Reshuffling Vector Multiplication Question

Question

In a proof for a problem i'm working on, there is a step in the solutions:

$\vec{a} \vec{a}^T \vec{b} = \vec{b}^T \vec{a} \vec{a}$

$\vec{a}, \vec{b} \in R^2$ for this problem.

I verified this holds on paper and on matlab for $R^2$ and $R^3$, so I'm assuming it also holds for $R^m$. I dont remember ever learning about this, so was a bit surprised to see it.

Wondering: How is this motivated? Did I simply forget about this result from my linear class? Does this result hold in the complex domain aswell?

Thanks

Answer 1

Suppose $a, b \in \mathbb{R^n}$. Then we have that $a^Tb$ and $b^Ta$ are both the same as the dot product $a \cdot b$ and $b \cdot a$. These are both scalars. So multiplying $a$ on the left and the right by this quantity is the same.

Answer 2

Take $a, b \in \mathbb R^n$. Since $a^\top b \in \mathbb R$ (hence $a^\top b = b^\top a$), you have $\mathbb R^n \ni aa^\top b = a(a^\top b) = (a^\top b)a = (b^\top a)a$.

EDIT : In the complex numbers, it is also true that $a^\top b = b^\top a$, but it is not how we define the usual dot product (see What is the dot product of complex vectors?)