[proofplan] The proof uses the [invariance of the Killing form](/theorems/3808) to compare $\kappa([h,x],y)$ with $\kappa(x,[h,y])$ for $x$ and $y$ lying in root spaces. Since $h$ acts on $\mathfrak{g}_{\alpha}$ and $\mathfrak{g}_{\beta}$ by the scalars $\alpha(h)$ and $\beta(h)$, invariance forces $(\alpha+\beta)(h)\kappa(x,y)=0$ for every $h \in \mathfrak{h}$. If $\alpha+\beta$ is not the zero functional, this gives the desired orthogonality. Finally, an element of $\mathfrak{h}$ orthogonal to $\mathfrak{h}$ is also orthogonal to every root space, hence to all of $\mathfrak{g}$; the nondegeneracy of the Killing form on the semisimple Lie algebra then forces that element to be zero. [/proofplan] [step:Use invariance of the Killing form to relate the two root eigenvalues] Let $x \in \mathfrak{g}_{\alpha}$, let $y \in \mathfrak{g}_{\beta}$, and let $h \in \mathfrak{h}$. Since $x \in \mathfrak{g}_{\alpha}$ and $y \in \mathfrak{g}_{\beta}$, the defining root-space identities give \begin{align*} [h,x] &= \alpha(h)x, & [h,y] &= \beta(h)y. \end{align*} The Killing form is invariant, so for all $a,b,c \in \mathfrak{g}$ one has \begin{align*} \kappa([a,b],c)=\kappa(a,[b,c]). \end{align*} Applying this with $a=h$, $b=x$, and $c=y$ gives \begin{align*} \kappa([h,x],y)=\kappa(h,[x,y]). \end{align*} Applying the same invariance identity with $a=x$, $b=h$, and $c=y$ gives \begin{align*} \kappa([x,h],y)=\kappa(x,[h,y]). \end{align*} Since $[x,h]=-[h,x]$, we obtain \begin{align*} \kappa([h,x],y)=-\kappa(x,[h,y]). \end{align*} Substituting the root-space identities into both sides yields \begin{align*} \alpha(h)\kappa(x,y) &= \kappa(\alpha(h)x,y) \\ &= \kappa([h,x],y) \\ &= -\kappa(x,[h,y]) \\ &= -\kappa(x,\beta(h)y) \\ &= -\beta(h)\kappa(x,y). \end{align*} Therefore \begin{align*} (\alpha(h)+\beta(h))\kappa(x,y)=0 \end{align*} for every $h \in \mathfrak{h}$. [guided] We want to prove that $x$ and $y$ are orthogonal under $\kappa$. The only information we have about $x$ and $y$ is how elements of $\mathfrak{h}$ bracket with them, so we insert an arbitrary $h \in \mathfrak{h}$ and use invariance of the Killing form. Because $x \in \mathfrak{g}_{\alpha}$ and $y \in \mathfrak{g}_{\beta}$, the maps \begin{align*} \operatorname{ad}_h|_{\mathfrak{g}_{\alpha}}: \mathfrak{g}_{\alpha} &\to \mathfrak{g}_{\alpha}, & u &\mapsto [h,u], \end{align*} and \begin{align*} \operatorname{ad}_h|_{\mathfrak{g}_{\beta}}: \mathfrak{g}_{\beta} &\to \mathfrak{g}_{\beta}, & v &\mapsto [h,v], \end{align*} act by scalar multiplication with eigenvalues $\alpha(h)$ and $\beta(h)$ respectively. Thus \begin{align*} [h,x] &= \alpha(h)x, & [h,y] &= \beta(h)y. \end{align*} The Killing form is invariant: for all $a,b,c \in \mathfrak{g}$, \begin{align*} \kappa([a,b],c)=\kappa(a,[b,c]). \end{align*} Applying this once with $(a,b,c)=(h,x,y)$ gives \begin{align*} \kappa([h,x],y)=\kappa(h,[x,y]). \end{align*} Applying it again with $(a,b,c)=(x,h,y)$ gives \begin{align*} \kappa([x,h],y)=\kappa(x,[h,y]). \end{align*} Since the Lie bracket is skew-symmetric, $[x,h]=-[h,x]$. Hence \begin{align*} -\kappa([h,x],y)=\kappa(x,[h,y]), \end{align*} or equivalently \begin{align*} \kappa([h,x],y)=-\kappa(x,[h,y]). \end{align*} Now substitute the two eigenvalue identities: \begin{align*} \alpha(h)\kappa(x,y) &= \kappa(\alpha(h)x,y) \\ &= \kappa([h,x],y) \\ &= -\kappa(x,[h,y]) \\ &= -\kappa(x,\beta(h)y) \\ &= -\beta(h)\kappa(x,y). \end{align*} Moving