[proofplan] We use a two-dimensional reduction argument: pick any vector $u \neq 0$, use non-degeneracy to find a partner $v$ with $\omega(u, v) = 1$, and form the plane $W_1 = \mathrm{span}\langle u, v \rangle$. On $W_1$ the form $\omega$ is the standard $2 \times 2$ skew block by skew-symmetry. We then split $V = W_1 \oplus W_1^{\perp_\omega}$, where the symplectic complement $W_1^{\perp_\omega}$ is shown to satisfy (a) $W_1^{\perp_\omega} \cap W_1 = \{0\}$, (b) it has codimension exactly $2$, and (c) the restriction of $\omega$ to it is again non-degenerate. Induction on dimension produces a basis of $W_1^{\perp_\omega}$ bringing $\omega|_{W_1^{\perp_\omega}}$ to standard block form; concatenating with $\{u, v\}$ yields the desired global basis. The even-dimensional conclusion follows because the induction terminates only at dimension $0$ (the only dimension on which a non-degenerate skew form can exist vacuously), never at dimension $1$ (where every skew form is zero, violating non-degeneracy). [/proofplan]

[step:Set up the induction on $\dim V$ and dispose of the base case] We argue by induction on $n := \dim_{\mathbb{R}} V < \infty$. **Base case $n = 0$.** The statement is vacuous: $V = \{0\}$ has even dimension $0$ and the empty basis puts $\omega$ in the empty block-diagonal form. Non-degeneracy holds vacuously. **Impossibility at $n = 1$.** If $\dim V = 1$, every skew-symmetric bilinear form on $V$ is zero: for any $u \in V$, $\omega(u, u) = -\omega(u, u)$ forces $\omega(u, u) = 0$, and by bilinearity $\omega$ vanishes on the whole of $V$ (every vector is a scalar multiple of $u$). A zero form is degenerate. Hence no non-degenerate skew form exists on a $1$-dimensional space, so the hypothesis of the theorem excludes $n = 1$. **Inductive hypothesis.** Assume the theorem holds for every vector space of dimension strictly less than $n$, with $n \ge 2$. We prove it for $V$ of dimension $n$. [/step]

[step:Pick a non-zero vector $u$ and use non-degeneracy to produce a partner $v$ with $\omega(u, v) = 1$] Since $n \ge 2 > 0$, we may pick any $u \in V$ with $u \neq 0$. By [non-degeneracy](/page/Non-Degenerate%20Bilinear%20Form) of $\omega$, the linear functional \begin{align*} \omega(u, \cdot): V &\to \mathbb{R} \\ w &\mapsto \omega(u, w) \end{align*} is non-zero. Hence there exists $v_0 \in V$ with $\omega(u, v_0) = c \neq 0$. Define $v := v_0 / c$. Then \begin{align*} \omega(u, v) = \omega(u, v_0 / c) = \omega(u, v_0) / c = 1. \end{align*} The vectors $u, v$ are linearly independent: if $v = \lambda u$ for some $\lambda \in \mathbb{R}$, then $\omega(u, v) = \lambda\, \omega(u, u) = 0$ by skew-symmetry ($\omega(u, u) = -\omega(u, u)$ gives $\omega(u, u) = 0$), contradicting $\omega(u, v) = 1$. Hence $W_1 := \mathrm{span}\langle u, v \rangle$ is a $2$-dimensional subspace. On $W_1$ the Gram matrix of $\omega$ in the ordered basis $(u, v)$ is \begin{align*} \begin{pmatrix} \omega(u, u) & \omega(u, v) \\ \omega(v, u) & \omega(v, v) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \end{align*} using skew-symmetry $\omega(u, u) = \omega(v, v) = 0$ and $\omega(v, u) = -\omega(u, v) = -1$. So $\omega|_{W_1}$ is already in standard $2 \times 2$ block form. In particular $\omega|_{W_1}$ is non-degenerate. [/step]

