equations of unbounded operator $T$ 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.
 A: 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
$$
     \left[\begin{array}{c} -Tx \\ x \end{array}\right]+\left[\begin{array}{c} y\\ T^*y\end{array}\right]=\left[\begin{array}{c} a \\ b \end{array}\right]
$$
