# 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?

• please specify on which space does your linear operator act? – GregVoit Mar 31 '15 at 7:03
• @GregVoit Done sorry, it is sl_2 to sl_2 – Topologist's student Mar 31 '15 at 7:08
• Well, in that case your vector space basis is a matrix basis. Matrix space is a vector space. Your operator acts on $X\in sl(2,\mathbb{C})$, therefore $X$ is a matrix – GregVoit Mar 31 '15 at 7:13
• @GregVoit Thank you, that is what I expected – Topologist's student Mar 31 '15 at 7:13

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 $A$.

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 $\mathbb R^3$.