Linear Algebra: Linear Transformations

[Hurricane3]

Homework Statement

let T: R$$^{3}$$ -> R$$^{3}$$ 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.

Homework Equations

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

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

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

 

[EnumaElish]

For addition v has to be an arbitrary vector (you assumed v = 0.)

 

[HallsofIvy]

Homework Statement

let T: R$$^{3}$$ -> R$$^{3}$$ 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.

Homework Equations

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

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

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) )

I would put one more step in here:
= c(x (subscript 1) , 0 , 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)

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)

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

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], $x_1$.