[proofplan] The proof uses the standard structure theorem for Cartan subalgebras of finite-dimensional semisimple Lie algebras over algebraically closed fields of characteristic $0$: such a Cartan subalgebra is abelian, consists of semisimple elements, and equals its own zero weight space. Since the commuting operators $\operatorname{ad}(h)$ are diagonalizable, they are simultaneously diagonalizable on $\mathfrak{g}$, giving a direct sum of common eigenspaces. The zero common eigenspace is exactly $\mathfrak{h}$, and the roots span $\mathfrak{h}^*$ because any element of $\mathfrak{h}$ annihilated by all roots would lie in the center of the semisimple Lie algebra. [/proofplan]

[step:Use semisimplicity to make the Cartan subalgebra toral]We use the following standard structure result for Cartan subalgebras in semisimple Lie algebras over algebraically closed fields of characteristic $0$: if $\mathfrak{g}$ is finite-dimensional and semisimple, and $\mathfrak{h}\subset\mathfrak{g}$ is a Cartan subalgebra, then $\mathfrak{h}$ is abelian, every element of $\mathfrak{h}$ is semisimple in the adjoint representation, and \begin{align*} \{x\in\mathfrak{g} : [h,x]=0 \text{ for every } h\in\mathfrak{h}\} = \mathfrak{h}. \end{align*} (citing a result not yet in the wiki: Cartan subalgebras of semisimple Lie algebras are toral and self-centralizing) Thus, for every $h\in\mathfrak{h}$, the [linear map](/page/Linear%20Map) \begin{align*} \operatorname{ad}(h):\mathfrak{g}&\to\mathfrak{g}\\ x&\mapsto [h,x] \end{align*} is diagonalizable over $k$.[/step]

[guided]The point of the semisimplicity hypothesis is that Cartan subalgebras behave like maximal diagonal subalgebras. We use the standard theorem saying that, over an algebraically closed field of characteristic $0$, a Cartan subalgebra $\mathfrak{h}$ of a finite-dimensional semisimple Lie algebra $\mathfrak{g}$ is abelian and consists of adjoint-semisimple elements. In concrete terms, this means that for each $h\in\mathfrak{h}$ the operator \begin{align*} \operatorname{ad}(h):\mathfrak{g}&\to\mathfrak{g}\\ x&\mapsto [h,x] \end{align*} is diagonalizable over $k$. The same structure theorem also gives the self-centralizing property \begin{align*} \{x\in\mathfrak{g} : [h,x]=0 \text{ for every } h\in\mathfrak{h}\} = \mathfrak{h}. \end{align*} This identity will later identify the zero root space with $\mathfrak{h}$. Without this input, simultaneous diagonalization would still produce a zero eigenspace, but it would not identify that eigenspace with the Cartan subalgebra.[/guided]

[step:Simultaneously diagonalize the adjoint action of $\mathfrak{h}$]For $h_1,h_2\in\mathfrak{h}$, the Jacobi identity and $[h_1,h_2]=0$ give \begin{align*} [\operatorname{ad}(h_1),\operatorname{ad}(h_2)] = \operatorname{ad}([h_1,h_2]) = 0. \end{align*} Thus the family \begin{align*} \{\operatorname{ad}(h):h\in\mathfrak{h}\}\subset \operatorname{End}_k(\mathfrak{g}) \end{align*} is a commuting family of diagonalizable linear operators on the finite-dimensional $k$-[vector space](/page/Vector%20Space) $\mathfrak{g}$. Since $k$ is algebraically closed, the standard simultaneous diagonalization theorem for commuting diagonalizable operators applies. Hence $\mathfrak{g}$ decomposes as a direct sum of common eigenspaces: \begin{align*} \mathfrak{g} = \bigoplus_{\alpha\in\mathfrak{h}^*} \mathfrak{g}_{\alpha}, \end{align*} where only finitely many summands are nonzero, and where \begin{align*} \mathfrak{g}_{\alpha} = \{x\in\mathfrak{g}:[h,x]=\alpha(h)x\text{ for every }h\in\mathfrak{h}\}. \end{align*}[/step]

