[proofplan] We split into two cases depending on whether $B$ has a vector $u$ with $B(u,u) \neq 0$. If so, $v = u$ is anisotropic. If every vector is isotropic, the symmetry of $B$ and the assumption $\operatorname{char}(k) \neq 2$ allow us to construct an anisotropic vector as the sum $v = u + w$ of any pair satisfying $B(u,w) \neq 0$, which must exist since $B$ is nonzero. [/proofplan]

[step:Reduce to the totally isotropic case]Since $B$ is nonzero, there exist vectors $u, w \in V$ with $B(u,w) \neq 0$. If $B(u,u) \neq 0$ for some $u \in V$, set $v = u$ and we are done: $B(v,v) = B(u,u) \neq 0$. It remains to treat the case where $B(u,u) = 0$ for every $u \in V$.[/step]

[guided]Since $B$ is nonzero, there exist vectors $u, w \in V$ with $B(u,w) \neq 0$. We consider two cases based on whether any vector $u \in V$ satisfies $B(u,u) \neq 0$. **Case 1.** If $B(u,u) \neq 0$ for some $u \in V$, set $v = u$. Then $B(v,v) = B(u,u) \neq 0$, and $v$ is anisotropic. **Case 2.** Suppose $B(u,u) = 0$ for every $u \in V$. Every vector is isotropic, but $B$ is not the zero form: the pair $u, w$ with $B(u,w) \neq 0$ still exists. In Case 2, we will construct an anisotropic vector as $v = u + w$. Bilinearity expands $B(u+w, u+w)$ into four terms, and the isotropy hypothesis kills the diagonal terms $B(u,u)$ and $B(w,w)$. Symmetry collapses the two cross-terms into $2\,B(u,w)$, and $\operatorname{char}(k) \neq 2$ guarantees $2\,B(u,w) \neq 0$. It remains to carry out this computation explicitly.[/guided]

[step:Construct an anisotropic vector from a pair with nonzero cross-term]Assume $B(u,u) = 0$ for all $u \in V$. Since $B \neq 0$, choose $u, w \in V$ with $B(u,w) \neq 0$. Define $v := u + w$. Expanding by bilinearity and symmetry of $B$: \begin{align*} B(v,v) &= B(u + w,\, u + w) \\ &= B(u,u) + B(u,w) + B(w,u) + B(w,w) \\ &= 0 + B(u,w) + B(u,w) + 0 \\ &= 2\,B(u,w). \end{align*} Here we used the symmetry $B(w,u) = B(u,w)$ and the hypothesis $B(u,u) = B(w,w) = 0$. Since $\operatorname{char}(k) \neq 2$, the element $2 \in k$ is a unit ($2 \neq 0$ in $k$), so $2\,B(u,w) \neq 0$. Therefore $B(v,v) \neq 0$, and $v$ is anisotropic.[/step]

[guided]We assume every vector is isotropic: $B(u,u) = 0$ for all $u \in V$. Since $B$ is nonzero, we can find $u, w \in V$ with $B(u,w) \neq 0$. The idea is to form a linear combination of $u$ and $w$ and use bilinearity to produce a nonzero diagonal value. Define $v := u + w$. Expanding $B(v,v)$ by bilinearity (distributing over both arguments): \begin{align*} B(v,v) &= B(u + w,\, u + w) \\ &= B(u,u) + B(u,w) + B(w,u) + B(w,w). \end{align*} Now we substitute the known values. By hypothesis, $B(u,u) = 0$ and $B(w,w) = 0$ (since every vector is isotropic). By symmetry of $B$, we have $B(w,u) = B(u,w)$. This gives: \begin{align*} B(v,v) = 0 + B(u,w) + B(u,w) + 0 = 2\,B(u,w). \end{align*} Why does $2\,B(u,w) \neq 0$? Two things must hold: $B(u,w) \neq 0$ (which is how we chose $u$ and $w$) and $2 \neq 0$ in the field $k$. The second condition is exactly the hypothesis $\operatorname{char}(k) \neq 2$: in a field of characteristic $2$, we have $2 = 0$, and the entire argument collapses. This is the precise point where the characteristic assumption is consumed. Since both factors are nonzero, $B(v,v) = 2\,B(u,w) \neq 0$, and $v = u + w$ is the desired anisotropic vector. What goes wrong in characteristic $2$? If $\operatorname{char}(k) = 2$, then symmetry gives $B(w,u) = B(u,w)$, but $B(u,w) + B(u,w) = 2\,B(u,w) = 0$. Every vector remains isotropic, and indeed there exist nonzero symmetric bilinear forms in characteristic $2$ that are totally isotropic.[/guided]