(Exercise from Lay, Lay, McDonald.) Consider an invertible linear transformation $f:R^n\to R^n$ such that $f(u)=f(v)$ for a pair of distinct vectors $u,v$. Can it map $R^n$ onto $R^n$?
I suppose the answer is no. My reasoning is that since $f(u)=f(v)$, it is not one-to-one. But being one-to-one is the same is being onto for finite-dimensional vector spaces. Am I right?
Invertibility is equivalent to injectivity is equivalent to trivial kernel.
You can't have $f(u)=f(v)$ for distinct $u,v$ and invertibility.
The exercise makes no sense.