1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Column space and kernel

  1. May 2, 2008 #1
    1. The problem statement, all variables and given/known data

    If col (A) is column space of A and ker(A) null space of A
    with ker(A) = {Ax = 0}
    and ker(A') = {A'y = 0}

    2. Relevant equations

    Consider the (3x2) matrix :

    A = [1,2 ; 3,4 ; 5,6] (matlab syntax)

    Show that

    col(A) = c1 * [1,0,-1]' + c2 * [0,1,2]'

    3. The attempt at a solution

    Find c1 and c2, starting from Col(A) definition.

    ---

    We tried in this way :

    Col(A) = [c1, c2, -c1+c2]'

    Ax=Col(A)
    Then:
    [1,2 ; 3,4 ; 5,6]*[x1 ; x2]=[c1 ; c2 ; -c1+c2]

    rank(A)=2
    rank(A|Col(A))=2
    and the system is soluble.

    With Cramer I have:
    x1=c2-2c1
    x2=-1/2*(c2-3c1)

    Ax=0 for ker(A)
    and i do the system
    1st eq: c2-2c1=0
    2nd eq: c2-3c1=0

    But here the solution is meaningless.

    How can I solve this exercise?
     
    Last edited: May 2, 2008
  2. jcsd
  3. May 2, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I didn't see any definition of Col(A), and I don't see what ker has to do with anything. The columns of your matrix are just [1,3,5] and [2,4,6]. So I would say that the definition of col(A) is a1*[1,3,5]+a2*[2,4,6]. The exercise is to show that space is the same as c1*[1,0,-1]+c2*[0,1,2]. How about finding the value of c1 and c2 corresponding to [1,3,5] and [2,4,6], would that help?
     
  4. May 2, 2008 #3
    It is similar to my idea but how could i solve a1*[1,3,5]'+a2*[2,4,6]'=c1*[1,0,-1]'+c2*[0,1,2]'.... expecially because neither c1,c2 and a1,a2 are known.
     
  5. May 2, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Solve [1,3,5]=c1*[1,0,-1]'+c2*[0,1,2]'. Why are you putting primes (') on things? Now do the same thing for [2,4,6].
     
  6. May 2, 2008 #5
    " ' " is the transposition from row to column vector.
    So those are column vectors. We are trying.. thanks we'll write you soon
     
  7. May 2, 2008 #6
    I tried this:
    [1 ; 3 ; 5] + [2 ; 4 ; 6] = c1*[1 ; 0 ; -1] + c2*[0 ; 1 ; 2]
    and the solutions are
    c1=3 & c2=7 (confirmed by -c1+2c2=11)

    could it be right?
     
  8. May 2, 2008 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No. Please turn off matlab and think about the problem. You have two spaces A=a1*[1,3,5]+a2*[2,4,6] and C=c1*[1,0,-1]+c2*[0,1,2]. You want to show they are the same. If you could show that [1,3,5] is in C and [2,4,6] is in C, wouldn't that show that a1*[1,3,5]+a2*[2,4,6] is in C?? So that makes A a subset of C. Now can you show C is a subset of A?? If so then that makes C=A. That's a bit more work than you actually have to do, if you know something about linear independence and dimension you really don't have to go both ways.
     
  9. May 2, 2008 #8
    mmm... I think I give up thanks anyway for your help
     
  10. May 2, 2008 #9
    but we studied linear indipendence
     
  11. May 2, 2008 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You should still show that [1,3,5] and [2,4,6] are linear combinations of [1,0,-1] and [0,1,2]. If you know that the two pairs are linearly independent then you don't have to check the reverse.
     
  12. May 3, 2008 #11
    [1,3,5] linear combination of [1,0,-1] ? ... mmmm
     
  13. May 3, 2008 #12
    well [1,3,5]' = c1*[1 ; 0 ; -1]' + c2*[0 ; 1 ; 2]'
    makes c1 =1 c2= 3 and 2c2-c1 = 5

    but [2 ; 4 ; 6]' = c1*[1 ; 0 ; -1]' + c2*[0 ; 1 ; 2]'
    makes c1=2 c2=4 and 2c2-c1=6

    so which are the right values of c1 and c2??
     
    Last edited: May 3, 2008
  14. May 3, 2008 #13
    Pleasee
     
  15. May 3, 2008 #14
    this thing
    seems impossible...
    consider that we aren't doing this for a maths course but econometrics and we have never done those things in maths before... so if someone could please help us more.... we have been thinking about many hours and we aren't sure what we should do about this...
     
    Last edited: May 3, 2008
  16. May 3, 2008 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [1, 3, 5] and [2, 4, 6] are two different vectors. The are different linear combinations of [1, 0, -1] and [0, 1, 2]. c1= 1, c2= 3 for [1, 3, 5], while c1= 2, c2= 4 for [2, 4, 6].

    That's it. Your're done! You have shown that [1, 3, 5] and [2, 4, 6] (and so any linear combination of them) can be written as linear combinations of [1, 0, -1] and [0, 1, 2].
     
  17. May 3, 2008 #16
    so the question of the exercise was which are the values of c1 and c2 so the values of c1 are 1 or 2 and the values of c2 are 3 or 4?
     
  18. May 3, 2008 #17
    or maybe it's like this [1,3,5]'+[2,4,6]'=c1[1,0,-1]'+c2[0,1,2]'
    sum of coloumn vectors and it makes c1=3 and c2=7
     
  19. May 3, 2008 #18

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You said the problem was
    That makes no sense at all. The left hand side of your equation, "col(A)", is a vector space (the column space of A), not a vector, but the right hand side is a single vector. They are not equal for any c1 or c2!
     
  20. May 3, 2008 #19
    sorry starting from space of column definition
     
  21. May 3, 2008 #20
    also instead of c in the real exercise there's y upside down λ

    but if you say that that before was the right solution I trust you...
    Thanks a lot to everyone :-D I'll let you know
     
    Last edited: May 3, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Column space and kernel
  1. Column space (Replies: 4)

  2. Column space (Replies: 4)

  3. Basis for kernel space (Replies: 3)

  4. Column Space problem (Replies: 4)

Loading...