the right-hand term to the left gives \begin{align*} (\alpha(h)+\beta(h))\kappa(x,y)=0. \end{align*} This identity holds for every $h \in \mathfrak{h}$. [/guided] [/step] [step:Choose an element of the Cartan subalgebra detecting $\alpha+\beta$] Assume $\alpha+\beta \ne 0$ as a functional in $\mathfrak{h}^*$. Then there exists $h_0 \in \mathfrak{h}$ such that \begin{align*} (\alpha+\beta)(h_0)\ne 0. \end{align*} Applying the identity from the previous step with $h=h_0$ gives \begin{align*} (\alpha(h_0)+\beta(h_0))\kappa(x,y)=0. \end{align*} Since $\alpha(h_0)+\beta(h_0)=(\alpha+\beta)(h_0)\ne 0$, division in $\mathbb{C}$ gives \begin{align*} \kappa(x,y)=0. \end{align*} Because $x \in \mathfrak{g}_{\alpha}$ and $y \in \mathfrak{g}_{\beta}$ were arbitrary, this proves \begin{align*} \kappa(\mathfrak{g}_{\alpha},\mathfrak{g}_{\beta})=0 \end{align*} whenever $\alpha+\beta \ne 0$. [/step] [step:Use invariance again to make the radical vector orthogonal to every nonzero root space] Let $z \in \mathfrak{h}$ satisfy \begin{align*} \kappa(z,h)=0 \end{align*} for every $h \in \mathfrak{h}$. We prove directly that $z$ is orthogonal to every nonzero root space, without treating $0$ as a root. Let $\alpha \in \mathfrak{h}^*$ be a nonzero root, and let $x \in \mathfrak{g}_{\alpha}$. Since $\alpha$ is a nonzero linear functional on $\mathfrak{h}$, there exists $h_{\alpha} \in \mathfrak{h}$ such that \begin{align*} \alpha(h_{\alpha}) \ne 0. \end{align*} The Cartan subalgebra $\mathfrak{h}$ is abelian, so $[z,h_{\alpha}]=0$. The root-space identity for $x \in \mathfrak{g}_{\alpha}$ gives \begin{align*} [h_{\alpha},x]=\alpha(h_{\alpha})x. \end{align*} Using invariance of the Killing form with $a=z$, $b=h_{\alpha}$, and $c=x$, we obtain \begin{align*} 0 &= \kappa([z,h_{\alpha}],x) \\ &= \kappa(z,[h_{\alpha},x]) \\ &= \kappa(z,\alpha(h_{\alpha})x) \\ &= \alpha(h_{\alpha})\kappa(z,x). \end{align*} Since $\alpha(h_{\alpha}) \ne 0$, division in $\mathbb{C}$ gives \begin{align*} \kappa(z,x)=0. \end{align*} Thus $z$ is orthogonal to $\mathfrak{h}$ by hypothesis and to every nonzero root space by the direct invariance argument above. [/step] [step:Use the root space decomposition and nondegeneracy of $\kappa$ on $\mathfrak{g}$] By the Cartan root space decomposition for a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$, the [vector space](/page/Vector%20Space) $\mathfrak{g}$ decomposes as the direct sum \begin{align*} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi}\mathfrak{g}_{\alpha}, \end{align*} where $\Phi \subset \mathfrak{h}^*$ denotes the set of nonzero roots. From the previous step, $\kappa(z,w)=0$ for every $w \in \mathfrak{h}$ and for every $w \in \mathfrak{g}_{\alpha}$ with $\alpha \in \Phi$. By bilinearity of $\kappa$ and the direct-sum decomposition, it follows that \begin{align*} \kappa(z,w)=0 \end{align*} for every $w \in \mathfrak{g}$. Since $\mathfrak{g}$ is semisimple, Cartan's criterion for semisimplicity gives that the Killing form $\kappa$ is nondegenerate on $\mathfrak{g}$. Therefore the only element of $\mathfrak{g}$ orthogonal to all of $\mathfrak{g}$ is $0$, so $z=0$. We have shown that the radical of $\kappa|_{\mathfrak{h}\times\mathfrak{h}}$ is zero. Hence the restricted [bilinear form](/page/Bilinear%20Form) \begin{align*} \kappa|_{\mathfrak{h}\times\mathfrak{h}}: \mathfrak{h}\times\mathfrak{h}\to \mathbb{C} \end{align*} is nondegenerate. [/step]