Matrices and Invertible Linear Transformations

[war485]

Homework Statement

How do I know if this linear transformation is invertible or not?

T: [ x ] ---> [ 2y ]
[ y ] [ x-3y ]

(I also uploaded a small .bmp file to represent this if this looks too ugly)

The Attempt at a Solution

Well, I thought maybe it could be represented by a transformation matrix T
[ 0 2 ]
[ 1 -3 ]

So then I just took the inverse of T, which I got as
[ 1.5 1 ]
[ 0.5 0 ]

So does that mean that in the question, it is invertible? Because if it is, I'm getting the impression that these R² to R² linear transformations are invertible. Would this impression be correct?

 

[HallsofIvy]

Yes, a linear transformation is invertible if and only if a matrix representing it in some basis is invertible. And that is the true if and only if the determinant is non-zero so just observing that 0(-3)- (1)(2)= -2 is sufficient.


Of course, from the basic definition a function is invertible if and only if it is "one-to-one" and "onto" so you could do this:
Suppose f[(x,y])= f(]x',y']). Then [2y, x- 3y]= [2y', x'- 3y'] so 2y= 2y' and y= y'. Then x- 3y= x' -3y'= x'- 3y so x= x'. The function is "one-to-one".

If [u, v] is any vector in R², if f([x,y])= [u, v], then 2y= u and x- 3y= v. From the first equation, y= u/2 so x- 3y= x- (3/2)u= v and x= (3/2)u+ v. Yes, there exist [x, y] such that f([x,y])= [u, v] for any [u,v] so f is "onto".