If $Q$ is a nonnegative operator on a Hilbert space $U$, show that $\langle Q^{1/2}u_0,Q^{1/2}v_0\rangle_{Q^{1/2}U}=\langle u_0,v_0\rangle_{Q^{1/2}U}$

Question

Let

• $U$ be a $\mathbb R$-Hilbert space
• $Q\in\mathfrak L(U)$ be nonnegative and self-adjoint
• $Q^{-1/2}$ denote the pseudoinverse of $Q^{1/2}$, i.e. $$Q^{-1/2}:=\left(\left.Q^{1/2}\right|_{\left(\ker Q^{1/2}\right)^\perp}\right)^{-1}$$
• $U_0:=Q^{1/2}U$ be equipped with $$\left\langle u_0,v_0\right\rangle_{U_0}:=\left\langle Q^{-1/2}u_0,Q^{-1/2}v_0\right\rangle_U\;\;\;\text{for all }u_0,v_0\in U_0$$

Let $u_0,v_0\in U_0$. I want to show that $$\left\langle Q^{1/2}u_0,Q^{1/2}v_0\right\rangle_{U_0}=\left\langle u_0,v_0\right\rangle_{U_0}.\tag1$$

Let $\operatorname P_{\left(\ker Q^{1/2}\right)^\perp}$ denote the orthogonal projection from $U$ onto $\left(\ker Q^{1/2}\right)^\perp$. Then, $$Q^{-1/2}Q^{1/2}=\operatorname P_{\left(\ker Q^{1/2}\right)^\perp}\tag2$$ and hence (since an orthogonal projection is self-adjoint and idempotent) $$\left\langle Q^{1/2}u_0,Q^{1/2}v_0\right\rangle_{U_0}=\left\langle\operatorname P_{\left(\ker Q^{1/2}\right)^\perp}u_0,v_0\right\rangle_U.\tag3$$

However, I don't see how this helps to prove $(1)$. So, what can we do?

Answer 1

Your equality is not true, even if $Q$ is invertible. For instance take $$ Q=\begin{bmatrix} 1&0\\0&4\end{bmatrix},\ \ u_0=v_0=\begin{bmatrix} 1\\2\end{bmatrix}. $$ Then $$ Q^{1/2}=\begin{bmatrix} 1&0\\0&2\end{bmatrix}. $$ You have $$ \langle Q^{1/2}u_0,Q^{1/2}v_0\rangle_{U_0}=\langle Q^{-1/2}Q^{1/2}u_0,Q^{-1/2}Q^{1/2}v_0\rangle_U=\langle u_0,v_0\rangle_U=1+4=5, $$ while $$ \langle u_0,v_0\rangle_{U_0}=\langle Q^{-1/2}u_0,Q^{-1/2}v_0\rangle_U =\left\langle \begin{bmatrix} 1\\1\end{bmatrix},\begin{bmatrix}1\\1\end{bmatrix}\right\rangle=2. $$