proving that $\mathcal{B}^M(F)$ is a subalgebra of $\mathcal{B}(F)$ Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ (i.e. $M^*=M$ and $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$.
I want to show that $\mathcal{B}^M(F)$ is a subalgebra of $\mathcal{B}(F)$, where
$$\mathcal{B}^M(F)=\left\{S\in \mathcal{B}(F):\,\,\,\text{Im}(S^{*}M)\subseteq \text{Im}(M)\right\}.$$
Thank you.

There is no need for $M$ to be positive. 
For any $x\in F$, $S,T\in B^M(F)$, $\alpha\in\mathbb C$, 
$$
(\alpha S+T)^*Mx=\bar\alpha S^*Mx+T^*Mx\in MF
$$
since $MF$ is a subspace. 
And, sincd $S^*Mx=My$ for some $y$
$$
(ST)^*Mx=T^*(S^*Mx)=T^*My\in MF. 
$$
So $B^M(F)$ is a subalgebra. It is not a $*$-subalgebra in general, even if $M$ is positive: if $M=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$, $S=\begin{bmatrix} 1&0\\1&0\end{bmatrix}$, then $S^*MF\subset MF$ (in fact, $S^*M=M$), but $SMF\not\subset MF$. 
