x = k B x
(B^-1)A x = k x
Which reduces to whether (B^-1)A has at least one eigenvalue. Taking [[0, 1], [-1, 0]], the answer is yes in C but no in R.
so e.g. a solution for real x for n=2 is {identity, [[0, -1], [1, 0]]}, right.
(conceptually, because your resulting system of eqns is: x1 = k x2, x2 = -k x1, right.)
In general, if some matrix isn't invertible then we also get a trivial solution.
If we're working over any algebraically closed field that works for all n.
If A_1 and A_2 are invertible, we can find an eigenvalue k of (A_1^-1)A_2, which gives us a solution.
Consider now the case R^n for odd n>1. The characteristic polynomial of (A_1^-1)A_2 is of degree n, so it has a real root and a real eigenvalue. The case for even n>2 uhh idk.
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where to find waifish pseuds...
wait waifish is gender neutral, right?
anyway im using the word because its funnier and comes off as less basic than saying one wants to fuck twinks and skinny women
where to find waifish pseuds...
wait waifish is gender neutral, right?
