An application to Douglas Theorem Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and
let $\mathcal{L}(E)$ be the algebra of all bounded linear operators from $E$ to $E$. We recall the Douglas Theorem:

Douglas Theorem: Let $T_1,T_2 \in \mathcal{L}(E)$. The equation $T_1X=T_2$ has solutions in $\mathcal{L}(E)$ if and only if $\mathcal{R}(T_2) \subseteq \mathcal{R}(T_1)$, 
  where $\mathcal{R}(T_1)$, $\mathcal{R}(T_2)$ are respectively the ranges of $T_1$ and $T_2$.

Let $M\in \mathcal{L}(E)^+$ (i.e. $M^*=M$ and $\langle Mx\;, \;x\rangle \geq0,\;\forall x\in E$). By Douglas theorem $MX=T^*M$ has solutions if and only if $\mathcal{R}(T^{*}M)\subseteq \mathcal{R}(M)$.
I want to construct an example in which $MX=T^*M$ hasn't any solutions (i.e. $\mathcal{R}(T^{*}M)\nsubseteq \mathcal{R}(M)$.
Thank you.

Take
$$
T=\begin{bmatrix} 0&1\\ 0&0\end{bmatrix}
,\ \ M=\begin{bmatrix} 1&0\\ 0&0\end{bmatrix}.
$$
Then $T^*M=I-M$.
