[proofplan] Let $W \leq V_n$ be a nonzero $G$-invariant subspace. We use the maximal-torus action $\rho_n(\mathrm{diag}(z, z^{-1}))x^{n-i}y^i = z^{n-2i} x^{n-i}y^i$ to extract a single basis monomial from any nonzero element of $W$ (the eigenvalues $z^{n}, z^{n-2}, \ldots, z^{-n}$ are distinct for generic $z \in S^1$, so a Vandermonde-style elimination by suitable linear combinations of $\rho_n(t)w$ kills all but one term). We then act by a fixed $90$-degree rotation in $\operatorname{SU}(2)$ to produce a vector with nonzero $x^n$ component, apply Step 1 again to extract $x^n$, and finally hit $x^n$ with a generic $\rho_n(g)$ to obtain $(ax + by)^n \in W$. Expanding by the binomial theorem and using a Vandermonde argument once more gives all basis monomials in $W$, so $W = V_n$. [/proofplan]

[step:Set up notation and the torus action on $V_n$] Recall $V_n$ is the $(n+1)$-dimensional space of homogeneous polynomials of degree $n$ in $x, y$, with basis $\{x^{n-i} y^i : 0 \leq i \leq n\}$. The action of $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \operatorname{SU}(2)$ on $f \in V_n$ is \begin{align*} \rho_n: \operatorname{SU}(2) \times V_n &\to V_n \\ (g, f)(x, y) &\mapsto f(g^{-1} \cdot (x, y)^\top) = f(dx - by, -cx + ay). \end{align*} For the diagonal element $t_z := \mathrm{diag}(z, z^{-1}) \in T \subseteq \operatorname{SU}(2)$ with $z \in S^1$, $t_z^{-1} = \mathrm{diag}(z^{-1}, z)$, so \begin{align*} \rho_n(t_z) (x^{n-i} y^i) = (z^{-1} x)^{n-i} (z y)^i = z^{-(n-i) + i} x^{n-i} y^i = z^{2i - n} x^{n-i} y^i. \end{align*} (The convention $z^{2i-n}$ vs $z^{n-2i}$ is a matter of taste; we work with $z^{n-2i}$ by replacing $z$ with $z^{-1}$, equivalently using $t_z = \mathrm{diag}(z^{-1}, z)$. The argument is unchanged.) Adopting the convention \begin{align*} \rho_n(t_z)(x^{n-i} y^i) = z^{n-2i} x^{n-i} y^i, \end{align*} the basis monomials are weight vectors with distinct weights $n, n-2, \ldots, -n$ for the torus $T$. Let $W \leq V_n$ be a nonzero $G$-invariant subspace. [/step]

[step:Extract a single basis monomial from any nonzero $w \in W$][claim:Any nonzero $w \in W$ has the property that some basis monomial $x^{n-k} y^k$ lies in $W$] [proof] Write $w = \sum_{j=0}^n r_j x^{n-j} y^j$ with $r_j \in \mathbb{C}$, and let $S := \{j : r_j \neq 0\} \neq \varnothing$. We argue by induction on $|S|$. *Base case $|S| = 1$.* Then $w = r_k x^{n-k} y^k$ for the unique $k \in S$, so $r_k^{-1} w = x^{n-k} y^k \in W$ (since $W$ is a subspace). *Inductive step $|S| \geq 2$.* Fix $i \in S$. Choose $z \in S^1$ such that the weights $\{z^{n-2j} : j \in S\}$ are pairwise distinct. Such a $z$ exists: the equations $z^{n-2j} = z^{n-2j'}$ with $j \neq j'$ in $S$ define $z^{2(j' - j)} = 1$, a finite set of roots of unity in $S^1$; the union over the finitely many pairs $(j, j') \in S \times S$ with $j \neq j'$ is finite, while $S^1$ is uncountable, so a $z$ outside this finite set exists. Form \begin{align*} w' := \rho_n(t_z) w - z^{n-2i} w = \sum_{j=0}^n r_j (z^{n-2j} - z^{n-2i}) x^{n-j} y^j. \end{align*} By $G$-invariance of $W$, $\rho_n(t_z) w \in W$, so $w' \in W$. The coefficient of $x^{n-i} y^i$ in $w'$ is $r_i (z^{n-2i} - z^{n-2i}) = 0$, so $i \notin \mathrm{supp}(w')$. For $j \in S \setminus \{i\}$, the coefficient is $r_j (z^{n-2j} - z^{n-2i}) \neq 0$ by distinctness of weights. Hence $\mathrm{supp}(w') = S \setminus \{i\}$, with $|\mathrm{supp}(w')| = |S| - 1$. If $|S| - 1 \geq 1$, then $w' \neq 0$ and the inductive hypothesis applies. If $|S| - 1 = 0$, then $w' = 0$ — but this happens only when $S = \{i\}$, which is the base case. In either case, $W$ contains a basis monomial. [/proof] [/claim][/step]

