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!

Linearly Independent Vectors

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Let [tex]A = \left[\begin{array}{ccccc} 2&0&-2&1&-3 \\ 1&1&-3&0&-2\\1&0&-1&-1&3 \end{array}\right][/tex]

    a) Find two linearly independent vectors u and v in R5 such that span{u,v}={[tex]x \in R^5[/tex] : Ax=0}.

    b) Find three linearly independent vectors u,v and w such that col(A) = span{u,v,w}.

    2. Relevant equations


    3. The attempt at a solution


    a) I row reduced A:

    [tex]\left[\begin{array}{ccccc} 1&0&-1&0&0 \\ 0&1&-2&0&-2\\0&0&0&1&-3 \end{array}\right][/tex]

    I'm not sure how to answer this question. I think the first and second column are linearly independent and if we add them we get the third column. I need some help, I don't know what to do from here.

     
  2. jcsd
  3. Mar 20, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    How about you start solving the equation Ax = 0?

    I.e. reduce
    [tex]
    A = \left[\begin{array}{ccccc|c} 2&0&-2&1&-3 &0 \\ 1&1&-3&0&-2 &0 \\1&0&-1&-1&3 &0 \end{array}\right][/tex]
    Then you can take, for example, x4 = r, x5 = s and write the solution as
    xi = xi(r, s)
     
  4. Mar 20, 2009 #3
    Hi

    I reduced A and I'm left with:

    [tex]= \left[\begin{array}{ccccc|c} 1&0&-1&0&0&0\\ 0&1&-2&0&-2&0 \\0&0&0&1&-3&0 \end{array}\right][/tex]

    The linear system corresponding to the row-reduced matrix is:

    x1-x3=0
    x2-2x3-2x5=0
    x4-3x5=0

    variables corresponding to leading 1's are x1, x2 and x4, the rest I think are free variables, solving for the leading variables:

    x1=x3
    x2=2x5+2x3
    x4=3x5


    Now, I don't know where to go from here. I'm not exactly sure how to represent the solutions for this. How does this help us to solve part (a)?
     
  5. Mar 20, 2009 #4

    Mark44

    Staff: Mentor

    x1, x2, and x4 can be found if you know or are given x3 and x5.

    Code (Text):

    x_1 = x_3
    x_2 = 2x_3 + 2x_5
    x_3 = x_3
    x_4 =         3x_5
    x_5 =          x_5

     
    So any vector (x1, x2, x3, x4, x5) can be written as a linear combination of two vectors. I'm hopeful that you will notice these two vectors lurking in the arrangement above.
     
  6. Mar 20, 2009 #5
    Thanks, I notice it now, x3 and x5 are the two linearly independent vectors so u=(-1,-2,0) & v=(0,-2,-3).

    Could you please explain to me part (b) of the question? I don't understand what the question wants :rolleyes:
     
  7. Mar 21, 2009 #6

    Mark44

    Staff: Mentor

    I don't think you understand the a part just yet. x3 and x5 are just plain old numbers, not vectors, let alone linearly independent vectors. And your vectors u and v have to be in R5, not R3 as the ones you show are.

    Take a look again at the code block I put in my last response. There are two vectors in R5 hiding in there on the right side. If you stare at it long enough, you might see them.
     
  8. Mar 21, 2009 #7
    Hint:

    in post #4 set the free variables x_3 and x_5 to constants c_1 and c_2.

    you can separate the resulting vector into the sum of two vectors, one containing the c_1's and the other containing only the c_2's
     
  9. Mar 21, 2009 #8
    I see it!

    [tex]\left[\begin{array}{ccccc} x_{3}\\2x_{3} + 2x_{5}\\x_{3}\\3x_{5}\\x_{5} \end{array}\right]=[/tex] x3[tex]\left[\begin{array}{ccccc} 1\\2\\1\\0\\0 \end{array}\right]+[/tex]x5[tex]\left[\begin{array}{ccccc} 0\\2\\0\\3\\1 \end{array}\right][/tex]

    Is this what you meant? so u=(1,2,1,0,0) and v=(0,2,0,3,1), both in R5

    Now, could you please explain how to solve part (b)?
     
  10. Mar 21, 2009 #9
    good!

    b is much easier than a

    think about what the column space means. they want you to find the 3 linearly independent vectors in the given matrix which constitute the basis of the column space. You already did the first step in this by finding the row reduced matrix. What can you pull from that new matrix that might tell you about the column space?
     
  11. Mar 21, 2009 #10

    Mark44

    Staff: Mentor

    Here's your row-reduced matrix:
    [tex]\left[\begin{array}{ccccc} 1&0&-1&0&0 \\ 0&1&-2&0&-2\\0&0&0&1&-3 \end{array}\right][/tex]
    For part b, you want three linearly independent vectors in R3, that span the columns of the original matrix. IOW, you want three linearly independent vectors u, v, w, in R3 such that any column vector in your matrix is a linear combination of u, v, and w.

    You can get those three vectors by inspection in your reduced matrix.
     
  12. Mar 21, 2009 #11
    If I had to guess I'd say the first, second and the fourth column which have pivot elements ( leading 1's) & they are linearly independent and the other two vectors can be written in terms of them. Am I right?
     
  13. Mar 22, 2009 #12

    Mark44

    Staff: Mentor

    Right.
     
  14. Mar 23, 2009 #13
    Thanks Mark. I have another question (my last question) about the same matrix, it looks similar to the part (b) but it's confusing:

    Find linearly-independent vctors {u1, u2,...,um} such that: row(A)=span{u1, u2,...,um}
    Justify that vectors are linearly independent.


    How can I find this row vectors by looking at the reduced matrix?
     
  15. Mar 23, 2009 #14
    Finding the Row Space is similar to the Column Space. First you find the transpose of the original matrix, and perform your row reductions to find the pivots as with before
     
  16. Mar 23, 2009 #15

    Mark44

    Staff: Mentor

    Here's your reduced matrix (taken from what you posted earlier)
    [tex]= \left[\begin{array}{ccccc} 1&0&-1&0&0\\ 0&1&-2&0&-2 \\0&0&0&1&-3 \end{array}\right][/tex]

    The rows in this reduced matrix span the rowspace of A, and are clearly (to my eye at least) linearly independent. Call these vectors u1, u2, and u3. To justify that they are linearly independent, show that the equation c1*u1 + c2*u2 + c3*u3 = 0 has only one solution for the constants c1, c2, and c3.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linearly Independent Vectors
Loading...