Question:
For every pair of linearly independent, $n\times n $, skew-Hermitian matrices $A, B\in \mathfrak{su}(n)$ with $[A,B]\neq 0$, does there exist a basis that spans $A$ and $B$ over the real numbers but which also satisfies the commutation relations of $\mathfrak{su}(2)$ (or $\mathfrak{u}(2)$)? If so, how do we construct it?
I feel like there is a simple answer that I am not seeing.
What I have tried:
- Finding a basis $\{e_1,e_2\}$ for $\mathrm{span}(\{A,B\})$ alone can be done via Gram-Schmidt orthogonalization, but this basis will clearly not span $\mathfrak{su}(2)$.
- Realize that $\mathfrak{su}(n) = \mathfrak{h}\oplus\mathfrak{h_+}\oplus\mathfrak{h_-}$. I can pick $A$ to be in $\mathfrak{h}$ and then expand $B$ in a basis for $\mathfrak{h}$, $\mathfrak{h_+}$ and $\mathfrak{h_-}$. I think there exists a basis where $B$ only has support on a single element of both $\mathfrak{h}_\pm$, but I can't show that this is the case. If I did however, I think those 3 elements could all be chosen so that they satisfy $\mathfrak{su}(2)$'s commutation relations.
It appears that you are asking the following:
Is it true that for every pair of noncommuting elements $A, B\in su(n)$, the linear span $Span(\{A,B\})$ is contained in a subalgebra isomorphic to $su(2)$?
The answer to this question is negative (unless $n=2$ of course). Namely, take two elements $A, B\in su(n)$ which generate $su(n)$ (i.e. the smallest Lie subalgebra of $su(n)$ containing $A, B$ equals the entire $su(n)$). For the existence of such $A, B$, see here. Clearly, such $A, B$ cannot commute, otherwise they generate a commutative subalgebra.
Since $A, B$ generate $su(n)$ and $su(n)$ is not isomorphic to $su(2)$ (unless $n=2$), we conclude that $\{A, B\}$ is not contained in a subalgebra isomorphic to $su(2)$.