How to prove $v_1v_2^Tv_2=v_2v_2^Tv_1$?

Question

How can I prove $$ v_1v_2^Tv_2=v_2v_2^Tv_1 $$ where $v_1$ and $v_2$ are all of shape $(1, n)$. The original problem is from this paper, where Eq.4 in page 3 is defined as But I derive something like $$ \alpha R(\tau')\nabla_{\theta'}\log\pi_{\theta'}(\tau')(R(\tau)\nabla_\theta\log\pi_\theta(\tau)^T\nabla_\theta\log\pi_\theta(\tau)) $$ where $\alpha$, $R(\tau')$ and $R(\tau)$ are scalars. In fact I'm not so sure why I transpose the first $\nabla_\theta\log\pi_\theta(\tau)$, I do so simply because I think there should be a matrix. I hope someone could help me clarify it. Thanks in advance.

Answer 1

It is in general not true that $v_1 v_2^T v_2 = v_2 v_2^T v_1$. Try for example with $v_1 =(1,0,0), v_2=(0,0,1)$.