Prove that $\operatorname{trace}(T^*T) = \|Tu_1\|^2+\cdots+ \|Tu_n\|^2$ Let V be an inner product space with   orthonormal basis  $(u_1,...,u_n)$, 

Show that 
  $$\operatorname{trace}(T^*T) = \|Tu_1\|^2+\cdots+ \|Tu_n\|^2$$

I was able to solve this in The case $u_i$'s are eigenvectors of the operator $T$. How can do this for any other orthonormal basis of V? 

In the $u_i$ basis, we have the matrix of $T^\ast T$:
$[(T^\ast T)_{ij}] = \langle u_i, T^\ast T u_j \rangle = \langle Tu_i, Tu_j \rangle; \tag 1$
thus
$(T^\ast T)_{ii}  = \langle Tu_i, Tu_i \rangle = \Vert Tu_i \Vert^2. \tag 2$
In any basis, the trace is the sum of the diagonal entries of the matrix of $T^\ast T$ in that basis; thus
$\operatorname{trace}(T^\ast T) = \sum_1^n (T^\ast T)_{ii} = \sum_1^n \Vert Tu_i \Vert^2, \tag 3$
is proved per request.
The issue of whether $\{u_i\}$ forms an eigenbasis or not does not enter into the above demonstration.  It is important, however, the $\{u_i\}$ is orthonormal; otherwise, the dual basis to $\{u_i\}$, that is the
$\theta_j:V \to \Bbb F, \; \theta_j(u_k) = \delta_{jk}, \tag 4$
where $\Bbb F$ is the base field (most likely either $\Bbb R$ or $\Bbb C$), will not satisfy
$\theta_j(v) = \langle u_j, v \rangle, \; v \in V;\tag 5$
since in this more general case
$(T^\ast T)_{ij} = \theta_i(T^\ast Tu_j), \tag 6$
(1) will not necessarily bind.
