I am curries whether the following binomial theorem is correct: \begin{align} \left( \text{Trace} (v-u)(v-u)^T \right)^n =\sum { n \choose i} \text{Trace}(vv^T)^{n-i} \text{Trace}(uu^T)^i \end{align} If not how to properly expand $\left( \text{Trace} (v-u)(v-u)^T \right)^n$?
I searched through my linear algebra reference but couldn't find anything.
Thank you
Assume $\vec{v}$ and $\vec{u}$ are in $\mathbb{R}^m$. Since the trace of an outer product is an inner product, then the L.H.S. is $$\left( \sum^m_{i=1} (v_i-u_i)^2 \right)^n= \left( \sum^m_{i=1} (v_i^2+u_i^2-2u_i v_i) \right)^n$$ The R.H.S is $$\left( \sum^m_{i=1} (v_i^2+u_i^2) \right)^n$$
The equality of the expressions are true iff $\vec{u}$ and $\vec{v} $ and perpendicular, i.e. $\sum^m_{i=1} u_i v_i =0$