Preservation of a scalar product under the action of an isomorphism Introduce the following assumptions and definitions,

*

*$\Omega\subset \mathbb{R}^3$ a bounded Lipshitz domain,

*$(L^2(\Omega))^3$ the space of vector square integrable functions,

*$M$ a $3\times 3$ real symmetric positive definite matrix (so $M$ is invertible),

*$A: (L^2(\Omega))^3\to (L^2(\Omega))^3$ a bounded self adjoint operator which satisfies:
$$
AM\neq MA
$$
and
$$
 0\leq (u,Au)\leq \| u\|^2 ,\;\forall u\in (L^2(\Omega))^3,
$$
where $(\cdot,\cdot)$ is the $L^2$ scalar product and $\|\cdot\|$ is the associated norm.
Then, I want to check if we have or not the following property:
$$
 0\leq (u,(M^{-1}AM)u)\leq \| u\|^2 ,\;\forall u\in (L^2(\Omega))^3.
$$
I know that the operator $M^{-1}AM$ defined on $(L^2(\Omega))^3$ is spectrally equivalent to $A$ but I don't know how to use it to check the above inequalities.

 A: This already fails in the scalar case. Consider
$$
A=\begin{bmatrix} 0&0\\0&1\end{bmatrix},\qquad\qquad M=\begin{bmatrix} 1&1\\1&2\end{bmatrix}.
$$
Both are selfadjoint and invertible, and $\|A\|=1$, so $(u,Au)\leq \|u\|^2$. We have
$$
M^{-1}AM=\begin{bmatrix} 2&-1\\-1&1\end{bmatrix}\begin{bmatrix} 0&0\\0&1\end{bmatrix} \begin{bmatrix} 1&1\\1&2\end{bmatrix}
=\begin{bmatrix} -1&-2\\1&2\end{bmatrix}.
$$
Then
$$
\bigg(\begin{bmatrix} 0\\1\end{bmatrix} ,M^{-1}AM\begin{bmatrix} 0\\1\end{bmatrix} \bigg)=\bigg(\begin{bmatrix} 0\\1\end{bmatrix},\begin{bmatrix} -2\\2\end{bmatrix}\bigg)=2>1=\bigg\|\begin{bmatrix} 0\\1\end{bmatrix}\bigg\|^2.
$$

Edit: controlling $M$ is not enough.
Let
$$
A=\begin{bmatrix} 0&0\\0&1\end{bmatrix},\qquad\qquad M=\begin{bmatrix} 1/2&1/4\\1/4&1/2\end{bmatrix}.
$$
Then $\|M\|=3/4$ and
$$
M^{-1}AM=\frac13\begin{bmatrix} -1&-2\\ 2&4\end{bmatrix},
$$
and
$$
\bigg(\begin{bmatrix} 0\\1\end{bmatrix} ,M^{-1}AM\begin{bmatrix} 0\\1\end{bmatrix} \bigg)=\bigg(\begin{bmatrix} 0\\1\end{bmatrix},\begin{bmatrix} -2/3\\4/3\end{bmatrix}\bigg)=\frac43>1=\bigg\|\begin{bmatrix} 0\\1\end{bmatrix}\bigg\|^2.
$$
