If a positive operator $y$ has the same kernel as $cy$, what can we conclude about the kernel of $c$?

Question

Let us consider the equation $x=cy$ in $B(H)$. Assume that:

1. $y$ is a positive operator.
2. $x$ and $y$ have the same null space.
3. Ker($y$) is contained in Ker($c$).

Can we conclude that Ker($y$)=Ker($c$)?

Answer 1

Not true as stated. Let $y$ be the operator on $\ell_2$ that maps each sequence $(t_j)_{j\in\mathbb{N}}$ to $(2^{-j}t_j)_{j\in\mathbb{N}}$. This is a positive operator with zero kernel. Let $c$ be the orthogonal projection onto the orthogonal complement of the vector $z = (1,1/2,1/3,\dots)$. Since $z\notin \operatorname{ran} y$, it follows that $cy$ also has zero kernel. On the other hand, $\ker c$ is nontrivial.

The statement is true under the additional assumption that $y$ has closed range. In this case $\operatorname{ran} y = (\ker y)^\perp$. Hence, if $\ker c$ contains $\ker y$ properly, then it also contains a vector from $\operatorname{ran}y$, which contradicts the assumption $\ker (cy)=\ker y$.