Verify is that $\overline{R(V + W)}= N(U)^{\perp}$, and that $V = W = 0$ on $N(U)$

Question

Let $H$ Hilbert space, $U ∈ \mathcal{L}(H )$ be such that $U^* = U$ and $(Ux,x) \geq 0$, in this case, exists $V ∈ \mathcal{L}(H )$ such that $V^* = V$, $(Vx,x) \geq 0$, and $V^2 = U$, $V$ is called square root of $U$. I am trying to determine a uniqueness of this root. For this if $W$ is any operator such that $W^* = W$, $(Wx,x) \geq 0$, and $W^2 = U$, I managed to prove that $V$ commutes with $W$, then we see that $(V − W) \circ (V + W) = 0$ and we deduce that $V = W$ on $R(V + W)$. Therefore $V = W$ on $\overline{R(V + W)}$. what I'm not able to verify is that $\overline{R(V + W)}= N(U)^{\perp}$, and that $V = W = 0$ on $N(U)$. I'm not figuring out how to relate these terms from their definitions, any tips on how to do it?

Answer 1

Assume that $y\in N(U)$ such that $Uy=0$, then $(Uy,y)=(V^2y,y)=(Vy,Vy)=0$, and the same holds for $W$. This shows that $V=W=0$ on $N(U)$. On the other hand, it is quite obvious that $U=0$ on $N(V)$ and $N(W)$, proving that $N(U)=N(V)=N(W)$.

For the second part, we use that $\overline{R(V+W)}=N(V+W)^\perp $. Knowing that $V$ and $W$ are positive and self-adjoint we have $(Vx,x)=0$ iff $x\in N(V)=N(U)$ and the same for $W$, which can be seen for example by the existence of a positive self-adjoint square root of $V$ (not necessarily unique). It follows then from positivity of $V$ and $W$ that $Vz\neq -Wz$ for any $z\notin N(U)$ such that $N(V+W) =N(V)=N(W)=N(U)$. Hence, $\overline{R(V+W)}=N(V+W)^\perp=N(U)^\perp$.