Let $X\in \mathbb{R}^{m,n}$ and $A\in \mathbb{R}^{n,m}$, $b\in \mathbb{R}^{m}$.
How do we solve the following maximization problem?
$$\text{maximize}_{||X||=1} ||AXAb||_2$$
My thought was that we can select $X=U\Lambda V^{T}$ where $U$ and $V$ are the left and right singular vectors of $A$ and $\Lambda = diag(1, 0,...,0)$. But I think this solution seems like it doesn't care about the vector $b$.
Any suggestion will be appreciated.
This should be easy. If $Ab=0$, any $X$ with unit norm will do. Suppose $Ab\ne0$. By the submultiplicativity of the induced $2$-norm, we have $\|AXAb\|\le\|A\|\|X\|\|Ab\|=\|A\|\|Ab\|$ and the upper bound is attained at $X=vw^T$, where $v$ is a unit right singular vector corresponding to the largest singular value of $A$ and $w=Ab/\|Ab\|$.