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Use the classification of finite-dimensional sl2 representations. In the irreducible representation $V(m)$, choose $v_j=f^j{v}_0$, $0\le j\le m$, sowith vectors beyond the endpoints interpreted as zero. The invariant form on an irreducible sl2 module isIts anti-diagonal entries are nonzero, so it is nondegenerate. Lie-invariant bilinear form invariance under $h$ follows from the sum of the two weights. For $f$, the two potentially nonzero terms are $(-1{)}^{j+1}$ and $(-1{)}^j$. For $e$, their coefficients coincide when $j+k=m+1$, and their signs are opposite. These checks prove invariance under the generators of the sl2 Lie algebra.

Interchanging $j,k$ multiplies the form by $(-1{)}^m$, givingEquivalently it is a symmetric bilinear form in odd integer dimension and an alternating bilinear form in even number dimension. The Weyl complete reducibility theorem expresses any finite-dimensional representation as a direct sum of these irreducibles. Give each summand the displayed form and make different summands orthogonal. The resulting form is invariant and nondegenerate. On a reducible representation with both parities, this orthogonal sum need not itself be symmetric or alternating; the dichotomy in the preceding part required irreducibility.