Is it enough for linear transformation to be injective to be invertible? Lets say I have a linear transformation $T$. On what conditions I will have this transformation inverse $T^{-1}$? 
And what is the difference between isomorphic and invertible transformations?
Thank you in advance.
 A: If $T:V\to W$ is a linear map between finite-dimensional vector spaces of the same dimension, then $T$ is invertible if and only if it is injective. If $V$ and $W$ have different dimensions then $T$ cannot be invertible, but may be injective.
If the vector spaces are not finite-dimensional then the above can fail: consider $V=\mathbb{R}^{\infty}$ and
$$ T((x_1,x_2,\dots))=(0,x_1,x_2,\dots) $$
A linear map $T:V\to W$ is an isomorphism if there is a linear map $U:W\to V$ such that $TU$ is the identity on $W$ and $UT$ is the identity on $V$. It turns out that the inverse of a linear map (if it exists) is also linear, hence the notions of isomorphism and invertibility are the same for vector spaces.
A: The question is what do you mean by invertible?
If you deal with a finite dimensional vector space $V$ and a transformation $f:V\rightarrow V$, then yes: injectivity is equivalent to isomorphy, by the rank theorem. 
If you deal with infinite dimensional vector spaces you first have to ask yourself what you want from your transformation? Do you want it to be bounded - in the case of a Banach space - or do you just want a linear map. For if the map is not surjective you can only define an inverse on the image of the transformation. Sometimes this is just what you need, but rather often you need the map to be surjective to have an everywhere defined inverse.
