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> [!proposition] Proposition. ([[root space decomposition of sln(C)]])
> For $\mathfrak{g}=\mathfrak{sl}_{n}(\mathbb{C})$ and the [[Cartan subalgebra]] $\mathfrak{t}$ of [[diagonal matrix|diagonal matrices]], define $e_{i} \in \mathfrak{t}^{*}$ by $e_{i} (\begin{bmatrix}
> t_{1} & & \\
> & \ddots & \\
> & & t_{n}
> \end{bmatrix}) = t_{i}.$
> Then the [[root space decomposition of a Lie algebra|root space decomposition]] of $\mathfrak{g}=\mathfrak{sl}_{n}(\mathbb{C})$ is given by $\mathfrak{g}=\mathfrak{t} \oplus \bigoplus_{i \neq j} \mathfrak{g}_{e_{i}- e_{j}}$
> where $e_{i}-e_{j} \in \mathfrak{t}^{*}$ are the [[simultaneously diagonalizable|simultaneous eigenvalues]] corresponding to simultaneous [[eigenvector|eigenvectors]] $E_{ij}$ (in other words, the standard basis of the [[special linear Lie subalgebra]] is a [[simultaneously diagonalizable|simultaneous]] [[eigenbasis]] for the adjoint action of $\mathfrak{t}$).
>[!proof]- Proof. ([[root space decomposition of sln(C)]])
>
> Examine [[adjoint representation|adjoint action]] of $\mathfrak{t}$. Given $t \in \mathfrak{t}$, compute $\begin{align}
> t \cdot E_{ij} &= [t, E_{ij}] \\
> &= t_{i} E_{ij} - t_{j} E_{ij} \\
> & \big( e_{i}(t) - e_{j}(t) \big) E_{ij}.
> \end{align}$
> And $\begin{align}
> t \cdot (E_{i i}- E_{i+1, i+1}) & = t \cdot E_{i i} - t \cdot E_{i+1, i+1} \\
> &= 0 .
> \end{align}$
> This gives $\mathfrak{g}_{0}=\mathfrak{t}$ a simult. eigenbasis of the $n-1$ basis vectors $E_{ii}-E_{i+1,i+1}$ and otherwise get $n^{2}-n$ root spaces $\mathfrak{g}_{e_{i}-e_{j}}$, $i \neq j$, corresponding to the basis vectors $E_{ij}$.
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####
- [[special linear Lie subalgebra]]
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
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> GROUP BY Tag
> ```
> [!frontlink]
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> GROUP BY Tag
> ```