[guided]Because $\mathfrak{h}$ is abelian, the adjoint operators attached to its elements commute. Indeed, for $h_1,h_2\in\mathfrak{h}$, the commutator of the two endomorphisms $\operatorname{ad}(h_1)$ and $\operatorname{ad}(h_2)$ is computed by the Jacobi identity: \begin{align*} [\operatorname{ad}(h_1),\operatorname{ad}(h_2)] = \operatorname{ad}([h_1,h_2]). \end{align*} Since $[h_1,h_2]=0$, this gives \begin{align*} [\operatorname{ad}(h_1),\operatorname{ad}(h_2)] = 0. \end{align*} We already know that each $\operatorname{ad}(h)$ is diagonalizable over $k$. Therefore the family \begin{align*} \{\operatorname{ad}(h):h\in\mathfrak{h}\} \end{align*} is a commuting family of diagonalizable endomorphisms of the finite-dimensional vector space $\mathfrak{g}$. The simultaneous diagonalization theorem says that such a family admits a basis of common eigenvectors. Equivalently, $\mathfrak{g}$ is the direct sum of the common eigenspaces \begin{align*} \mathfrak{g}_{\alpha} = \{x\in\mathfrak{g}:[h,x]=\alpha(h)x\text{ for every }h\in\mathfrak{h}\}, \end{align*} where $\alpha$ ranges over the linear functionals on $\mathfrak{h}$ that occur as joint eigenvalues. Therefore \begin{align*} \mathfrak{g} = \bigoplus_{\alpha\in\mathfrak{h}^*}\mathfrak{g}_{\alpha}. \end{align*} Only finitely many summands are nonzero because $\mathfrak{g}$ is finite-dimensional.[/guided]

[step:Identify the zero root space with the Cartan subalgebra]By definition, \begin{align*} \mathfrak{g}_0 = \{x\in\mathfrak{g}:[h,x]=0\text{ for every }h\in\mathfrak{h}\}. \end{align*} The self-centralizing property of Cartan subalgebras in semisimple Lie algebras gives \begin{align*} \mathfrak{g}_0=\mathfrak{h}. \end{align*} Separating the zero functional from the nonzero functionals with nonzero eigenspace gives \begin{align*} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha\in\Phi}\mathfrak{g}_{\alpha}. \end{align*} This proves the root space decomposition.[/step]

[guided]The zero functional $0\in\mathfrak{h}^*$ gives the common eigenspace \begin{align*} \mathfrak{g}_0 = \{x\in\mathfrak{g}:[h,x]=0\text{ for every }h\in\mathfrak{h}\}. \end{align*} This is exactly the centralizer of $\mathfrak{h}$ in $\mathfrak{g}$. The structure theorem for Cartan subalgebras in semisimple Lie algebras says that this centralizer is $\mathfrak{h}$ itself: \begin{align*} \mathfrak{g}_0=\mathfrak{h}. \end{align*} The remaining nonzero common eigenspaces are indexed precisely by \begin{align*} \Phi = \{\alpha\in\mathfrak{h}^*\setminus\{0\}:\mathfrak{g}_{\alpha}\neq\{0\}\}. \end{align*} Thus the simultaneous eigenspace decomposition becomes \begin{align*} \mathfrak{g} = \mathfrak{g}_0 \oplus \bigoplus_{\alpha\in\Phi}\mathfrak{g}_{\alpha} = \mathfrak{h} \oplus \bigoplus_{\alpha\in\Phi}\mathfrak{g}_{\alpha}. \end{align*} Because the decomposition came from simultaneous diagonalization, the sum is direct.[/guided]

[step:Show that the roots span the dual of the Cartan subalgebra]Let \begin{align*} V = \operatorname{span}_k(\Phi)\subseteq\mathfrak{h}^* \end{align*} be the $k$-linear span of the roots. Suppose $V\neq\mathfrak{h}^*$. Since $\mathfrak{h}$ is finite-dimensional, there exists a nonzero element $h_0\in\mathfrak{h}$ such that \begin{align*} \alpha(h_0)=0 \end{align*} for every $\alpha\in\Phi$. For $x\in\mathfrak{h}$, we have $[h_0,x]=0$ because $\mathfrak{h}$ is abelian. For $x\in\mathfrak{g}_{\alpha}$ with $\alpha\in\Phi$, the definition of $\mathfrak{g}_{\alpha}$ gives \begin{align*} [h_0,x]=\alpha(h_0)x=0. \end{align*} Using the direct sum decomposition \begin{align*} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha\in\Phi}\mathfrak{g}_{\alpha}, \end{align*} it follows that $[h_0,y]=0$ for every $y\in\mathfrak{g}$. Define the center of $\mathfrak{g}$ by \begin{align*} Z(\mathfrak{g}) = \{z\in\mathfrak{g}:[z,y]=0\text{ for every }y\in\mathfrak{g}\}. \end{align*} Then $h_0\in Z(\mathfrak{g})$. Since $\mathfrak{g}$ is semisimple, its center is zero. Therefore $h_0=0$, contradicting the choice of $h_0\neq 0$. Hence $V=\mathfrak{h}^*$, so $\Phi$ spans $\mathfrak{h}^*$.[/step]

