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11:05 PM
Hm, does somebody know a matrix (or a method to generate one) of size about ~100/1000 that is real symmetric and has a few negative eigenvalues?
 
Was referring to your compass and straightedge starred post @noɥʇʎPʎzɐɹC
 
@KajHansen hi
 
Hey
 
would you take the time to teach me for free? just asking ;)
 
I do that already to some extent, lol
 
11:12 PM
i am really spoonfed by googling solutions I don't deserve haha
 
im back, but only momentarily
 
@meow-mix still no mouth hehe, how is the day?
 
good; i made it into spanish club (100 on spanish final last year)
 
@meow-mix sincere congratulations on that
 
thanks :)
i'm going to eat dinner now, bye!
 
11:15 PM
good meal
 
11:27 PM
Let $F$ be a field and $V$ a vectorspace over $F$.
Let $v_1,...,v_n\in V$ be vectors and let $f:F^n\to V$ the mapping, that maps $e_i$ to $v_i$ for $i=1,...,n$, where $e_1,...,e_n$ is the standard basis of $F^n$.

a) $f$ is injektiv $\iff$ $v_1,...,v_n$ is linearly independent.
b) $f$ is surjective $\iff$ $\langle v_1,...,v_n\rangle _F=V$.

I am stuck. I know all the terms, but appearantly not the implications. Could you give me a hint? @KajHansen
i work currently on a)
the standard base has vectors of the form: $(1_F,0_F,...)$ right?
 
@Null are the $e_i$ all basis vectors?
 
If it's injective, $f(x) = 0 \iff x = 0$
 
also, you spelled "injective" wrong. german form i presume :P
OOPS
 
That function argument $x$ can be written as a linear combo of $e_i$
 
sorry i didn't see
 
11:37 PM
obv german :d
 
e_1 = (1, 0, 0, ...) yeah
 
@Null is $f$ necessarily linear?
or can it be any function
 
@meow-mix I wrote the excercise exactly that way, so no, its not neccessarily linear
@KajHansen is that the big hint? ;)
 
ok, sorry
lol
 
@meow-mix np, i mean i typed it like i got it
 
11:40 PM
im not really sure, sorry
 
@KajHansen but the function doesn't have to be 0 any time or?
I think b) is in my reach
@meow-mix how was your meal?
 
@Null meh
 
Re @meow ... Howdy, @Kaj @Null
 
@TedShifrin i feel honored that you greet me. now I only have to avoid getting smacked :s
 
@TedShifrin i have a question
with the way you explained the conic to balarka
as a projective mapping from a pencil to another
are you talking about the mapping from $L_{[s,t]} \mapsto M_{[s.t]}$?
 
11:50 PM
@Null: That's hopeless.
Yes, @meow ... but you could put any projective transformation in there. I've forgotten what I did in the book.
I think I assumed the identity for ease of proof.
 
How do you say the following: a is mapped to b. b is???? by a?
 
Try proving that a linear transformation is lin. ind. $\iff \Big( f(x) = 0 \iff x = 0 \Big)$
From there, notice that if the v_k's aren't lin.ind., a nontrivial linear combination of them will give 0
 
@Null: b is "hit" by a, is how I like to say it colloquially. Officially, b is the image of a.
@Kaj: I hope you still remember our linear independence incantations :P
 
Oh, hmmm
@Ted, of course, how could I forget
Even if I wanted to...
 
@KajHansen i make b) first, because i feel that's in my reach^^
 
11:54 PM
I'm sure if you wanted to, you would.
 
These statements make perfect sense if $f$ is linear. Are you sure it isn't @Null ?
I have counterexamples to these if $f$ is not linear
 
Ah, @meow: See the discussion between Lemma 2.12 and Corollary 2.13.
BTW, @Kaj, if you'd like to try to help organize a lunch get-together in ATL on Jan 2 or so with chat folks from the environs (@Brody claims he's out of town, but it's just all my rolled eyes that scare him; @Fargle maybe) I'd be game :P
 
@KajHansen the function has no restrictions
 
@TedShifrin It's possible I could do that.
 
I have my (I hope) final cancer check-up the morning of Jan 3, @Fargle, but days are somewhat flexible.
 
11:59 PM
Gotcha. I'll figure stuff out.
 
@TedShifrin will i ever meet you? :P
we may never know
 
@Kaj will tell you I'm very scary, @meow :D
 
well, actually, we will
 

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