Basis 'vectors', basis 'matrices'? Let $\mathfrak{sl}_2$ be the vector space of $2\times 2$ traceless matrices. Let $A\in \mathfrak{sl}_2$ be a diagonal matrix. Define a linear operator:
$$\phi_A(X)=AX-XA$$
$\phi_A:sl_2\to sl_2$
What are the basis vectors of $X$? I think I have basis matrices, are these the same?
The basis I have is $$\left\{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix}\right\}$$
Are these basis 'vectors' or basis matrices, are they the same things?
 A: The elements of a vector space are called vectors.  It happens that in this case, your $3$-dimensional vector space $\mathfrak{sl}_2(\Bbb{C})$ is made of $2 \times 2$ matrices over $\Bbb{C}$.  Perhaps it is useful to name the vectors in your basis:
$$
e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},
\quad
h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},
\quad
f = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}.
$$
(These are standard names, by the way.)  Now, to represent your transformation $\phi_A: \mathfrak{sl}_2 \to \mathfrak{sl}_2$ in this basis, you have to fill in the entries of a $3 \times 3$ matrix in the usual way:  the columns will be $\phi_A(e)$, $\phi_A(h)$, and $\phi_A(f)$ expanded as linear combinations of $e$, $h$, and $f$.  The actual entries will, of course, depend on $A$.
A: There is no such thing as "a base vector of $X$" in your case. The only place $X$ appears in in your example is in the definition of $\phi_A$, so it is only a variable. Saying what the basis of $X$ is is like saying 

If $f(x)=x^2$, what are the prime divisors of $x$?

What you probably need is the basis of $sl_2$. Now, $sl_2$ is a 3 dimensional vector space (because it is the null-space of a rank-one operator on the space of all $2\times 2$ matrices, which is a 4 dimensional space), meaning that what you found is the basis of the vector space $sl_2$.
Now, are the elements of this basis (and, indeed, $sl_2$) vectors or matrices?
The answer is both. 


*

*Because $sl_2$ is a space of matrices with some property, all elements of $sl_2$ are, by definition, matrices.

*Because $sl_2$ is a vector space, you can also say that its elements are vectors. Note, however, that the vectors are vectors of a three-dimensional space, meaning that $sl_2$ is isomorphic to $\mathbb R^3$.

