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 $AA^*=\lambda BB^*$?
2026-03-30 11:57:49.1774871869
On the image of two bounded operators
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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.