Let $A, B \in B(\mathcal{H})$ such that $Im(A) = Im(B)$, where $Im(A)$ denotes the image of $A$. Is there any $\lambda>0$ such that $AA^*=\lambda BB^*$?

up vote 1 down vote accepted

This is not true for $H=\mathbb{R}^2$. Take $A=Id$ then $AA^\star = Id$, and take $B(x,y)=(2x,y)$ then $BB^\star (x,y) = (4x,y)$ which is not a constant multiplication of the identity.

  • Under which conditions on $A, B$ is there such this $\lambda$? – saeed Sep 14 at 14:04
  • @saeed Notice that $AA^\star = BB^\star$ implies that $AB^{-1}$ is unitary. (But this argument only works if we assume that our operators are invertible, otherwise I don't have an answer to your comment question) – Yanko Sep 14 at 14:12

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