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Homework Help: 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

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

    HallsofIvy

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    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].
     
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