Is any linear transformation with $\text{ker }(T)=\left\{\vec{0}\right\}$ an isomorphism? I'm thinking no; for instance, $\exists \left\{\left.T:V\rightarrow W\right|  \text{Im }(T)\neq W\right\}$. 
This seems counterintuitive, though. If such a $T$ with maximal rank exists, What would the target space $W$ look like?
 A: Consider the 0 dimensional subspace of $\mathbb{R}^2$, $\{\vec{0}\}$. Consider the map $T(\vec{0})=\vec{0}$. Clearly this isn't bijective.
A: No. A linear function is injective if, and only if, its kernel is $\{0\}$. It's easy to find examples where the surjectivity doesn't hold
A: Let $T:V\rightarrow W$. If you have $\dim V = \dim W$, then yes, it is bijective (isomorphism), because of the following known theorem:
$$\dim V = \dim \text{Im } T + \dim \text{Ker } T$$
Which implies that $\dim W = \dim V = \dim \text{Im } T$ given that $\text{Ker }(T)=\left\{\vec{0}\right\}$, which shows surjectivity.
If $\dim V \neq \dim W$, though, there are many examples of injective transformations that aren't surjective, therefore aren't isomorphisms.
EDIT: As Travis mentioned in a comment, what I said only applies when $V$ and $W$ have finite dimension. If the dimensions are infinite, we can use a broader theorem (which implies the former theorem I mentioned), stated below:
$$\text{Im } T \cong V/\text{ ker }T$$
Where $V_1\text{ }/\text{ }V_2$ denotes the quotient space of $V_1$ over $V_2$.
A transformation is injective if and only if $\text{Ker }(T)=\left\{\vec{0}\right\}$, including the case of infinite dimension. It is surjective if and only if $\text{Im } T = W$. Therefore, if you study $V/\text{ ker }T$, you can decide if $T$ is bijective or not.
There will be examples in which you'll have bijectivity with infinite dimension, and examples like Travis' one:
$$R: (a_1, a_2, a_3, \ldots) \mapsto (0, a_1, a_2, \ldots)$$
in which you won't have bijectivity (only injectivity).