[step:Define the symplectic complement $W_1^{\perp_\omega}$ and prove $V = W_1 \oplus W_1^{\perp_\omega}$]Define \begin{align*} W_1^{\perp_\omega} := \{ a \in V : \omega(a, u) = 0 \text{ and } \omega(a, v) = 0 \}. \end{align*} This is the simultaneous kernel of the two linear functionals $\omega(\cdot, u)$ and $\omega(\cdot, v)$ on $V$. **Codimension.** The map \begin{align*} \Phi: V &\to \mathbb{R}^2 \\ a &\mapsto (\omega(a, u), \omega(a, v)) \end{align*} is linear with kernel $W_1^{\perp_\omega}$. We claim $\Phi$ is surjective, hence $\dim W_1^{\perp_\omega} = n - 2$. Indeed, $\Phi(v) = (\omega(v, u), \omega(v, v)) = (-1, 0)$ and $\Phi(-u) = (-\omega(u, u), -\omega(u, v)) = (0, -1)$, and $(-1, 0)$ and $(0, -1)$ span $\mathbb{R}^2$. **Trivial intersection.** Suppose $w \in W_1 \cap W_1^{\perp_\omega}$, so $w = \alpha u + \beta v$ for some $\alpha, \beta \in \mathbb{R}$ and $\omega(w, u) = \omega(w, v) = 0$. Compute \begin{align*} \omega(w, u) = \alpha\, \omega(u, u) + \beta\, \omega(v, u) = -\beta, \qquad \omega(w, v) = \alpha\, \omega(u, v) + \beta\, \omega(v, v) = \alpha. \end{align*} Both vanishing force $\alpha = \beta = 0$, so $w = 0$. **Direct sum decomposition.** From $\dim W_1 + \dim W_1^{\perp_\omega} = 2 + (n-2) = n = \dim V$ and $W_1 \cap W_1^{\perp_\omega} = \{0\}$, we conclude \begin{align*} V = W_1 \oplus W_1^{\perp_\omega}. \end{align*}[/step]

[guided]We want to peel off the standard $2 \times 2$ block and apply induction to the complement. The right notion of complement here is the **symplectic complement**, not the Euclidean orthogonal complement (there is no inner product available). We define it as the simultaneous annihilator of $u$ and $v$ under $\omega$. **Why $\dim W_1^{\perp_\omega} = n - 2$?** Because $W_1^{\perp_\omega}$ is the kernel of a linear map $\Phi: V \to \mathbb{R}^2$, and we can **surject onto $\mathbb{R}^2$** only if non-degeneracy on $V$ translates into non-degeneracy of the pairing $V \times W_1 \to \mathbb{R}$. Computing: $\Phi(v) = (-1, 0)$ and $\Phi(-u) = (0, -1)$, so the image contains both standard basis vectors of $\mathbb{R}^2$. Hence $\Phi$ is surjective and $\dim \ker \Phi = n - 2$. **Why is the intersection $\{0\}$?** A vector $w \in W_1 \cap W_1^{\perp_\omega}$ must both (i) be a linear combination $\alpha u + \beta v$ and (ii) pair to zero with both $u$ and $v$. Using the Gram matrix already computed, the pairing values are $(-\beta, \alpha)$, forcing $\alpha = \beta = 0$. This is the algebraic manifestation of $\omega|_{W_1}$ being non-degenerate: no non-zero vector of $W_1$ is $\omega$-orthogonal to all of $W_1$. **Direct sum.** Two subspaces with intersection $\{0\}$ whose dimensions add to $\dim V$ give an internal direct sum: every $w \in V$ decomposes uniquely as $w_1 + w'$ with $w_1 \in W_1$, $w' \in W_1^{\perp_\omega}$.[/guided]

[step:Verify that $\omega$ restricted to $W_1^{\perp_\omega}$ is non-degenerate]We must check the hypothesis of the inductive hypothesis applied to $W_1^{\perp_\omega}$ equipped with the restricted form $\omega' := \omega|_{W_1^{\perp_\omega} \times W_1^{\perp_\omega}}$. **Skew-symmetry of $\omega'$.** Immediate, inherited from $\omega$. **Non-degeneracy of $\omega'$.** Let $a \in W_1^{\perp_\omega}$ with $\omega'(a, b) = 0$ for every $b \in W_1^{\perp_\omega}$. We show $a = 0$. By global non-degeneracy of $\omega$ on $V$, if $a \neq 0$ there exists $b_0 \in V$ with $\omega(a, b_0) \neq 0$. Decompose $b_0$ via the direct sum: $b_0 = w_1 + b$ with $w_1 \in W_1$ and $b \in W_1^{\perp_\omega}$. Then \begin{align*} \omega(a, b_0) = \omega(a, w_1) + \omega(a, b). \end{align*} Since $a \in W_1^{\perp_\omega}$, $\omega(a, u) = 0$ and $\omega(a, v) = 0$. Writing $w_1 = \alpha u + \beta v$: \begin{align*} \omega(a, w_1) = \alpha\, \omega(a, u) + \beta\, \omega(a, v) = 0. \end{align*} Hence $\omega(a, b_0) = \omega(a, b) = \omega'(a, b) = 0$ by our assumption on $a$, contradicting $\omega(a, b_0) \neq 0$. Therefore $a = 0$, and $\omega'$ is non-degenerate.[/step]

