Mathematics - 2016-12-09 (page 7 of 7)

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user86234

23:05

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?

Kaj Hansen

Was referring to your compass and straightedge starred post @noɥʇʎPʎzɐɹC

Null

@KajHansen hi

Kaj Hansen

Hey

Null

would you take the time to teach me for free? just asking ;)

Kaj Hansen

I do that already to some extent, lol

Null

23:12

i am really spoonfed by googling solutions I don't deserve haha

meow-mix

im back, but only momentarily

Null

@meow-mix still no mouth hehe, how is the day?

meow-mix

good; i made it into spanish club (100 on spanish final last year)

Null

@meow-mix sincere congratulations on that

meow-mix

thanks :)

i'm going to eat dinner now, bye!

Null

23:15

good meal

Null

23:27

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?

meow-mix

@Null are the $e_i$ all basis vectors?

Kaj Hansen

If it's injective, $f(x) = 0 \iff x = 0$

meow-mix

also, you spelled "injective" wrong. german form i presume :P

OOPS

Kaj Hansen

That function argument $x$ can be written as a linear combo of $e_i$

meow-mix

sorry i didn't see

Null

23:37

obv german :d

Kaj Hansen

e_1 = (1, 0, 0, ...) yeah

meow-mix

@Null is $f$ necessarily linear?

or can it be any function

Null

@meow-mix I wrote the excercise exactly that way, so no, its not neccessarily linear

@KajHansen is that the big hint? ;)

meow-mix

ok, sorry

lol

Null

@meow-mix np, i mean i typed it like i got it

meow-mix

23:40

im not really sure, sorry

Null

@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?

meow-mix

@Null meh

Ted Shifrin

Re @meow ... Howdy, @Kaj @Null

Null

@TedShifrin i feel honored that you greet me. now I only have to avoid getting smacked :s

meow-mix

@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 $L_{[s,t]} \mapsto M_{[s.t]}$?

Ted Shifrin

23:50

@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.

Null

How do you say the following: a is mapped to b. b is???? by a?

Kaj Hansen

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

Ted Shifrin

@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

Kaj Hansen

Oh, hmmm

@Ted, of course, how could I forget

Even if I wanted to...

Null

@KajHansen i make b) first, because i feel that's in my reach^^

Ted Shifrin

23:54

I'm sure if you wanted to, you would.

Kaj Hansen

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

Ted Shifrin

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

Null

@KajHansen the function has no restrictions