What would be the upper bound on the 2-norm (or any norm) of the following matrix product ? Please consider the smallest upper bound.
$\|\left(I+BA^T\right)\left(I+AA^T\right)^{-1}\|< ?$
where A and B are matrices and not necessarily square, and matrix I denotes the identity matrix. And we also know that $\|B\|<\|A\|$
You can use $$ I+BA^T=I+AA^T+(B-A)A^T. $$ Hence $$ \|(I+BA^T)(I+AA^T)^{-1}\|_2\leq 1+\|B-A\|_2\|A^T(I+AA^T)^{-1}\|_2. $$ By using the SVD of $A$, one can [from the fact that $x\mapsto x/(1+x^2)$ is for $x\geq 0$ bounded by $1/2$] get $$ \|A^T(I+AA^T)^{-1}\|_2\leq \frac{1}{2}, $$ so
You can of course use the triangle inequality and the assumption $\|B\|_2<\|A\|_2$ to get $$ \|(I+BA^T)(I+AA^T)^{-1}\|_2\leq 1+\|A\|_2. $$ (According to my random simulations, the norm can be larger than 1.)