I have a result of the form:
$r = A^\top B^\top B A$
Where $A_{[m \times 1]}$ and $B_{[n \times m]}$, so $A^\top A$ is a scalar.
I want to bound $r$ from below using some property of B, and reach the form:
$r \geq b \cdot (A^\top A)$, where $b$ is a scalar derived from the matrix $B$.
A simple example could be the case where $B$ is diagonal. Then if $b = \min_i B_{i,i}^2$ (the minimal squared value found in $B$), then the form above hold.
Is there anything I can use for the case $B$ is non diagonal? something that perhaps includes the $trace$ of $B$?
Without additional information, the best you can say is $M = B^TB$ is a positive semi-definite $n\times n$ matrix; this means $M$ may have an eigenvalue $0$.
In this case, if $A\neq 0$ is its associated eigenvector, we'd have $r = 0$, which would make $b=0$.
Notice that, as a linear transformation, $B: \Bbb R^m\longrightarrow \Bbb R^n$. This means that when $m>n$ it's guaranteed that $B$ fails to be injective, and hence that $M$ has nontrivial kernel, that is, $M$ has an eigenvalue $0$.
In other words, you could only hope for $b>0$ when $m\leqslant n$.
In general, $b$ would be $M$'s smallest eigenvalue (all of which are real and non-negative).