= Solution
A <highest-weight vector> is a nonzero vector $v_0$ satisfying $J_3v_0=jv_0$ and $J_+v_0=0$; its eigenvalue $j$ is the <highest weight>. Existence does not require the <Lie algebra representation> to be antihermitian. Since $V$ is a nonzero finite-dimensional complex <vector space>, $J_3$ has an eigenvector $v$ with some eigenvalue $\lambda$. The <commutator> with $J_+$ shows that each nonzero $J_+^kv$ has eigenvalue $\lambda+k$. Infinitely many such vectors would have distinct eigenvalues and be linearly independent, which is impossible. The last nonzero vector in this raising chain is therefore a <highest-weight vector> $v_0$.
Put $v_\ell=J_-^\ell v_0$. The <ladder operator> relations give $J_3v_\ell=(j-\ell)v_\ell$. The raising coefficient follows by expanding the commutator of $J_+$ with a power of $J_-$:
$$
\begin{aligned}
J_+v_\ell&=[J_+,J_-^\ell]v_0
=\sum_{r=0}^{\ell-1}J_-^rJ_3J_-^{\ell-1-r}v_0\\
&=\sum_{r=0}^{\ell-1}(j-\ell+1+r)v_{\ell-1}
=\frac{\ell(2j-\ell+1)}2v_{\ell-1}.
\end{aligned}
$$
The factor $1/2$ is essential: these <ladder operators> have $[J_+,J_-]=J_3$, rather than the convention with $2J_3$. The same finite-dimensionality argument gives a last nonzero lowered vector $v_r$, with $J_-v_r=v_{r+1}=0$. Applying $J_+$ to $v_{r+1}=0$ yields
$$
0=\frac{(r+1)(2j-r)}2v_r,
$$
so \b[$2j=r\in\mathbb Z_{\ge0}$]. All $v_0,\ldots,v_r$ are nonzero and have distinct eigenvalues. Their span is invariant under $J_3,J_+,J_-$ and therefore under the original three generators. Irreducibility of the <Lie algebra representation> makes this span equal to $V$. Hence
$$
\boxed{j\in\left\{0,\tfrac12,1,\ldots\right\},\quad \dim V=2j+1,\quad V=\operatorname{span}\{J_-^\ell|j,j\rangle:0\le\ell\le2j\}.}
$$
In particular, the initial construction produces the largest eigenvalue of $J_3$, since its full spectrum is now $j,j-1,\ldots,-j$. This establishes the <highest weight> without assuming beforehand that all eigenvalues are real or that $J_3$ is diagonalizable.
For the additional antihermitian <Lie algebra representation>, the given definitions imply $J_3^\dagger=J_3$ and $J_+^\dagger=J_-$. Thus distinct $J_3$ eigenvalues give orthogonal <weight vectors>. With $\|v_0\|=1$, write $N_\ell=\|v_\ell\|^2$. The adjoint relation and the raising coefficient give
$$
N_\ell=\langle v_{\ell-1},J_+J_-v_{\ell-1}\rangle
=\frac{\ell(2j-\ell+1)}2N_{\ell-1},\qquad N_0=1.
$$
Multiplying this recursion proves the <normalized highest-weight lowering formula> in the present normalization:
$$
\boxed{N_\ell=\frac{(2j)!\,\ell!}{2^\ell(2j-\ell)!},\qquad |j,m\rangle=\frac{J_-^{j-m}|j,j\rangle}{\sqrt{N_{j-m}}}.}
$$
Every factorial argument is an integer, even for half-integral $j$. The displayed states have unit norm, mutually distinct eigenvalues, and span $V$, so they form an <orthonormal basis>. Their phase convention also fixes the positive ladder amplitudes
$$
J_\pm|j,m\rangle=\sqrt{\frac{(j\mp m)(j\pm m+1)}2}\,|j,m\pm1\rangle,
$$
with the endpoint actions interpreted as zero. These amplitudes will determine the <Clebsch-Gordan coefficients>.
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