Is there a natural way to define an ordered basis for the vector space of $n\times n$ matrices over some field $K$?
$E^{ij}$ be the matrix which has the entry $1$ in $ij$ and $0$ in every other place. Then,
${ B }_{ 1 }=\{ { E }^{ 11 },{ E }^{ 12 },..,{ E }^{ 1n },{ E }^{ 21 },..{ ,E }^{ 2n },..{ E }^{ nn }\} \\ { B }_{ 2 }=\{ { E }^{ 11 },{ E }^{ 21 },..,{ E }^{ n1 },{ E }^{ 12 },..,{ E }^{ n2 },..{ E }^{ nn }\} $
I have seen that in some cases $B_1$ is clearly a better choice than $B_2$. Since the number of basis vectors is $n\times n$ is there a choice that can be made so that the situation can be made easier?
Actually, a base is almost independent from the order in which you place vectors. You notice it from the fact that, given a matrix $A \in \mathbb{R}^{n,n}$, then
$$ A = \sum_{i=1}^n \sum_{j=1}^n a_{ij} E^{i,j} = \sum_{j=1}^n \sum_{i=1}^n a_{ij} E^{i,j} $$
so the answer is no. You usually care more about the elements in a base, rather than their order