[guided]We now prove that the roots do not miss any direction in $\mathfrak{h}^*$. Define \begin{align*} V = \operatorname{span}_k(\Phi)\subseteq\mathfrak{h}^*. \end{align*} Assume, for contradiction, that $V\neq\mathfrak{h}^*$. Since $\mathfrak{h}$ is finite-dimensional, this means there is a nonzero vector $h_0\in\mathfrak{h}$ annihilated by every functional in $V$. In particular, \begin{align*} \alpha(h_0)=0 \end{align*} for every root $\alpha\in\Phi$. We show that $h_0$ commutes with every element of $\mathfrak{g}$. First, if $x\in\mathfrak{h}$, then \begin{align*} [h_0,x]=0 \end{align*} because $\mathfrak{h}$ is abelian. Second, if $x\in\mathfrak{g}_{\alpha}$ for some $\alpha\in\Phi$, then the defining property of the root space gives \begin{align*} [h_0,x]=\alpha(h_0)x. \end{align*} But $\alpha(h_0)=0$, so \begin{align*} [h_0,x]=0. \end{align*} The root space decomposition already proved says that every $y\in\mathfrak{g}$ has a unique decomposition \begin{align*} y = x_0+\sum_{\alpha\in\Phi}x_{\alpha}, \end{align*} where $x_0\in\mathfrak{h}$, $x_{\alpha}\in\mathfrak{g}_{\alpha}$, and all but finitely many $x_{\alpha}$ are zero. By bilinearity of the Lie bracket, \begin{align*} [h_0,y] = [h_0,x_0]+\sum_{\alpha\in\Phi}[h_0,x_{\alpha}] = 0. \end{align*} Define the center of $\mathfrak{g}$ by \begin{align*} Z(\mathfrak{g}) = \{z\in\mathfrak{g}:[z,y]=0\text{ for every }y\in\mathfrak{g}\}. \end{align*} Thus $h_0\in Z(\mathfrak{g})$. A semisimple Lie algebra has zero center. To justify this, define the solvable radical of $\mathfrak{g}$ to be the largest solvable ideal of $\mathfrak{g}$. If $z\in Z(\mathfrak{g})$, then the one-dimensional subspace $kz$ is an ideal because $[y,z]=0\in kz$ for every $y\in\mathfrak{g}$. It is abelian, hence solvable, so it is contained in the solvable radical of $\mathfrak{g}$. Semisimplicity means that this solvable radical is zero. Therefore $Z(\mathfrak{g})=\{0\}$, and so $h_0=0$. This contradicts the construction of $h_0$ as a nonzero element of $\mathfrak{h}$. Hence $V=\mathfrak{h}^*$.[/guided]

[step:Collect the diagonal action and the decomposition] The simultaneous diagonalization step shows that every operator $\operatorname{ad}(h)$ with $h\in\mathfrak{h}$ acts diagonally on the decomposition \begin{align*} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha\in\Phi}\mathfrak{g}_{\alpha}. \end{align*} Indeed, $\operatorname{ad}(h)$ acts by the scalar $0$ on $\mathfrak{h}=\mathfrak{g}_0$ and by the scalar $\alpha(h)$ on each root space $\mathfrak{g}_{\alpha}$. The preceding steps prove $\mathfrak{g}_0=\mathfrak{h}$, the displayed direct sum decomposition, diagonalizability of every $\operatorname{ad}(h)$, and spanning of $\mathfrak{h}^*$ by $\Phi$. [/step]