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

1. $v\ne0$ and
2. $Q(v)=0$

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.