Show $dim([\mathfrak{g_{\alpha}},\mathfrak{g_{-\alpha}}])=1$

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I have a problem by understanding a proof in the lecture of lie algebras. In fact we want to show that $[\mathfrak{g_{\alpha}},\mathfrak{g_{-\alpha}}]$ has dimension $1$.Where $\mathfrak{g}$ is a semisimple lie algebra and $\alpha$ is a root.

Let $x \in \mathfrak{g_{\alpha}}$, $y \in \mathfrak{g_{-\alpha}}$ and $h \in \mathfrak{h}$ where $\mathfrak{h}$ is a Cartan subalgebra. Since the Killing $B$ form is invariant we have

$B(h,[x,y])=B([h,x],y)=\alpha(h) B(x,y)$

Therefore we have $ker(\alpha) \subset [\mathfrak{g_{\alpha}},\mathfrak{g_{-\alpha}}]^{\bot} $ and we conclude that $dim([\mathfrak{g_{\alpha}},\mathfrak{g_{-\alpha}}])=1$ . Someone has an answer to this?

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The first argument with the Killing form shows that $\dim \, [\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}] \ge 1$, because $[x_{\alpha},x_{-\alpha}]$ is a nonzero element in $[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]$. A second Lemma, using Lie's Theorem shows that $[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]\cap \ker (\alpha)=0$. Then we have, with Cartan subalgebra $\mathfrak{h}$, \begin{align*} \dim \, [\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}] & = \dim ([\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]+\ker (\alpha)) + \dim ([\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]\cap \ker (\alpha)) - \dim \ker (\alpha) \\ & \le \dim \mathfrak{h} + 0 - (\dim \mathfrak{h} -1) \\ & = 1. \end{align*}