[guided]The point of the elimination $w' := \rho_n(t_z) w - z^{n - 2i} w$ is to use the distinctness of the torus weights to perform single-step Gaussian elimination in $W$. The torus diagonalises $V_n$ into one-dimensional weight spaces, but $W$ might be a "diagonal" sum of these in a basis-dependent sense; we use the weights to peel off coordinates one by one. Why does the choice of $z$ matter? If we picked $z$ with $z^{n-2j} = z^{n-2i}$ for some $j \in S$ with $j \neq i$, the elimination would also kill the $j$-th term and we would not control which terms survive. Choosing $z$ to make all weights $\{z^{n-2j} : j \in S\}$ distinct guarantees that exactly the $i$-th term is killed. Since $S \subseteq \{0, 1, \ldots, n\}$ is finite, only finitely many $z \in S^1$ are bad, and $S^1$ is uncountable, so a good $z$ always exists. The induction reduces $|S|$ by one at each step, eventually arriving at the base case where $W$ contains a single basis monomial outright.[/guided]

[step:Promote any basis monomial $x^{n-k} y^k \in W$ to $x^n \in W$]By Step 2, $W$ contains some basis monomial $x^{n-k} y^k$ with $0 \leq k \leq n$. We now show $x^n \in W$. Let \begin{align*} g_0 := \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}. \end{align*} Verify $g_0 \in \operatorname{SU}(2)$: $\det g_0 = \frac{1}{2}(1 \cdot 1 - (-1) \cdot 1) = 1$, and $g_0^* g_0 = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = I$. Compute $\rho_n(g_0)(x^{n-k} y^k)$. Since $g_0^{-1} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$, \begin{align*} \rho_n(g_0)(x^{n-k} y^k) = \left(\frac{x + y}{\sqrt{2}}\right)^{n - k} \left(\frac{-x + y}{\sqrt{2}}\right)^k = 2^{-n/2} (x + y)^{n-k} (y - x)^k. \end{align*} Expand using the binomial theorem: \begin{align*} 2^{-n/2}(x + y)^{n-k} (y - x)^k = 2^{-n/2} \sum_{p=0}^{n-k} \sum_{q=0}^{k} \binom{n-k}{p} \binom{k}{q} (-1)^q x^{p + q} y^{n - p - q}. \end{align*} The coefficient of $x^n y^0$ corresponds to $p + q = n$, $n - p - q = 0$, i.e., $p = n - k$, $q = k$: \begin{align*} \text{coef of } x^n = 2^{-n/2} \binom{n-k}{n-k} \binom{k}{k} (-1)^k = 2^{-n/2} (-1)^k \neq 0. \end{align*} Hence $\rho_n(g_0)(x^{n-k} y^k) \in W$ has nonzero $x^n$ coefficient. Applying Step 2 (the elimination procedure) to this nonzero element, we extract a single basis monomial. To extract specifically $x^n$, we run the elimination targeting all indices $j \neq 0$: for each such $j$, we kill the $j$-th coefficient using a torus element $t_{z}$ with weights distinct on the support. Iterating Step 2 reduces the support to a single index. We claim that index can be chosen to be $0$. To extract $x^n$ specifically: rather than rely on which monomial Step 2 finally produces, we modify the construction. Set $w_0 := \rho_n(g_0)(x^{n-k} y^k) \in W$. For $j \in \{1, \ldots, n\}$, choose $z_j \in S^1$ such that $z_j^{n-2 \cdot 0} = z_j^n \neq z_j^{n - 2j}$, equivalently $z_j^{2j} \neq 1$, and form \begin{align*} w_0 \mapsto \rho_n(t_{z_j}) w_0 - z_j^{n-2j} w_0, \end{align*} which kills the coefficient of $x^{n-j} y^j$ while preserving (with a nonzero scaling) the coefficient of $x^n$ since $z_j^n - z_j^{n-2j} \neq 0$. Iterating over $j = 1, 2, \ldots, n$ — at each stage choosing $z_j$ avoiding the finitely many bad values that would cause unwanted vanishings — produces an element of $W$ supported only on the index $0$. That element is a nonzero scalar multiple of $x^n$, hence $x^n \in W$.[/step]

