$\alpha$ is a root $\implies -\alpha$ is a root

Question

Let $\mathfrak{h}$ be a Cartan subalgebra of Lie algebra $\mathfrak{g}$.

I want to prove:

$\alpha\in \mathfrak{h}^*$ be a root of $\frak g$, $\implies$ so is $-\alpha$.

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Let $\mathfrak{h}$ have basis $\{h_1,\cdots,h_r\}$

I am not really sure how to prove this, I mean $\alpha$ being a root, means that $e_\alpha \in \frak g \setminus h$ have $ad_{h_i}(e_\alpha) = \alpha_i e_\alpha$ and we take the eigenvalues $\alpha_i$ and form the root:

$$(\alpha_1,\cdots,\alpha_r)$$

So I need to prove that:

$$(-\alpha_1,\cdots,-\alpha_r)$$

is a root, which feels like I need to show that $e_{-\alpha}$ is a vector in $\frak g\backslash h$, which I don't know how to really 'show'.

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Is this as simple as: $ad_{h_i}(-e_\alpha)=[h_i,-e_\alpha]=-[h_i,e_\alpha]=-\alpha_i e_\alpha$ for each $h_i$, $i\in \{1,\cdots,r\}$:

$$(-\alpha_1,\cdots,-\alpha_r)$$

As desired?

Answer 1

It is easy to show it using the fact that:

If $\alpha,\beta \in \Delta,\ \alpha+\beta\neq 0$ then $K(e_\alpha,e_\beta)=0$

where $\Delta$ are the roots and $K$ the killing form.