Question

Give an example of a non-zero alternating 2-linear form w on 3-dimensional space such that w(x1, x2) =0 for some set of linearly independent vector x1,x2
Please, help

Answer 1

@hartnn @Kainui @kirbykirby

Answer 2

Hm I am not sure what an alternating 2-linear form is :S

Answer 3

bilinear.

Answer 4

Hm I'm sorry my brain is too tired right now for linear algebra :( I can't think clearly .

Answer 5

a better question is why are you doing homework on a friday night

Answer 6

@bh6180 I have a bunch of homework due on Tuesday. Although I started my homework right after the lecture and in weekend, I rarely have it done before due date. Friday?? what is special?

Answer 7

@amistre64

Answer 8

does w: (x1,x2) to 0, for all x1,x2 in the set if linearlly independant vectors? if so, isnt that a kernel maybe? these are just some guesses that i have no clue about

Answer 9

@ikram002p I need understand the stuff . I am ok if the stuff is all letters, like w (x1,x2) = u (x2,x3) v (x1) +.....
but really lost when apply to particular case like this

Answer 10

i have the solution here, but don't understand the logic on it.

Answer 11

My particular problem is n = 3 and k =2.

Answer 12

ohk , ill review it :O

Answer 13

humm i checked the txt book :O
but sry have no clue abt it , seems i forget it :'(

Answer 14

It's ok, friend. I appreciate any trying. :)

Answer 15

it seems the example is given in the solution ... what do you want more?

Answer 16

I want to know why the last column of det is (1,1,0)

Answer 17

and how to defined \(Alt_2 (R^3)\)?

Answer 18

it is define that way ... all you need is to check weather W(x1, x2) is bilinear or not.

Answer 19

\( Alt_2 (\Bbb R^3) \) is probably your functional space.

Answer 20

It says K \(\in Alt_2\) does it imply that there is \(Alt_1\)?

Answer 21

no .. Alt_1 is no longer bilinear form. for bilinear you need two elements. Alt_1 is your regular linear map.

Answer 22

If I read the solution, I can (accompanied with your support!!:)) understand it. However, if I don't have the solution, I don't know how to construct the function K

Answer 23

Trivially, I have just standard basis of R^3, say x1,x2,x3
how about y1,y2,y3? where do they come from? from dual space of R^3?

Answer 24

define\( K: \Bbb R^3 \times \Bbb R^3 \to \Bbb R \) by
$$K(w_1, w_2) = \det (w_1, w_2, (1,1,0)^T)$$
determinant of this thing. one extra element is added to it. keeping one, you can show the map is linear for other element.
and probably Alt_1 does not exist. IDK
http://en.wikipedia.org/wiki/Alternatization

Answer 25

Is there any way else to defined K? It is always = det ( something depends on the given condition?)

Answer 26

The idea that I have, is to try and modify the cross product. The cross product is bilinear and alternating, so it should be possible to make a slight alteration and get what we want. For example, we could define for two vectors \(x=(x_1,x_2,x_3),y=(y_1,y_2,y_3)\) the map\[x\oplus y=a_1\vec{i}\]where \(x\times y=a_1\vec{i}+a_2\vec{j}+a_3\vec{k}\) is the traditional cross product. I think this will be a bilinear alternating form that is not identically 0.

Answer 27

well ... it seems it works for R^3.

Answer 28

so? a1= x2y3-x3y2
a2= x1y3-x3y1
a3=x1y2-x2y1
then?

Answer 29

SxS is always zero

Answer 30

You shouldn't need to prove anything down from the dirty definitions like that. Alternating and bilinearity follow from the fact that the cross product itself is alternating and bilinear. It's non-zero because \(\vec{j}\oplus\vec{k}=\vec{i}\). However, \(\vec{i}\oplus\vec{j}=\vec0\).

Answer 31

\(\bigotimes\) or direct sum?

Answer 32

I'm just using \(\oplus\) as a symbol for an operation. Also, there's a slight error in my original definition. If we want this to be a form, it needs to map vectors in \(\mathbb{R}^3\) to \(\mathbb{R}\), and not more vectors. So instead of taking \(a_1\vec{i}\), just take the coefficient \(a_1\).

Answer 33

If you don't like \(\oplus\), you can use \(*\) or \(\otimes\) or \(\cdot\), etc.

