$T$ is densely defined closed operator on $\mathcal{H}$ which is a Hilbert space. Prove that $\forall a,b\in H$,the equations \begin{equation} \label{eq:1} \left\{ \begin{aligned} -Tx+y &= a \\ x+T^{\star}y&=b \end{aligned} \right. \end{equation} have unique solution $x\in D(T),y\in D(T^\star)$. $T^\star$ is the adjoint operator of $T$. $D(T)$ is the space where $T$ is defined.
I have proved the uniqueness. And if $a\in D(T^\star),b\in D(T)$, I can just solve it just by taking $T$ on the second equation and plus the first one. But for $\forall a,b\in H$, I have some troubles. Any idea will help, thank you.
Because $T$ is closed and densely-defined, then so is $T^*$. Furthermore, $$ \langle Tx,y\rangle-\langle x,T^*y\rangle = 0,\;\; x\in\mathcal{D}(T),y\in\mathcal{D}(T^*), $$ which may be written in terms of an orthogonality condition on $\mathcal{H}\times\mathcal{H}$: $$ \left\langle \left[\begin{array}{c}-Tx \\ x\end{array}\right], \left[\begin{array}{c}y \\T^*y \end{array}\right]\right\rangle = 0, \;\;\;x\in\mathcal{D}(T)\;\; y\in\mathcal{D}(T^*). $$ These subspaces are orthogonal complements of each other, which gives $$ X\times X = \left\{\left[\begin{array}{c}-Tx \\ x\end{array}\right]+ \left[\begin{array}{c}y \\T^*y \end{array}\right] : x\in\mathcal{D}(T),\;y\in\mathcal{D}(T^*)\right\}. $$ Therefore, given $a,b\in\mathcal{H}$, there exists a unique $x\in\mathcal{D}(T)$ and $y\in\mathcal{D}(T^*)$ such that $$ \left[\begin{array}{c} -Tx \\ x \end{array}\right]+\left[\begin{array}{c} y\\ T^*y\end{array}\right]=\left[\begin{array}{c} a \\ b \end{array}\right] $$