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".

 

[Architeuthis Dux]

Yes, your impression is correct. The linear transformation T is invertible because its transformation matrix has a non-zero determinant, which is a necessary and sufficient condition for invertibility. Your calculation of the inverse matrix is also correct. Keep in mind that not all linear transformations are invertible, so it is important to check the determinant or the existence of an inverse before assuming invertibility.