Question

If y(x)=ξ1+ξ2+ξ3
whenever x=(ξ1,ξ2,ξ3) is a vector in C^3, then y is a linear functional on C^3; find a basis of the subspace consisting of all those vectors x, for which[x,y]=0
Please, help

Answer 1

@Spacelimbus

Answer 2

What is the notation \( [x,y]=0\) supposed to mean? A closed interval that is equal to zero? Wouldn't seem very clear to me.

Answer 3

it's just y(x) =0
Follow definition, I start from:
1/ choose the base: b1 (1,0,0) , b2(0,1,0) and b3(0,0,1)
2/ find dual base by y (b1) = alpha
y (b2) = beta
y (b3) = gamma

Answer 4

\(x = (\xi_1, \xi_2,\xi_3\))
y is a linear functional,\( [x, y ]= [\xi_1,y] +[\xi_2,y] +[\xi_3,y]\)

Answer 5

I have to find out y (\(\xi_1)\) = .... such that [x,y]= 0

Answer 6

I don't understand the concept of annihilators in vector space. Any suggestion, please

Answer 7

I have never seen this notation before, who introduced such a thing?

I do understand that if \(y\) is linear then
\[\large y( \xi_1+\xi_2 + \xi_3)= y(\xi_1) + y(\xi_2) + y(xi_3) \]
due to the fact that \(y\) is linear, but I don't see the correlation between \(\xi_1\) (which is the entry of a vector? Or a vector itself?)

Answer 8

this is for annihilator.

Answer 9

Is this for linear algebra or for functional analysis / complex analysis? Seems like this is something I will do next term (if even).

Answer 10

This course is Theoretical Linear Algebra

Answer 11

I recommend you to forward this question to http://www.math.stackexchange.com

Answer 12

since it leaves the theories of elementary linear algebra, otherwise I guess the book wouldn't introduce dual bases and dual spaces at page 20+

Answer 13

I understand the dual spaces and know how to work on dual bases. This problem is from annihilator.
Btw, you said this is functional analysis, right?

Answer 14

I can't really say what this belongs to, I have had a course in Linear Algebra 1 and 2, and didn't confront annihilators there, I once saw annihilators in a minor course of function analysis. But it is well possible that it's part of theoretical linear algebra.

At some points Analysis and Linear Algebra intersect and use results of one another which might just happen here. It seems to be an interesting topic, but I yet have to read into it. Usual you study first linear algebra 1-2 before you go into functional analysis, but a theoretical course about linear algebra might be possible at some Universities.

Answer 15

@Spacelimbus they answer me exactly what I did in group work. However, all other students didn't agree with me . That fact made me doubt myself
hehehe. just null space
y(x) = x1 + x2+x3 =0 --> x1 = -x2-x3
so, the kernel basis is \(\left[\begin{matrix}-x2-x3\\x2\\x3\end{matrix}\right] =x2\left[\begin{matrix}-1\\1\\0\end{matrix}\right] +x3\left[\begin{matrix}-1\\0\\1\end{matrix}\right] \)
so, the basis of the subspace consisting of all x for that [x,y]=0 is\[ \left[\begin{matrix}-1\\1\\0\end{matrix}\right], \left[\begin{matrix}-1\\1\\0\end{matrix}\right]\]

Answer 16

That makes a lot more sense now, to me as well, thanks for sharing. the above vectors have the quality that they satisfy \(y(x)=0 \) for all scalars. The notation seems very new to me, haven't encountered it before yet.

Answer 17

Again, thanks for the link

Answer 18

you're very welcome.