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!

Linear Algebra - Spanning / Matrix Equation

  1. Sep 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Let B be 4x4 matrix:

    [ 1 3 -2 2 ]
    [ 0 1 1 -5 ]
    [ 1 2 -3 7 ]
    [ -2 -8 2 -1 ]

    a) Do the columns of B span R^4?
    b) Does the equation Bx = y have a solution for each y in R^4?

    Sorry for the crappy matrix, its a 4x4, R^4 is the funky double R for 'Reals'

    2. Relevant equations

    A) So, started out with row reduction. To find B spans R^4, B must have pivot positions in every row.

    I won't go through all the row reduction steps, but I came out to a matrix looking like so:

    [ 1 3 -2 2 ]
    [ 0 1 1 -5 ]
    [ 0 0 0 0 ]
    [ 0 0 0 -7 ]

    There are only pivot positions in row 1 + 2, so no, B does NOT span R^4.

    B) Bx = y does NOT have solutions for each y in R^4. This is due to (what we called in class Theorem 4) with the logically equivalent assumptions that:

    1) For each y in R^4, Bx = y has solution
    2) Each y (element of R^4) is linear combo of A columns
    3) Columns of A span R^4
    4) B has pivot in every row

    Since B does not span R^4 and does not have pivots in every row, hence forth the statement that Bx = y has solutions for y in R^4 is incorrect.


    ...Does this look right to anyone?
     
  2. jcsd
  3. Sep 4, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Looks ok to me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?