A vector $v$ (an element of $V$) in a quadratic space $(V,Q)$ is isotropic if

1. 1.

$v\neq 0$ and

2. 2.

$Q(v)=0$.

Otherwise, it is called anisotropic. A quadratic space $(V,Q)$ is isotropic if it contains an isotropic vector. Otherwise, it is anisotropic. A quadratic space $(V,Q)$ is totally isotropic if every one of its non-zero vector is isotropic, or that $Q(V)=0$.

Similarly, an isotropic quadratic form is one which has a non-trivial kernel, or that there exists a vector $v$ such that $Q(v)=0$. The definitions for that of an anisotropic quadratic form and that of a totally isotropic quadratic form should now be clear from the above discussion (anisotropic: $\operatorname{ker}(Q)=0$; totally isotropic: $\operatorname{ker}(Q)=V$).

Examples.

• Consider the quadratic form $Q(x,y)=x^{2}+y^{2}$ in the vector space $\mathbb{R}^{2}$ over the reals. It is clearly anisotropic since there are no real numbers $a,b$ not both $0$, such that $a^{2}+b^{2}=0$.

• However, the same form is isotropic in $\mathbb{C}^{2}$ over $\mathbb{C}$, since $1^{2}+i^{2}=0$; the complex numbers are algebraically closed.

• Again, using the same form $x^{2}+y^{2}$, but in $\mathbb{R}^{3}$ over the reals , we see that it is isotropic since the $z$ term is missing, so that $Q(0,0,1)=0^{2}+0^{2}=0$.

• If we restrict $Q$ to the subspace consisting of the $z$-axis ($x=y=0$) and call it $Q_{z}$, then $Q_{z}$ is totally isotropic, and the $z$-axis is a totally isotropic subspace.

• The quadratic form $Q(x,y)=x^{2}-y^{2}$ is clearly isotropic in any vector space over any field. In general, this is true if the coefficients of a diagonal quadratic form $Q$ consist of $1,-1,0$ ($0$ is optional) and nothing else.