Finding the null space of symmetric matrix generated by outer product

Question

Let $p, q \in \mathbb{R}^n$ such that $\|p\| = \|q\| = 1$ and define $$A = pq^T + qp^T$$ Find the null space of $A$.

I am not having very much luck. I have managed to show that $p + q$ and $p - q$ are eigenvectors, but I can't seem to connect those two. I know that since $A$ is generated by an outer product, it is of rank one. That implies that the rank of the null space is $n - 1$, but I can't seem to get any further. Can anyone help?

Answer 1

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 :

1. $p,q$ are linearly independant : in this case the Cauchy-Schwarz inequality implies that both eigenvalues above are non-zero, and the null space has dimension $n-2$.
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 $$0=Aw=\pm 2qq^Tw\Rightarrow q^Tw=0.$$

Answer 2

I saw a very similar question from my textbook. In that question, it defines A as a nxn symmetric matrix. With this additional piece of information, this question becomes very easy. First of all, we can prove N(A) = N(A_TA), since A is symmetric, A_TA = I. N(A) = N(I) = {0}. Similarly, we have rank(A) = rank(A_T*A) = rank(I) = n