Suppose $A$ and $B$ are square matrices. I would like to find the trace or upper and lower bounds on the trace of $ABA^{T}$, where the lower diagonal and diagonal elements of $B=(b_{ij})$ are all zeros, for instance,
$$B = \begin{bmatrix} 0 & b_{1} & b_{2} & b_{3} \\ 0 & 0 & b_{1} & b_{2} \\ 0 & 0 & 0 & b_{1} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$
and $A=(a_{ij})$ is a general matrix (in the sense that it is not diagonal and all elements are different from zero) assumed to be full rank. Thanks a lot!
Here is the answer I already gave:
let $$A=\pmatrix{k&1&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\}$$
It is a full rank matrix if $k\neq 0$.
$$\tag{1}ABA^T=\pmatrix{kb_1&kb_1&b_1+kb_2&b_2+kb_3\\0&0&b_1&b_2\\0&0&0&b_1\\0&0&0&0\\}$$
If $b_1\neq 0$, this trace can take any nonzero value, in particular can be arbitrarily large.
Now, the non zero condition on all entries. In a nutshell, it suffices to add a convenient $\varepsilon\neq0$ to all entries of matrix $A$: this will modify the trace in an arbitrary small amount. In this way, the result stays the same: the trace can be arbitrarily large.
Remark: The computation in $(1)$ can be done very easily by block computations.
See also interesting answers to almost the same question (Simple way to find the trace or the bound of the trace of $ ABA^T$?)