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A quadratic form $q$ over a field $k$ is an anisotropic quadratic form if $q(v)=0$ forces $v=0$. An isotropic quadratic form instead has a nonzero zero vector. This is a property of the form together with its coefficient field, rather than just the displayed polynomial.^[ref]

For example, $q(x,y)=x^2-2y^2$ is anisotropic over $\mathbb{Q}$. If a nonzero rational solution had $y=0$, then $x=0$ too; otherwise $x/y$ would be a rational square root of two. Over $\mathbb{R}$, however, $(\sqrt{2},1)$ is a nonzero zero vector. The coordinate change $u=x+\sqrt{2}y$, $v=x-\sqrt{2}y$ gives $q=uv$, a hyperbolic plane of a quadratic form.

This example also separates anisotropy from positive definiteness: the rational form takes the positive value $1$ and the negative value $-2$, despite having no nonzero rational zero. In a Witt decomposition, an anisotropic remainder is what remains after all hyperbolic planes have been split off. Its Witt index is zero. In characteristic two, one must test vanishing of the quadratic form itself; properties of the associated polar bilinear form alone do not supply the same test.

Bill Casselman writes research and expository mathematics connecting algebraic structure with concrete calculations. His author-hosted Introduction to quadratic forms by Bill Casselman explains quadratic forms over general fields, including the distinctions required in characteristic two. It develops decomposition results using totally isotropic subspaces and the hyperbolic plane.^[ref]

For example, over $\mathbb{R}$ the quadratic form $Q(x,y,z)=xy+z^2$ separates into a hyperbolic plane on the first two coordinates and an anisotropic quadratic form on the last. The $x$-axis consists of isotropic vectors, but it is not invisible to the whole form: its associated bilinear form pairs $(1,0,0)$ nontrivially with $(0,1,0)$. Pairing such complementary directions produces the hyperbolic part of an orthogonal direct sum. The example helps distinguish a vector on which a quadratic form vanishes from a vector in the radical of its associated bilinear form.^[ref]

For a finite-dimensional nonsingular quadratic form $q$ over a field of characteristic different from two, a Witt decomposition is an orthogonal splittingwhere $H(u,v)=uv$ is a hyperbolic plane of a quadratic form and $q_{\mathrm{an}}$ is an anisotropic quadratic form. The integer $r$ is the Witt index. Both $r$ and the isometry class of the anisotropic part are determined by $q$, although the particular subspaces realizing the splitting need not be unique.^[ref]

Over $\mathbb{Q}$, take $q(x,y,z)=x^2+y^2-z^2$. Set $x=(u+v)/2$, $z=(u-v)/2$ and $y=w$. Thenso $q\cong H\perp ⟨1⟩$. The one-dimensional remainder is anisotropic and the Witt index is one. This is an isometry of quadratic spaces, stronger than merely expressing the underlying vector space as a direct sum.

The construction starts from an isotropic vector, finds a partner with nonzero polar pairing, and splits off the hyperbolic plane they generate. Repeating the process ends when no nonzero isotropic vector remains. The decomposition makes the isotropic directions and the residual arithmetic obstruction visible separately.

The Witt index of a nondegenerate quadratic form $q$ on a finite-dimensional vector space $V$ is the largest dimension of a totally isotropic subspace: a linear subspace $U$ for which $q(u)=0$ for every $u\in U$. In characteristic different from two, the Witt decomposition writeswhere $H$ is a hyperbolic plane of a quadratic form, $q_{\mathrm{an}}$ is an anisotropic quadratic form, and $r$ is the Witt index. The anisotropic part has no nonzero vector on which the form vanishes.^[ref]

For the real form $x_1^2+x_2^2-x_3^2-x_4^2$, the subspace spanned by $(1,0,1,0)$ and $(0,1,0,1)$ is totally isotropic. Projection onto the two negative coordinates is injective on any totally isotropic subspace, giving the matching upper bound two. More generally, a real form of signature $(a,b)$ has Witt index $\min (a,b)$.

The base field matters: $x^2+y^2$ has index zero over $\mathbb{R}$ but index one over $\mathbb{C}$, using the line spanned by $(1,i)$. In characteristic two, vanishing of the associated polar form on $U$ alone is insufficient; the condition remains $q{|}_U=0$. Over a finite field, the two nondegenerate types in dimension $2m$ have Witt indices $m$ and $m-1$.^[ref]