Finding the null space of symmetric matrix generated by outer product
First notice that any vector $w$ orthogonal to both $p$ and $q$ is the null space of $A$, since $$Aw=p(q^Tw)+q(p^Tw)=0.$$ Thus the null space has dimension at least $n-2$. Since the eigenvectors $p\pm q$ correspond to eigenvalue $p^Tq\pm 1$, at least one of the two eigenvalues is non-zero, so the nullspace has dimension at most $n-1$.
Now we have two cases to cover :
- $p,q$ are linearly independant : in this case the Cauchy-Schwarz inequality implies that both eigenvalues above are non-zero, and the null space has dimension $n-2$.
- $p,q$ are not linearly independant : then since they have the same norm we must have $p=\pm q$, thus $\langle p,q\rangle$ has dimension $1$, and its orthogonal complement has dimension $n-1$.
In both case, the null space is the orthogonal complement of $\langle p,q\rangle$ because one is included in the other and their dimensions agree.
You can even see the result more easily, without even considering eigenvalues and eigenvectors. The first equation shows that $\langle p,q\rangle^{\perp}\subset Ker A$. For the reverse inclusion, just notice that if $p,q$ are linearly independant then $$0=Aw=pq^Tw+qp^Tw\Rightarrow q^T w =0=p^Tw,$$and if they aren't then the dimensions agree (as explained above); or you can notice that $p=\pm q$ and thus $A=\pm 2 qq^T$, and thus $$0=Aw=\pm 2qq^Tw\Rightarrow q^Tw=0.$$