Question

Define a non zero functional y on C^3 such that x_1= (1,1,1) and x_2= (1,1,-1), then [x1,y]=[x2,y]=0 and find all such functionals
I don't know how to start
Please, help

Answer 1

\[x_1=(\xi_1,\xi_2, \xi_3)~~and~~[x_1,y]=\xi_1-\xi_2 =0\]
\[x_2=(\eta_1,\eta_2,\eta_3)~~and~~[x_2,y]=\eta_1+\eta_3=0\]
then, \([x_1,y]=[x_2,y]=0\)
From the method above, I can have
1/ above functional
2/\([x_1,y] =\xi_1-\xi_3\)
\([x_2,y]=\eta_1-\eta_2\)
3/\([x_1,y] =\xi_2-\xi_3\)
\([x_2,y]=\eta_2+\eta_3\)
Am I on the right way? Pleaaaaaaaaaaaase

Answer 2

@amistre64 @phi

Answer 3

pfft, i dont even know how to start with that either

Answer 4

Thanks for respond @amistre64

Answer 5

is this really linear algebra

Answer 6

abstract algebra

Answer 7

oh i better run lol i remember @ikram002p took this last sem

Answer 8

then, I am waiting for him. :)

Answer 9

im here

Answer 10

whts the question

Answer 11

@beastonastick am I on the right way?

Answer 12

yes

Answer 13

if so, how to prove the linear functional is correct?

Answer 14

hold on

Answer 15

it wont let me put the file down

Answer 16

ill work on it tomoro..
@ganeshie8 btw i took it last year spring :D

Answer 17

I approach the problem on another way:
[x1,y]=y(1,1,1) = a1+a2+a3 =0
[x2,y]=y(1,1,-1) = a1+a2-a3 =0
solve for them, I have a3=0, a1=-a2
so, y is defined by [x,y] = a1-a2
another is [x,y]= a2-a1
@ikram002p I confused every thing. I consider dropping the class :(
The most difficult thing is my prof symbolizes the concept without illustrating by numeric examples. I ....am tired with the course!!!

Answer 18

hmm it simple course
oly this functional need to work on u just need to solve
more of examples then ull get the idea, the rest of it
is like playing with number and groups or so , dnt give up

Answer 19

ok lets try this problem using function A onto B
or in ur case X onto Y

Answer 20

\[\phi:X \rightarrow Y\]
is onto if for each y in Y
there is at least one x in X
such that
\[\phi (x)=y\]

Answer 21

so nw we need to know what kind of function is onto on
\[C^3 \rightarrow C\]
s.t
y(x1)=y(x2)
the rest is as u did lol