An application to Douglas Theorem

Question

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.

Answer 1

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

Answer 2

Take $\mathcal{H}= \mathbb{C}^2$ with the standard scalar product and choose the operators given by the following matrices

$$ M = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \qquad T = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$ We can see that $T^*M=M-\begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}$.