Question abour linear maps proofs As example, consider the following statement: Let $V$ and $W$ be finite-dimensional with $\dim V\ge\dim W\ge 2$. Show that $A=\{T\in\mathcal{L}(V,W):T\ \text{not surjective}\}$ is not a subspace of $\mathcal{L}(V,W)$.
To prove the statement, we can pass by considering a basis of $V$ and $W$ and construct linear applications "sending" the basis of $V$ to $W$. My question is the following:
Why do we pass by it? Why we can't construct linear maps considering only elements of each vector space (to prove this type of statements)? My question might be vague, but I don't really understand why considering basis's it proves, let's say, the "general" case? Of course, knowing the basis of vector space we can construct each element of that vector space. But I am quite confused about the fact that we "send" a basis to another one... If someone could give a clear explanation of this, I would really appreciate it. Thank you in advance!
 A: WLOG, take $V = k^n$ and $W = k^m$ where $k$ is the underlying field, with basis vectors $e_i$ as usual.
Consider the linear maps $f_i : V \to W$ defined by $f_i(e_i) = e_i$ and $f_i(e_j) = 0$ for all $j \neq i$. These maps are not surjective because the image of $f_i$ is the 1-dimensional vector space spanned by $e_i$, and $\dim(W) \geq 2 > 1$, so the image of $f_i$ cannot be all of $W$.
Note: the maps $f_i$ are only well-defined because $n \geq m$. For suppose that $n < m$. Then one defining equation for $f_m$ is $f_m(e_m) = e_m$. But $e_m$ is not a basis vector of $k^n$, since $n < m$. I was advised to make my use of the assumption $\dim(V) \geq \dim(W)$ explicit, so I have done so.
But now consider $\sum\limits_{i = 1}^m f_i$. This map is indeed surjective.
Thus, the set of non-surjective maps is not closed under addition and is not a subspace of $\mathcal{L}(V, W)$.
Your question is why we can assume, WLOG, that $V = k^n$ and $W = k^m$. This is because every $j$-dimensional vector space is isomorphic to $k^j$.
Why do we specify a basis? Because it makes the math easier. Once we pick the basis, it's easy to define linear maps in terms of the basis. Once we can specify maps in terms of the basis, it's easy to find non-surjective linear maps whose combination is surjective.
How would you know from looking at the problem that you should use a basis? The fact that the problem specifies "finite-dimensional" is a dead giveaway. "Finite-dimensional" means "has a finite basis". We can't use the fact that the vector spaces are finite-dimensional unless we somehow make use of their finite bases.