[guided]The matrix $g_0$ is the "$45$-degree rotation" in $\operatorname{SU}(2)$ — it sends $x \mapsto (x + y)/\sqrt{2}$ and $y \mapsto (-x + y)/\sqrt{2}$ in the natural action on linear forms (after inverting). Why this matrix? Because we need an element of $\operatorname{SU}(2)$ whose action on $V_n$ does *not* preserve the monomial basis: such an element mixes coordinates and forces $x^{n-k} y^k$ into a polynomial with a nonzero $x^n$ term. Why the binomial expansion? We need to verify that the coefficient of $x^n$ in the expansion of $(x + y)^{n-k}(y - x)^k$ is nonzero. The coefficient comes from picking $x$ from each factor: $(x)^{n-k} \cdot (-x)^k = (-1)^k x^n$. The sign is harmless since it does not affect the property "coefficient of $x^n$ is nonzero". The second elimination round is just Step 2 applied with target $0$. The reason this works is that Step 2 only requires the *target* index to be in the support — nothing prevents us from running it specifically against the $x^n$ coordinate, provided that coordinate is nonzero (which we just verified).[/guided]

[step:Show every $(ax + by)^n \in W$, then deduce $W = V_n$]By Step 3, $x^n \in W$. We act on $x^n$ by a generic element of $\operatorname{SU}(2)$. Fix $a, b \in \mathbb{C}$ with $|a|^2 + |b|^2 = 1$ and $a, b \neq 0$. The matrix \begin{align*} g := \begin{pmatrix} a & -\bar{b} \\ b & \bar{a} \end{pmatrix} \end{align*} has $\det g = a \bar{a} + b \bar{b} = |a|^2 + |b|^2 = 1$ and $g^* g = \begin{pmatrix} \bar{a} & \bar{b} \\ -b & a \end{pmatrix}\begin{pmatrix} a & -\bar{b} \\ b & \bar{a} \end{pmatrix} = \begin{pmatrix} |a|^2 + |b|^2 & 0 \\ 0 & |a|^2 + |b|^2 \end{pmatrix} = I$, so $g \in \operatorname{SU}(2)$. (Every element of $\operatorname{SU}(2)$ has this form.) Compute $g^{-1} = g^* = \begin{pmatrix} \bar{a} & \bar{b} \\ -b & a \end{pmatrix}$. Then \begin{align*} \rho_n(g)(x^n) = (\bar{a} x + \bar{b} y)^n. \end{align*} Replacing $(\bar{a}, \bar{b})$ by $(a', b')$ — any pair of nonzero complex numbers with $|a'|^2 + |b'|^2 = 1$ — we obtain $(a' x + b' y)^n \in W$ for all such $(a', b')$. Expand by the binomial theorem: \begin{align*} (a' x + b' y)^n = \sum_{i=0}^n \binom{n}{i} (a')^{n - i} (b')^i x^{n - i} y^i. \end{align*} For each $i$, the coefficient $\binom{n}{i} (a')^{n-i} (b')^i$ is nonzero (since $\binom{n}{i} \geq 1$ and $a', b' \neq 0$). [claim:All basis monomials $x^{n-i} y^i$ ($0 \leq i \leq n$) lie in $W$] [proof] We show this by a Vandermonde argument. Choose $n + 1$ pairs $(a'_\ell, b'_\ell)$, $\ell = 0, 1, \ldots, n$, all in $\operatorname{SU}(2)$-admissible form ($|a'_\ell|^2 + |b'_\ell|^2 = 1$, $a'_\ell, b'_\ell \neq 0$), such that the matrix \begin{align*} M_{\ell, i} = \binom{n}{i} (a'_\ell)^{n-i} (b'_\ell)^i, \quad 