[proofplan] We prove directly that the adjoint map is a Lie algebra homomorphism. First, the bilinearity of the Lie bracket shows that each $\operatorname{ad}_x$ is a $k$-linear endomorphism of $\mathfrak g$ and that $x \mapsto \operatorname{ad}_x$ is itself $k$-linear. Then we compute the commutator $[\operatorname{ad}_x,\operatorname{ad}_y]$ on an arbitrary element $z \in \mathfrak g$ and use the Jacobi identity to identify it with $\operatorname{ad}_{[x,y]}$. [/proofplan]

[step:Verify that the adjoint map is a well-defined linear map into $\mathfrak{gl}(\mathfrak g)$] Fix $x \in \mathfrak g$. Define \begin{align*} \operatorname{ad}_x: \mathfrak g &\to \mathfrak g \\ y &\mapsto [x,y]. \end{align*} Since the Lie bracket $[\cdot,\cdot]: \mathfrak g \times \mathfrak g \to \mathfrak g$ is $k$-bilinear, the map $\operatorname{ad}_x$ is $k$-linear in the variable $y$. Hence $\operatorname{ad}_x \in \operatorname{End}_k(\mathfrak g) = \mathfrak{gl}(\mathfrak g)$. It remains in this step to check that \begin{align*} \operatorname{ad}: \mathfrak g &\to \mathfrak{gl}(\mathfrak g) \\ x &\mapsto \operatorname{ad}_x \end{align*} is $k$-linear. Let $a,b \in k$ and let $x_1,x_2,y \in \mathfrak g$. By bilinearity of the bracket in its first argument, \begin{align*} \operatorname{ad}_{a x_1 + b x_2}(y) &= [a x_1 + b x_2, y] \\ &= a[x_1,y] + b[x_2,y] \\ &= \bigl(a\operatorname{ad}_{x_1} + b\operatorname{ad}_{x_2}\bigr)(y). \end{align*} Since the two endomorphisms agree on every $y \in \mathfrak g$, \begin{align*} \operatorname{ad}_{a x_1 + b x_2} = a\operatorname{ad}_{x_1} + b\operatorname{ad}_{x_2}. \end{align*} Thus $\operatorname{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak g)$ is a well-defined $k$-[linear map](/page/Linear%20Map). [/step]

[step:Use the Jacobi identity to prove preservation of Lie brackets]Let $x,y \in \mathfrak g$. We must prove that \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)} = \operatorname{ad}_{[x,y]}. \end{align*} Because both sides are $k$-linear maps $\mathfrak g \to \mathfrak g$, it is enough to evaluate them on an arbitrary element $z \in \mathfrak g$. By the definition of the commutator bracket on $\mathfrak{gl}(\mathfrak g)$, \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) &= (\operatorname{ad}_x \circ \operatorname{ad}_y)(z) - (\operatorname{ad}_y \circ \operatorname{ad}_x)(z) \\ &= \operatorname{ad}_x([y,z]) - \operatorname{ad}_y([x,z]) \\ &= [x,[y,z]] - [y,[x,z]]. \end{align*} Using antisymmetry of the Lie bracket, we have $[y,[x,z]] = -[[x,z],y]$, so \begin{align*} [x,[y,z]] - [y,[x,z]] = [x,[y,z]] + [[x,z],y]. \end{align*} The Jacobi identity applied to $x,y,z \in \mathfrak g$ gives \begin{align*} [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0. \end{align*} Since $[z,x] = -[x,z]$ and $[z,[x,y]] = -[[x,y],z]$, this identity is equivalent to \begin{align*} [x,[y,z]] + [[x,z],y] = [[x,y],z]. \end{align*} Therefore \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) = [[x,y],z] = \operatorname{ad}_{[x,y]}(z). \end{align*} Since this holds for every $z \in \mathfrak g$, \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)} = \operatorname{ad}_{[x,y]}. \end{align*}[/step]

[guided]We want to prove that $\operatorname{ad}$ preserves the Lie bracket. The bracket in the target Lie algebra $\mathfrak{gl}(\mathfrak g)$ is not the original bracket on $\mathfrak g$; it is the commutator bracket of endomorphisms. Thus, for $x,y \in \mathfrak g$, the required identity is \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)} = \operatorname{ad}_{[x,y]}. \end{align*} Both sides are $k$-linear maps from $\mathfrak g$ to $\mathfrak g$. To prove two such maps are equal, we evaluate them on an arbitrary element $z \in \mathfrak g$. By definition of the commutator bracket in $\operatorname{End}_k(\mathfrak g)$, \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) &= (\operatorname{ad}_x \circ \operatorname{ad}_y)(z) - (\operatorname{ad}_y \circ \operatorname{ad}_x)(z). \end{align*} Now use the definition of each adjoint operator. Since $\operatorname{ad}_y(z) = [y,z]$ and $\operatorname{ad}_x(z) = [x,z]$, we get \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) &= \operatorname{ad}_x([y,z]) - \operatorname{ad}_y([x,z]) \\ &= [x,[y,z]] - [y,[x,z]]. \end{align*} The goal is to transform this expression into $\operatorname{ad}_{[x,y]}(z) = [[x,y],z]$. This is exactly where the Jacobi identity enters. First rewrite the second term using antisymmetry: \begin{align*} [y,[x,z]] = -[[x,z],y]. \end{align*} Therefore \begin{align*} [x,[y,z]] - [y,[x,z]] = [x,[y,z]] + [[x,z],y]. \end{align*} The Jacobi identity for the triple $x,y,z \in \mathfrak g$ states that \begin{align*} [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0. \end{align*} Since $[z,x] = -[x,z]$, bilinearity gives \begin{align*} [y,[z,x]] = -[y,[x,z]]. \end{align*} Equivalently, using antisymmetry again, $-[y,[x,z]] = [[x,z],y]$. Also, since $[z,[x,y]] = -[[x,y],z]$, the Jacobi identity becomes \begin{align*} [x,[y,z]] + [[x,z],y] - [[x,y],z] = 0. \end{align*} Rearranging gives \begin{align*} [x,[y,z]] + [[x,z],y] = [[x,y],z]. \end{align*} Combining this identity with the commutator computation yields \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) = [[x,y],z] = \operatorname{ad}_{[x,y]}(z). \end{align*} Because $z \in \mathfrak g$ was arbitrary, the two endomorphisms agree on every element of $\mathfrak g$. Hence \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)} = \operatorname{ad}_{[x,y]}. \end{align*}[/guided]

[step:Conclude that the adjoint map is a representation] The first step proves that $\operatorname{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak g)$ is $k$-linear. The second step proves that it preserves Lie brackets: \begin{align*} \operatorname{ad}_{[x,y]} = [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)} \end{align*} for all $x,y \in \mathfrak g$. Therefore $\operatorname{ad}$ is a Lie algebra homomorphism from $\mathfrak g$ to $\mathfrak{gl}(\mathfrak g)$, which is precisely a representation of $\mathfrak g$ on the [vector space](/page/Vector%20Space) $\mathfrak g$. This proves the theorem. [/step]