If we take a linear transformation $T$ from $R^n$ to $R^n$ and assume that it is not one-one. Can we find what exactly its rank is?
If it is one-one then $\ker T =\{0\}$ and the matrix of the L.T. must be non singular and therefore rank will be $n$. But if it is not one-one then rank must be less than $n$. But can we get the exact rank of the matrix of L.T.?
Yes. You can. The kernel of a linear transformation measures how far the transformation is from being injective. If $\ker{T}=\{0\}$ then it is injective, and as the dimension of the kernel increases, it becomes less and less injective, i.e. you will have more degrees of freedom in solutions of the system.
You can calculate the dimension of the kernel of a transformation by solving the system $Ax=0$ where $A$ is a matrix that represents your linear transformation $T$ in a basis. Since any two representations of a linear transformation are conjugate to each other, i.e. $B=P^{-1}AP$ for some $P$ (a change of basis), it doesn't matter what representation of $T$ you choose. The space of solutions to $Ax=0$ is your kernel and the dimension of this space is the dimension of the kernel of $T$.
Also, for finite dimensional vector spaces, when $T: \mathbb{R}^n \to \mathbb{R}^n$, if $\ker{T}=\{0\}$ then the rank-nullity theorem tells us that the range of $T$ is all of $\mathbb{R^n}$ (surjectivity of $T$) and otherwise, when $\ker{T} \neq \{0\}$ the range cannot be all of $\mathbb{R}^n$ (failure in surjectivity) because $$\dim(\ker{T})+\dim(\mathrm{rank}{T})=n$$
Therefore, your system will fail to have a solution for some vectors $\vec{b} \in \mathbb{R}^n$ and how much it fails (failure in surjectivity) in this particular case, again depends on the kernel of $T$.
But what I said earlier (namely the first two paragraphs) is true about any linear transformation $T: \mathbb{R}^m \to \mathbb{R}^n$.