# Determining invertibility

## Homework Statement

We have a linear transformation: T(x,y)=(x-2y,y,3x+4y). Determine whether T is invertible.

## The Attempt at a Solution

the basis of the image is (1,0,3) and (-2,1,4). I really have no clue how to proceed from here. I mean i can't even write (1,0) and (0,1) as an linear combination of the image vectors since they dont have the same dimension.

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HallsofIvy
Homework Helper
I assume you mean T as a function from R2 to the image of T (so it is necessarily "onto"). Then the only thing you need to show is that T is one-to-one. Suppose T(x, y)= T(x', y'). T(x,y)= (x-2y,y,3x+4y)= T(x',y')= (x'-2y',y',3x'+4y'). That is, x- 2y= x'- 2y', y= y', 3x+4y= 3x'+4y'. Can you show that x= x', y= y'?

Added after thought: The fact that your two basis vectors for the image, (1, 0, 3) and (-2, 1, 4) are independent shows that your transformation is from a two dimensional space onto a two dimensional space and so is one-to-one and invertible.

Mark44
Mentor
Aren't your image basis vectors just T(1,0) and T(0,1)?
Your basis for the image of T is a two-dimensional subspace of R3, and is a plane through the origin. If v is any vector in R3 that is not on this plane, can there be some vector in R2 that maps to v? I.e., is there some vector u in R2 such that T(u) = v, where v is as described above?