$f:V\rightarrow W$ and $dim(v) = dim(w)$. Prove that if $ker(f) = \left\{ O \right\}$, the transformation is going to be bijective:
My answer:
If I have $ker(f) = \left\{ O \right\}$ the transformation $f$ is going to be injective. So I just need to check for it being surjective.
But I know that the dimensions of the domain and codomain of my transformation are the same, so every vector in the domain is going to be mapped to just one vector in the codomain, and when you do that, you cover all the domain, because their dimension is the same, that implies that this trasformation is going to be onto too, concluding, its bijective.
Am I correct?
Best Answer
Hint
Look at the Rank-Nullity theorem.