Answer 34

I am not so sure about the multilinear alternating form ... but for your particular case, according to your pdf file, ,,, if u and v be two linearly independent vectors in R^3, just take another vector in the subspace of these two vectors. and define the map by it's determinant.

Answer 35

can we have K : R^3 to R^2? is it alternating still?

Answer 36

You mean from \(\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R}^2\)? I don't think you can have a form that does that since forms need to map to fields.

Answer 37

@experimentx
I have to prove the map K is alternating one AND, k-linear form on 3-dimension (the text book said that)
So, if I defined K like what you say, the very first step is proving those conditions

Answer 38

@KingGeorge so, nothing to do with the basis? we don't have to translate x = b1x1+b2x2 +b3x3, and y = c1x1+c2x2+c3x3,
then define K(x,y) =K(b1x1+b2x2 +b3x3, y) = b1K(x1,y) +...........

Answer 39

x1,x2,x3, is standard basis on R^3, any vector on R^3 can be represented by linear combination of them. so that I have x , y like what I did above

Answer 40

I go from definition of k -linear form
and construct
K (b1x1+b2x2+b3x3, c1y1+c2y2+c3y3) = b1c1K(x1,x1) +b1c2K(x1,x2) +b1c3K(x1,x3)+b2c1K(x2,x1)+.....
and any K(x_i,xj)=0 (given condition)
so that K (x,y) = b1c1K(x1,x1) +b2c2K(x2,x2) +b3c3K(x3,x3)
and stttttttttuuuuuuck.

Answer 41

To prove the map that I gave is alternating and bilinear, let \(\vec{x},\vec{y},\vec{z}\in\mathbb{R^3}\) and \(c\in\mathbb{R}\). Then let\[\vec{x}\times\vec{y}=a_1\vec{i}+a_2\vec{j}+a_3\vec{k}\]Then\[(c\vec{x})\times\vec{y}=c(\vec{x}\times\vec{y})=c(a_1\vec{i}+a_2\vec{j}+a_3\vec{k}),\]so \((c\vec{x})\oplus\vec{y}=ca_1=c(\vec{x}\oplus\vec{y})\). By bilinearity of the cross product, this means that \(\vec{x}\oplus(c\vec{y})=c(\vec{x}\oplus\vec{y})\) as well.

Answer 42

Also, if\[\vec{z}\times\vec{y}=b_1\vec{i}+b_2\vec{j}+b_3\vec{k}\]then\[(\vec{x}+\vec{z})\times\vec{y}=\vec{x}\times\vec{y}+\vec{z}\times\vec{y}=(a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}.\]So \[(\vec{x}+\vec{z})\oplus\vec{y}=a_1+b_1=\vec{x}\oplus\vec{y}+\vec{z}\oplus\vec{y}.\]Again since the cross product is bilinear we also have that\[\vec{x}\oplus(\vec{y}+\vec{z})=\vec{x}\oplus\vec{y}+\vec{x}\oplus\vec{z}.\]So this new product is bilinear.

Answer 43

To show it's alternating, we know that \[-\vec{x}\times\vec{y}=\vec{y}\times\vec{x}=-a_1\vec{i}-a_2\vec{j}-a_3\vec{k}\]So \[\vec{y}\oplus\vec{x}=-a_1=-\vec{x}\oplus\vec{y}.\]So the map is also alternating. Thus, it's an alternating bilinear form. Since \(\vec{j}\oplus\vec{k}=1\) it's not the zero form, but \(\vec{i},\vec{j}\) are two linearly independent vectors and \(\vec{i}\oplus\vec{j}=0\). So it satisfies what you need to it.

Answer 44

I got it. Thank you so much @KingGeorge and @experiment . Completely understood. :)

Answer 45

@experimentx hihihi.not @experiment

Answer 46

lol ... I did nothing more than to create confusion lol.

Answer 47

I can fly now, hehehe...

Answer 48

Thinking of examples for these kinds of things is something I definitely need practice on. I can do the theory, but not the examples...

Answer 49

I still have other problem , I will post and need you guys' brains :) Please

Answer 50

it seems that in what i wrote above " you need to just take another vector OTHER than in the subspace of these two vectors. and define the map by it's determinant."
Determinant is alternating and multilinear map.
Helps me too. haha
B(u, v) = -B(v, u)
ad B(u, u) = 0.

Answer 51

gotta go too ... see you around.