Lower bound for inner product between "close" vectors

141 Views Asked by At

I am trying to find a lower bound of the following form:

Given a vector $\left\Vert \boldsymbol{u}\right\Vert ^{2}\le1$ and given two non-expansion operators $\boldsymbol{A,P}$ (i.e., $\left\Vert \boldsymbol{A}\boldsymbol{u}\right\Vert _{2}\le1,\left\Vert \boldsymbol{P}\boldsymbol{u}\right\Vert _{2}\le1$) such that $\boldsymbol{P}$ is an (idempotent) projection matrix ($\boldsymbol{P}^{*}\boldsymbol{P}=\boldsymbol{P}$), if it holds that $$1-\epsilon \le\left\Vert \boldsymbol{A}\boldsymbol{u}\right\Vert ^{2} \\ 1-\epsilon \le\left\Vert \boldsymbol{P}\boldsymbol{u}\right\Vert ^{2}$$ for some $\epsilon\in\left(0,\frac{1}{2}\right)$, then we have that

$$\left(\boldsymbol{u}^{*}\boldsymbol{P}\boldsymbol{A}\boldsymbol{u}\right)^{2}\ge\,?$$


My intuition is that if $\epsilon$ is very close to $0$, then $\boldsymbol{A}\boldsymbol{u}, \boldsymbol{P}\boldsymbol{u}$ should be "close" with a high inner product.

I am also willing to restrict $\epsilon$ to a smaller interval if it helps.

Feeling like I'm missing something simple.

1

There are 1 best solutions below

1
On BEST ANSWER

This isn't true, because you've only bounded the norm, which can't say anything about angles, which is what inner products are about. Let's take $u = (1,0)$ in $\mathbb{R}^2$. Then we can take $A$ to be a rotation matrix by any angle $\vartheta$ that you like, and $P$ can be the projection matrix onto the $x-$axis. Then $||Au|| = ||Pu|| = 1$, so your conditions hold for any $\epsilon$. On the other hand, these things can have any inner product you want, by choosing $\vartheta$ accordingly.