Confusion on onto transformations

Question

Now transformation of $\mathbb{R}^n$ to $\mathbb{R}^m$ for $n > m$ is an onto function.(Or at least is what I'm understanding).

By definition, for an onto transformed vector b there is always a pre-image of b for which there exists a unique A^-1 , let the pre-image be x

Again, By definition of onto functions(or transformations), two vectors can map to one vector. So, let a != b be the two vectors.

But if there's a unique A^-1 but there are two vectors that map to the same b, how are both of the pre-images found?

Answer 1

I am a bit confused by the question too... However, I try to answer as well as I manage.

Now transformation of $\mathbb R^n$ to $\mathbb R^m$ for $n > m$ is obviously an onto function.

It does not have to be. For example the linear mapping $\mathbb R^2$ to $\mathbb R^1$ that sends everything to zero is not onto. That transformation is described by matrix $\pmatrix{0 & 0}$.

By definition, for an onto transformed vector b there is always a pre-image of b for which there exists a unique A-1 , let the pre-image be x

By definition of onto mappings, every element in $\mathbb R^m$ has a preimage, but not necessarily unique. That is not even possible if $n>m$ and the mapping is supposed to be reasonable (for example linear, or continuous).

Again, By definition of onto functions(or transformations), two vectors can map to one vector. So, let a != b be the two vectors.

That can happen. But just the fact that it can happen does not necessarily mean that it will happen. However, it is true in our case of $n>m$, so that part is essentially correct.

But if there's a unique $A^{-1}$ but there are two vectors that map to the same $b$, how are both of the pre-images found?

There is not an inverse mapping, since the pre-images are not unique. For any $b$ there are not just two pre-images but infinitely many of them. You can find them by solving the equation $Ax = b$ using, say, Gaussian elimination.