0 \leq \ell, i \leq n, \end{align*} is invertible. Concretely, take $a'_\ell = \cos\theta_\ell$, $b'_\ell = \sin\theta_\ell$ for distinct $\theta_\ell \in (0, \pi/2)$, $\ell = 0, \ldots, n$. The matrix $M$ factors as $M = D \cdot V$, where $D = \mathrm{diag}(\binom{n}{0}, \ldots, \binom{n}{n})$ (invertible since the binomials are positive) and $V$ has entries $V_{\ell, i} = (\cos \theta_\ell)^{n-i} (\sin\theta_\ell)^i = (\cos\theta_\ell)^n (\tan\theta_\ell)^i$. Factoring $(\cos\theta_\ell)^n$ from row $\ell$: \begin{align*} \det V = \prod_{\ell = 0}^n (\cos\theta_\ell)^n \cdot \det(\tan\theta_\ell)^i_{\ell, i}. \end{align*} The remaining matrix is a Vandermonde matrix in the variables $\tan\theta_0, \ldots, \tan\theta_n$, with determinant $\prod_{0 \leq \ell < \ell' \leq n} (\tan \theta_{\ell'} - \tan\theta_\ell) \neq 0$ since $\tan$ is injective on $(0, \pi/2)$ and the $\theta_\ell$ are distinct. Hence $\det V \neq 0$ and $\det M \neq 0$. Set $w_\ell := (a'_\ell x + b'_\ell y)^n \in W$ for $\ell = 0, \ldots, n$. The relation \begin{align*} w_\ell = \sum_{i=0}^n M_{\ell, i} \, x^{n-i} y^i \end{align*} expresses $w_\ell$ in the monomial basis. Since $M$ is invertible, we can solve for the basis monomials: there exist coefficients $\beta_{\ell, i}$ with $x^{n-i} y^i = \sum_\ell \beta_{\ell, i} w_\ell$. Each $w_\ell \in W$ and $W$ is a subspace, so $x^{n-i} y^i \in W$ for every $i$. [/proof] [/claim] The basis monomials span $V_n$, so $W = V_n$.[/step]

[guided]The strategy in this final step is to use a *single* highest-weight vector $x^n \in W$ and exploit the orbit of $x^n$ under all of $\operatorname{SU}(2)$. The orbit contains every $(a' x + b' y)^n$, which when expanded gives a polynomial with $n + 1$ distinct monomials, weighted by binomial coefficients and pure powers of $a', b'$. Why the Vandermonde argument? Because we want to show that the linear span of the orbit $\{(a' x + b' y)^n\}$ in $V_n$ is all of $V_n$. This is equivalent to showing that the natural map from the symmetric power $\operatorname{Sym}^n(\mathbb{C}^2) \cong V_n$ is "spanned by pure tensors $v^{\otimes n}$" — a classical fact in invariant theory. Concretely, choosing $n+1$ generic pairs $(a'_\ell, b'_\ell)$ produces $n+1$ vectors in $V_n$ whose change-of-basis matrix to the monomial basis is essentially Vandermonde, hence invertible. Once invertible, the span is all of $V_n$ by elementary linear algebra. This completes the proof: $W$ is nonzero $\Rightarrow$ $W$ contains a basis monomial (Step 2) $\Rightarrow$ $W$ contains $x^n$ (Step 3) $\Rightarrow$ $W$ contains every $(a' x + b' y)^n$ (Step 4) $\Rightarrow$ $W = V_n$. By the [Definition of Irreducibility](/theorems/2406), $V_n$ is irreducible.[/guided]