1. Limited time only! Sign up for a free 30min personal 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!

Find a basis of a linear system

  1. Apr 1, 2015 #1
    Find a basis for the solution space of the linear system

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

    I created a matrix (not augmented, will be 0 on right side no matter what row operations) and brought it to reduced echelon form. x2 and x4 were free variables and I set them to the variables t and s respectively. I found that:

    x1 = t - (1/5)s
    x3 = (-2/5)s

    All of this I'm fairly certain I did right, I just wanted to make sure I am stating the basis correctly because this is the first problem I've done where I've found a basis for a linear system. Stating all the x's in terms of s and t I get (forgive weird formatting not sure how to do matrices on here):

    [x1]\\\\\\[1]\\\\\\\[-1/5]
    [x2]\\=\t\[1]\+\s\[0]
    [x3]\\\\\\[0]\\\\\\\[-2/5]
    [x4]\\\\\\[0]\\\\\\\[1]

    Forming the 2 dimensional basis:
    [1] [-1]
    [1] [0]
    [0] [-2]
    [0], [5]
     
  2. jcsd
  3. Apr 1, 2015 #2

    Mark44

    Staff: Mentor

    You can check this for yourself. If A is the matrix of coefficients of your system above, and x1 and x2 are your two basis vectors, verify that Ax1 = 0 and that Ax2 = 0. If either of these doesn't come out to 0, then definitely you have made a mistake. If they both come out to zero, then A times any linear combination of the two vectors will also be zero.

    Edit: you have a sign error in your second basis vector.
     
    Last edited: Apr 1, 2015
  4. Apr 1, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You have four equation in three unknowns so one would initially expect the solution space to be 4- 3= 1 dimensional. Of course, if the equations are not independent that may not be true. However, I believe these are independent so your answer cannot be correct.
     
  5. Apr 1, 2015 #4

    Mark44

    Staff: Mentor

    The three equations are dependent. The OP has a sign error in one of his basis vectors.
     
  6. Apr 1, 2015 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Oops. Thanks.
     
  7. Apr 1, 2015 #6

    Mark44

    Staff: Mentor

    Happens to us all...
     
  8. Apr 2, 2015 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    From latexhelp:
     
    Last edited by a moderator: May 7, 2017
  9. Apr 2, 2015 #8

    Mark44

    Staff: Mentor

    I like the bmatrix environment, which uses brackets around the matrix rather than big parentheses.

    https://www.physicsforums.com/file:////begin%7Bpmatrix [Broken]}
    1 & 2 & 3 & 4\\
    a & b & c & d\\
    x & y & z & w
    \end{bmatrix}


    This yields
    $$
    \begin{bmatrix}
    1 & 2 & 3 & 4\\
    a & b & c & d\\
    x & y & z & w
    \end{bmatrix}
    $$

    The vmatrix environment is useful for rendering determinants.
    \begin{vmatrix]
    1 & 2 & 3\\
    a & b & c\\
    x & y & z
    \end{vmatrix}
    which renders as
    $$
    \begin{vmatrix}1 & 2 & 3\\
    a & b & c\\
    x & y & z
    \end{vmatrix}$$
     
    Last edited by a moderator: May 7, 2017
  10. Apr 2, 2015 #9
    Thanks everyone! I noticed the sign error, and when multiplying my basis vectors by the original matrix I get a null matrix in both cases. I am just wondering, in back-checking your answer to say positively that your basis is correct, is this the only step you need to perform?
     
  11. Apr 3, 2015 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I seem to remember you also need to show the basis vectors are independent, which in this case is a piece of cake.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find a basis of a linear system
  1. Help! Find a basis. (Replies: 1)

Loading...