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 and linearly independence

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data
    I have three vectors in R^(2x2):

    (1 0 , 0 1) (That is "1 0" horizontal first line, and "0 1" horizontal second line), (0 1, 0 0) and (0 0, 1 0).

    I have to determine if they are linear independent or not. I know how to do it in R^(2x1), but not in R^(2x2). What's the method?
     
  2. jcsd
  3. Dec 8, 2007 #2

    radou

    User Avatar
    Homework Helper

    Are these matrices?
     
  4. Dec 8, 2007 #3

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's the same thing

    If

    a( (1,0),(0,1) ) + b( (0,1),(0,0) ) + c( (0,0), (1,0) ) = 0 (0 being ( (0,0),(0,0) )
    then

    ( (a,b),(c,a) ) = ( (0,0),(0,0) ) so comparing positions, we get that a=b=c=0

    (this assumes I interpreted R^(2x2) correctly, but I think I did)
     
  5. Dec 8, 2007 #4
    Yes, these are matrices.

    So the example is linearly independent, because they cannot be written as a linear combination?

    If we take a new example: a(1,0)(0,1) + b(0,1)(0,0) + c(2,3)(0,2) = 0.

    Then we get:

    (a + 2c, b+3c)(0, a + 2c) = 0 so it is linearly dependent?
     
  6. Dec 8, 2007 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You almost "tricked" me- I misread "linearly dependent" as "linearly independent! Just setting the linear combination equal to 0, and adding is not sufficient. You have to specifically show that this can be satisfied without a, b, c all being 0. Here, you should also show that a+ 2c= 0, and b+ 3c= 0 are true if a= -2c and b= -3c. In particular, if c= 1, a= -2, b= -3, you have
    [tex]-2\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]- 3\left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right]+ \left[\begin{array}{cc}2 & 3 \\0 & 2\end{array}\right]= \left[\begin{array}{cc}0 & 0 \\ 0 & 0 \end{array}\right][/tex]
    Because it is possible to get the zero matrix without the coefficients all being 0, they are dependent. That is, by the way, equivalent to saying one of the matrices can be written as a linear combination of the other two- here, just move the first two matrices to the right of the equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear algebra and linearly independence
Loading...