[guided]For induction we need to know that the restricted form is again non-degenerate, so the inductive hypothesis applies. This is a small but essential verification — not automatic from non-degeneracy of the ambient form. **Strategy.** Suppose $a \in W_1^{\perp_\omega}$ pairs to zero with every element of $W_1^{\perp_\omega}$. We must show $a = 0$. **Use global non-degeneracy.** If $a \neq 0$, global non-degeneracy on $V$ gives $b_0 \in V$ with $\omega(a, b_0) \neq 0$. **Decompose $b_0$.** Using the direct sum $V = W_1 \oplus W_1^{\perp_\omega}$, write $b_0 = w_1 + b$ uniquely with $w_1 \in W_1$ and $b \in W_1^{\perp_\omega}$. **The $W_1$-component contributes nothing.** Since $a \in W_1^{\perp_\omega}$, by definition $\omega(a, u) = \omega(a, v) = 0$, so $\omega(a, w_1) = 0$ by linearity for any $w_1 \in W_1$. **The $W_1^{\perp_\omega}$-component contributes nothing by hypothesis.** $\omega(a, b) = 0$ because $a$ was assumed to annihilate $W_1^{\perp_\omega}$. **Contradiction.** So $\omega(a, b_0) = 0 + 0 = 0$, contradicting $\omega(a, b_0) \neq 0$. Hence $a = 0$, i.e. $\omega'$ is non-degenerate. This argument shows more generally that on any symplectic space, the symplectic complement of a **symplectic subspace** is again symplectic (this is not true for isotropic subspaces).[/guided]

[step:Apply the inductive hypothesis to $W_1^{\perp_\omega}$ and assemble the basis] The space $W_1^{\perp_\omega}$ has dimension $n - 2 < n$ and carries a non-degenerate skew-symmetric bilinear form $\omega'$ by the previous step. By the inductive hypothesis, $\dim W_1^{\perp_\omega}$ is even, and there exists a basis $e_3, f_3, e_5, f_5, \dots, e_{2k-1}, f_{2k-1}$ of $W_1^{\perp_\omega}$ (with $2k = n$) in which the Gram matrix of $\omega'$ is the direct sum of standard $2 \times 2$ blocks: \begin{align*} \omega'(e_{2j-1}, f_{2j-1}) = 1, \quad \omega'(f_{2j-1}, e_{2j-1}) = -1, \quad \text{all other pairings } 0. \end{align*} Setting $e_1 := u$ and $f_1 := v$, the ordered basis \begin{align*} e_1, f_1, e_3, f_3, e_5, f_5, \dots, e_{2k-1}, f_{2k-1} \end{align*} is a basis of $V$ (concatenation of bases of the two factors of $V = W_1 \oplus W_1^{\perp_\omega}$). **Gram matrix on the full basis.** The Gram matrix block-decomposes along $W_1 \oplus W_1^{\perp_\omega}$: cross-pairings $\omega(e_1, e_{2j-1})$, $\omega(e_1, f_{2j-1})$, $\omega(f_1, e_{2j-1})$, $\omega(f_1, f_{2j-1})$ for $j \ge 2$ all vanish because $e_{2j-1}, f_{2j-1} \in W_1^{\perp_\omega}$ pair to zero with $u = e_1$ and $v = f_1$ by definition of $W_1^{\perp_\omega}$. Within $W_1$ the Gram matrix is the first standard block; within $W_1^{\perp_\omega}$ it is the direct sum of the remaining blocks by the inductive hypothesis. Hence the Gram matrix of $\omega$ on the full basis is \begin{align*} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \oplus \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \oplus \dots \oplus \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \end{align*} as claimed. **Evenness of $\dim V$.** By the inductive hypothesis $\dim W_1^{\perp_\omega} = n - 2$ is even, so $n$ is even. This completes the induction and the proof. [/step]