Linearly Independent Linear Transformations I am currently studying some theories of single linear transformations. I feels like I understant 99% of it, but there is still one thing that I have not been able to resolve. My book explains it by noting that $M_n(\mathbb{F})$, for a field $\mathbb{F}$ is an $n^2$ dimensional vector space. They then go on to say that if $\lbrace A_1, ... A_{n^2}\rbrace$ is a basis for  $M_n(\mathbb{F})$, then the linear transformations $\lbrace T_1, ..., T_{n^2} \rbrace$, whose matrices with respect to $\lbrace v_1, ..., v_n \rbrace$ are $A_1, ... A_{n^2}$ form a basis of $L(V,V)$, the set of all linear transformations from a vector space $V$ to $V$. 
They then explain the linear transformations  $T_{ij} \in L(V,V)$ given by $T_{ij}(v_j)=v_i$, $T_{ij}(v_k)=0, k \neq j$, are a basis of $L(V,V)$ over $\mathbb{F}$. 
I have two main questions: (1) how do we know that $T_{ij}$ are linearly independent? and (2) this may perhaps answer 1, what does $T_{ij} \in L(V,V)$ mean, in the manner they have described it? For some reason, this notion/concept behind it is not making sense to me. I understand that the basis for $M_n(\mathbb{F})$ are the matrices with 1 in the $i,j$ position, but how does this translate to the linear transformations given by $T_{ij}$?
Any useful explanation would be much appreciated. 
When I am reading the notation  for describing $T_{ij}$, I cannot help but to think that since there is one and only one linear transformation such that $T(v_i)=v_i$, that we have repetitions of the same linear transformations as a basis of $L(V,V)$, and so the dimension of $L(V,V)$ would not be $n^2$. But this is clearly incorrect...
 A: Note that throughout this entire discussion $V$ is a $n$-dimensional vector space. Let $\{v_i\}_{i=1,...,n}$ be a basis for $V$, then any linear transformation $T \in L(V,V)$ is uniquely determined by its action on the basis (i.e. where it sends $v_1, ..., v_n$ to, this is true because of linearity).
Then for specified $i$ and $j$, $T_{ij}$ is just the transformation that sends $v_j \mapsto v_i$ and every other $v_k$ to $0$. It's easy to check that this is indeed a linear transformation. Loosely speaking, $T_{ij}$ just "picks out the basis vector $v_j$ and sends it to $v_i$".
You are right in observing that there is only one $T_{ij}$ such that $T_{ij}(v_i) = v_i$ (namely, $T_{ii}$), but since we choose $i=1, ..., n$ and $j =1, ...,n$ separately, there are $n \times n$ such $T_{ij}$. To see that they are linearly independent, suppose that
$$ \sum_{i,j \in \{1,...,n\}} a_{ij}T_{ij} = 0 $$
is the zero map. Fixing $j$, this maps $v_j \mapsto \sum_{i=1}^n a_{ij}v_i = 0$, hence by linear independence of the basis, $a_{ij} = 0$.
Note that in all of the above we did not think of matrices at all, but only in terms of basis vectors of $V$. Of course, in finite dimensions these two approaches are equivalent, with $T_{ij}$ represented by the matrix with $1$ in the $(i,j)$-th place and zeros everywhere else.
