Dimension Theorem and Understanding Nullspace and Range in regards to 1-1 and Onto My understanding of this topic is incredibly limited, and for now I'd just like some short explanations that might clear up the confusion I'm having with this topic:
From my understanding, given a linear map $T:V\to W$ where V is finite dimensional, Dimension Theorem says $\dim(N(T)) + \dim(R(T)) = \dim(V)$, where $N(T)$ is the nullspace of $T$ and $R(T)$ is the range of $T$. Given a basis, we can find the basis of the nullspace by finding where the basis gives the output of $\vec O_v$. 
One thing I don't understand is if we find a basis that gives us this, what is the dimension of $N(T)$? In order to find the dimension of the range, we just take the number of other basis that $\not=$ the basis for the nullspace. Once we have both of those, we add them up to get the dimension of $V$. 
 A: I am not sure what do you mean by "we can find the basis of the nullspace by finding where the basis gives the output of $\vec O_v$". The null-space
$$N(T)=\{ v \in V \mid T(v)=\vec 0_W \}$$
is a set of all vectors mapped by $T$ to zero vector; these vectors form a vector subspace of $V$. If we take basis $e_1,\ldots,e_k$ of $N(T)$ and extend it with vectors $e_{k+1},\dots,e_n$ to the basis of the whole space $V$, it turns out that
$$T(e_{k+1}),\ldots,T(e_n)$$
is a basis of $R(T)$. Indeed, these vectors span $R(T)$ since every vector $v \in R(T)$ can be written, for some scalars $\lambda_1,\ldots, \lambda_n$, as
$$ v = T(\lambda_1 e_1 + \cdots + \lambda_n e_n) = \lambda_{k+1} T(e_{k_1}) + \cdots + \lambda_n T(e_n)$$
since $T(e_1)=\cdots = T(e_k) = 0$. On the other hand, 
$$\lambda_{k+1} T(e_{k_1}) + \cdots + \lambda_n T(e_n) = T(\lambda_{k+1} e_{k+1} + \cdots + \lambda_n e_n) = \vec 0_W$$ implies $\lambda_1 = \cdots = \lambda_n = 0$, otherwise $\lambda_{k+1} e_{k+1} + \cdots + \lambda_n e_n$ would belong to $N(T)$. Thus vectors $T(e_{k+1}),\ldots,T(e_n)$ are also linearly independent.
A: Without knowing other information of a linear map $T:V\to W$ than linearity itself, one cannot tell what is the dimension of $N(T)$. 
In the proof of the rank-nullity theorem ("Dimension Theorem" you call), one does not find what is the dimension of $N(T)$.   What one does is assuming $\dim N(T)=m$ for some nonnegative integer $m$ and show that $\dim(R(T))=\dim(V)-m$.
The "Dimension Theorem" tells you a relation between $\dim N(T)$ and $\dim R(T)$ but not what specific numbers they are.
