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: Linear Transformations

  1. Nov 13, 2007 #1
    1. The problem statement, all variables and given/known data
    let T: R[tex]^{3}[/tex] -> R[tex]^{3}[/tex] be the mapping that projects each vector x = (x(subscript 1) , x(subscript 2) , x(subscript 3) ) onto the plane x(subscript 2) = 0. Show that T is a linear transformation.

    2. Relevant equations
    if c is a scalar...
    T(cu) = cT(u)

    T(u + v) = T(u) + T(v)


    3. The attempt at a solution

    Well I don't know if I proved the first condition correctly, but I have:

    T(cx) = T(c (x (subscript 1) , x (subscript 2) , x (subscript 3) )
    = T (c x (subscript 1) , c x (subscript 2) , c x (subscript 3) )
    = ( c x (subscript 1) , c 0 , c x (subscript 3) )
    = c T (x)

    Did I do this correctly??

    And same with the second condition.... I have:
    T ( x + 0 ) = T ( (x (subscript 1), x (subscript 2) , x (subscript 3) ) + ( 0 , 0 , 0 ) )
    = T ( x(subscript 1) , x (subscript 2) , x (subscript 3) ) + T ( 0 , 0 , 0)
    = ( x(subscript 1) , 0 , x (subscript 3) ) + ( 0 , 0 , 0)
    = T(x) + T(0)

    Did I do this correctly as well??

    And my conclusion is that this is a linear transformation....

    Oh and sorry about the x(subscript #)... i tried using the Latex Reference thing, but it shows the subscripts as superscripts....

    Thanks
     
  2. jcsd
  3. Nov 13, 2007 #2

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    For addition v has to be an arbitrary vector (you assumed v = 0.)
     
  4. Nov 14, 2007 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I would put one more step in here:
    = c(x (subscript 1) , 0 , x (subscript 3) )
    as EnumaElish said, you cannot assume that v is any particular vector- especially not 0.
    What if you have
    T ( x + y ) = T ( (x (subscript 1), x (subscript 2) , x (subscript 3) ) + (y (subscript 1), y (subscript 2) , y (subscript 3) )

    And you cannot assume that is
    T ( x(subscript 1) , x (subscript 2) , x (subscript 3) ) + T ( 0 , 0 , 0)
    or T ( x(subscript 1) , x (subscript 2) , x (subscript 3) ) + T (y (subscript 1), y (subscript 2) , y (subscript 3) )
    that's what you want to prove!

    Instead you should have
    T( x(subscript 1)+ y(subscript 1), x (subscript 2)+y(subscript 2) , x (subscript 3)+y(subscript 3))= (x(subscript 1)+ y(subscript 1), 0 , x (subscript 3)+y(subscript 3))
    = ( x(subscript 1), 0 , x (subscript 3))+ (y(subscript 1), 0 , y(subscript 3))
    = T( x(subscript 1),x (subscript 2), x (subscript 3))+ T(y(subscript 1),y(subscript 2) ,y(subscript 3))= T(x)+ T(y)

    A common way of doing subscripts without LaTex is x_1, x_2, etc. You can also use
    x[ sub ]1[ /sub ] without the spaces: x1 or, in LaTex, [ i t e x]x_1 [/ i t e x], [itex]x_1[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear Algebra: Linear Transformations
Loading...