= Solution
Use a real <Lie algebra> convention $[T_a,T_b]=f_{ab}{}^cT_c$. Its <Adjoint representation of a Lie algebra> acts on $\mathfrak g$ itself:
$$
\boxed{\operatorname{ad}_X(Y)=[X,Y].}
$$
Linearity is immediate. The <Jacobi identity> gives
$$
[\operatorname{ad}_X,\operatorname{ad}_Y]Z
=[X,[Y,Z]]-[Y,[X,Z]]=[[X,Y],Z],
$$
so $[\operatorname{ad}_X,\operatorname{ad}_Y]=\operatorname{ad}_{[X,Y]}$, the defining <Lie algebra representation> condition. In the basis $T_b$, its <matrix> entries are
$$
\boxed{(T_a^{\rm ad})^c{}_b=f_{ab}{}^c.}
$$
Thus the adjoint generators are the <structure constants> arranged as <matrices>. The representation has kernel equal to the <center of a Lie algebra>; it need not be faithful for an arbitrary algebra.
The <Adjoint representation of a Lie group> is $\operatorname{Ad}_g(Y)=gYg^{-1}$. It satisfies $\operatorname{Ad}_{gh}=\operatorname{Ad}_g\operatorname{Ad}_h$, preserves the identity, and sends inverses to inverse <matrices>. Differentiating $e^{tX}Ye^{-tX}$ gives $[X,e^{tX}Ye^{-tX}]$, so
$$
\boxed{\operatorname{Ad}_{e^X}=e^{\operatorname{ad}_X},\qquad e^{-X}Ye^X=e^{-\operatorname{ad}_X}Y.}
$$
Consequently the positive adjoint exponentials furnish the representation on elements $g=e^X$ and their products, and the intrinsic conjugation action defines it globally.
\b[The inverse-conjugation formula in the PDF needs a minus adjoint exponent with this standard definition.] Its first-order term is $Y-[X,Y]$, whereas $e^{+\operatorname{ad}_X}Y$ has first-order term $Y+[X,Y]$. For example $X=T_1,Y=T_2$ in a nonabelian algebra with $[T_1,T_2]\neq0$ already distinguishes the two. The <inverse conjugation and adjoint antirepresentations> identity explains the group-order issue as well: $F(g)=\operatorname{Ad}_{g^{-1}}$ satisfies $F(gh)=F(h)F(g)$. Retaining the printed inverse conjugation as a left action without this order reversal would not give an ordinary <group representation>.
The <Killing form> is the symmetric <bilinear form>
$$
\boxed{B(X,Y)=\operatorname{tr}(\operatorname{ad}_X\operatorname{ad}_Y).}
$$
To prove degeneracy for a non-<semisimple Lie algebra>, take its nonzero <solvable radical> and the last nonzero member $\mathfrak a$ of its <derived series of a Lie algebra>. This is a nonzero abelian <ideal of a Lie algebra>. For $x\in\mathfrak a$, $\operatorname{ad}_x$ maps $\mathfrak g$ into $\mathfrak a$ and kills $\mathfrak a$. Every $\operatorname{ad}_y$ preserves $\mathfrak a$. Hence $\operatorname{ad}_x\operatorname{ad}_y$ maps the full space into $\mathfrak a$ and has zero restriction there, so its trace is zero. Thus $B(x,y)=0$ for every $y$. This is the <abelian ideals lie in the radical of the Killing form> argument, and establishes a nonzero kernel and vanishing <determinant>.
For a compact real <semisimple Lie algebra>, the adjoint action is unitary in an invariant positive <inner product>. Its infinitesimal generators are skew-Hermitian, giving
$$
B(X,X)=-\operatorname{tr}(\operatorname{ad}_X^\dagger\operatorname{ad}_X)<0\qquad(X\neq0).
$$
Strictness follows because the adjoint kernel is the zero center. This explains the <compactness criterion from the Killing form>: “strictly negative” means negative-definite, not that every entry of its <matrix> is negative.
In the first three-generator example, take columns to be images of the basis $(X,Y,H)$. Direct use of the brackets gives
$$
\boxed{\operatorname{ad}_X=\begin{pmatrix}0&0&0\\0&0&2\\0&2&0\end{pmatrix},\quad
\operatorname{ad}_Y=\begin{pmatrix}0&0&2\\0&0&0\\-2&0&0\end{pmatrix},\quad
\operatorname{ad}_H=\begin{pmatrix}0&-2&0\\-2&0&0\\0&0&0\end{pmatrix}.}
$$
Taking traces of products yields the <Killing form for cyclic three-generator brackets>
$$
\boxed{[B]_{(X,Y,H)}=\operatorname{diag}(8,-8,8).}
$$
It is nondegenerate, hence semisimple by the preceding degeneracy result, but it has mixed signature and is not compact. An explicit realization is
$$
X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad
Y=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\quad
H=\begin{pmatrix}-1&0\\0&1\end{pmatrix},
$$
which span the real traceless two-by-two <matrices> and satisfy exactly these brackets.
Changing the sign of $[X,H]$ changes the first and third adjoint <matrices> to
$$
\operatorname{ad}_X=\begin{pmatrix}0&0&0\\0&0&-2\\0&2&0\end{pmatrix},\qquad
\operatorname{ad}_H=\begin{pmatrix}0&-2&0\\2&0&0\\0&0&0\end{pmatrix},
$$
while $\operatorname{ad}_Y$ is unchanged. Now
$$
\boxed{[B]=-8I_3,}
$$
the negative-definite compact case. The basis $X=-i\sigma_1$, $Y=-i\sigma_2$, $H=-i\sigma_3$ realizes it as the real <special unitary Lie algebra> $\mathfrak{su}(2)$.
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