Certain Subalgebra of $\mathfrak{so}(4,\Bbb{C})$ is Cartan 
Let $\frak{h} \subseteq \frak{so}(4,\Bbb{C})$ be the subalgebra consisting of matrices of the form $$\begin{pmatrix} 0 & a & 0 & 0 \\ -a & 0 & 0 & 0 \\ 0 & 0 & 0 & b \\ 0 & 0 & -b & 0 \end{pmatrix}.$$ Show that $\frak{h}$ is a Cartan subalgebra and find the corresponding root decomposition. 

For now I'm just trying to show that $\frak{h}$ so defined is a Cartan subalgebra. Clearly it is commutative; i.e., clearly $[x,y] = 0$ for all $x,y \in \frak{h}$ which is equivalent to $xy=yx$ for all $x,y \in \frak{h}$, because the Lie bracket is just the matrix commutator. 
Now I am trying to show that every element is semisimple; that is, the adjoint map $\text{ad } x : \frak{so}(4,\Bbb{C}) \to \frak{so}(4,\Bbb{C})$ is a diagonalizable operator for every $x \in \frak{h}$. One way of showing that this operator is diagonalizable is to show that the sum of the dimension of the eigenspaces is the same as $\dim \frak{so}(4,\Bbb{C})$. But that seems a little difficult. What is $\dim \frak{so}(4,\Bbb{C})$? How does one describe the elements of $\frak{so}(4,\Bbb{C})$; does it have a nice basis? 
I haven't really thought about showing the centralizer of $\frak{h}$ equals $\frak{h}$, because I don't really know how to describe the elements of $\frak{so}(4,\Bbb{C})$. 
 A: Let $x=x_{a,b}$ be the matrix
$$\begin{pmatrix}0&a&0&0\\-a&0&0&0\\0&0&0&b\\0&0&-b&0\end{pmatrix}.$$
Treat $x$ as an element of $\mathfrak{gl}(4,\mathbb{C})=\operatorname{End}_{\Bbb C}(\mathbb{C}^4)$.  Then notice that $x$ is diagonalizable (as a linear operator), and so it is a semisimple linear operator.  Use this to show that $\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x$ is a semisimple linear operator on $\mathfrak{gl}(4,\mathbb{C})$.  However, $\mathfrak{so}(4,\mathbb{C})$ is an invariant subspace of $\mathfrak{gl}(4,\mathbb{C})$ under $\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x$ (as $x\in\mathfrak{so}(4,\mathbb{C})$ and $\mathfrak{so}(4,\mathbb{C})$ is a Lie subalgebra of $\mathfrak{gl}(4,\mathbb{C})$), so $$\operatorname{ad}_{\mathfrak{so}(4,\mathbb{C})}x=\left.\left(\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x\right)\right|_{\mathfrak{so}(4,\mathbb{C})}.$$
This shows that $\operatorname{ad}_{\mathfrak{so}(4,\mathbb{C})}x$ is a semisimple linear operator on $\mathfrak{so}(4,\mathbb{C})$.
In fact, you should be able to see that the eigenvalues of $x$ are $\pm ai$ and $\pm bi$.  Therefore, the eigenvalues of $\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x$ are 


*

*$2ai,-2ai,2bi,-2bi$ each with multiplicity $1$

*$ai+bi,ai-bi,-ai+bi,-ai-bi$ each with multiplicity $2$, and 

*$0$ with multiplicity $4$.


Intersecting the eigenspaces of $\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x$ with $\mathfrak{so}(4,\mathbb{C})$, we see that the $6$-dimensional subspace $\mathfrak{so}(4,\mathbb{C})$ of $\mathfrak{gl}(4,\mathbb{C})$ consists of the following eigenspaces


*

*$1$-dimensional eigenspaces corresponding to the eigenvalues $ai+bi,ai-bi,-ai+bi,-ai-bi$, and

*$2$-dimensional eigenspace corresponding to the eigenvalue $0$.


The $2$-dimensional eigenspace corresponding to the eigenvalue $0$ is precisely $\mathfrak{h}$ itself.  This implies that the centralizer of $\mathfrak{h}$ is $\mathfrak{h}$ itself.  (Technically speaking, you are proving that $\mathfrak{h}$ is a maximal toral subalgebra of $\mathfrak{so}(4,\mathbb{C})$.  However, for finite-dimensional semisimple Lie algebras, Cartan subalgebras are identical to maximal toral subalgebras.) 
For simplicity, write $\mathfrak{g}=\mathfrak{so}(4,\mathbb{C})$.  Let $\alpha$ and $\beta$ be the linear functionals in $\mathfrak{h}^*=\operatorname{Hom}_{\Bbb C}(\mathfrak{h},\Bbb C)$ such that $\alpha(x_{a,b})=ai$ and $\beta(x_{a,b})=bi$.  For $\gamma\in\mathfrak{h}^*$, define
$\mathfrak{g}_{\gamma}$ to be the intersection of the eigenspaces of $x_{a,b}$ corresponding to the eigenvalue $\gamma(x_{a,b})$.  Then,
$$\mathfrak{g}=\mathfrak{h}\oplus \bigoplus_{s,t=\pm1}\mathfrak{g}_{s\alpha+t\beta}$$
with $\mathfrak{h}=\mathfrak{g}_0$.  Note that each $\mathfrak{g}_{s\alpha+t\beta}$ for $s,t=\pm 1$ is a $1$-dimensional subspace of $\mathfrak{g}$.
