Standard basis of a Matrix with identical entries. How would you represent a $\mathbb{R}$-Matrixspace which looks like this
$$\begin{bmatrix} a & b \\ b & a \end{bmatrix}$$
with standard basis
I can't think of anything else but
$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \quad \mbox{and} \quad 
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
but can we then call that a standard basis, because not all entries except one of them 
is 0?
 A: $$W = \left\{ \begin{bmatrix} a & b \\ b & a \end{bmatrix} \;\Big|\; a,b \in \mathbb{R} \right\}$$
is indeed a subspace of $M_2(\mathbb{R})=\mathbb{R}^{2 \times 2}$ the space of all $2 \times 2$ real matrices.
While $\mathbb{R}^{2 \times 2}$ has a "standard basis" of matrices... 
$$E_{11} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad
E_{12} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad
E_{21} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \quad \mbox{and} \quad
E_{22} = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$
...if you pick some random subspace (like your $W$) there is no general notion of a "standard basis". 
Your proposal of $E_{11}+E_{22}=I_2=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and
$E_{12}+E_{21}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ is a nice choice of basis. It's as "standard" as you can hope for -- but wouldn't really necessarily be called that.
In general, some vector spaces have a basis which is accepted as "the standard basis" but not all vector spaces have such a thing. Even in such a case, the way the basis is ordered (which effects coordinate vectors) isn't standardized.
For example: $\mathbb{R}^3$ has the standard basis ${\bf i}={\bf e}_1=(1,0,0)$, ${\bf j}={\bf e}_2=(0,1,0)$, and ${\bf k}={\bf e}_3=(0,0,1)$. In fact, the order ${\bf i}$ then ${\bf j}$ then ${\bf k}$ is the standard order.
On the other hand, $\mathbb{R}^{2 \times 2}$ has the standard basis of $E_{ij}$'s as mentioned above. But there is no universally accepted "standard" ordering for this basis.
Another common vector space with a standard basis is the algebra polynomials: $\mathbb{R}[x]$. Here the standard basis  is $\{1,x,x^2,\dots\}$.
But take some vector space like $W = \{ (a,b,c) \;|\; a+2b+3c=0 \}$ (a subspace of $\mathbb{R}^3$) and while $W$ has a basis (in fact infinitely many bases) it doesn't have a "standard basis".
