= Solution
For the finite-dimensional <Lie algebra representation> specified in the PDF, a <Lie-invariant bilinear form> satisfies
$$
B(xv,w)+B(v,xw)=0\qquad(x\in\mathfrak g,\ v,w\in V).
$$
The <dual Lie algebra representation> has action $(x\phi)(w)=-\phi(xw)$, so the map $T_B:V\to V^*$ defined by $T_B(v)=B(v,-)$ is an <intertwining operator>. If $V$ is <irreducible> and $B\ne0$, its <kernel> is zero. Since $V,V^*$ have equal finite <dimension>, $T_B$ is an <isomorphism>, and $B$ is a <nondegenerate bilinear form>.
For $B_1\ne0$, the composition $T_{B_1}^{-1}T_{B_2}$ is a representation endomorphism. Over an <algebraically closed field>, the <Schur lemma> makes it scalar, proving
$$
\boxed{B_2=\lambda B_1.}
$$
Transposing $B$ gives another <Lie-invariant bilinear form>. For $B\ne0$, write $B^T=\mu B$; transposing twice gives $\mu^2=1$. In <field characteristic> different from two, this means $\mu=1$ or $\mu=-1$. Thus the form is a <symmetric bilinear form> or an <alternating bilinear form>, respectively. The zero form has both properties. In the second case $B(v,v)=-B(v,v)$ forces $B(v,v)=0$, using the same characteristic assumption.
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