Let $A$ be a bounded linear operator on a complex Hilbet space $H$ such that $$A^2=I.$$
Why there exists a bounded operator $T$ such that $A$ will be equal to the operator matrix $\begin{pmatrix} I&T \\ 0&-I \end{pmatrix}$? Is it possible to find the expression of $T$?
This property is used in the proof of the following theorem in this paper
We can decompose $H$ as $$\tag1 H=\ker(A-I)\oplus \ker(A-I)^\perp.$$ Let $P$ be the orthogonal projection onto $\ker (A-I)$. For any $x\in H$ we have $Px\in \ker (A-I)$, so $$ APx=Px. $$ Thus $$\tag2AP=P.$$ If we write $A$ as a block matrix with respect to the decomposition $(1)$, we have $$ A=\begin{bmatrix} PAP&PA(I-P)\\ (I-P)AP&(I-P)A(I-P)\end{bmatrix}. $$ From $(2)$ we have that $PAP=P$, $(I-P)AP=0$, and $$\tag3 (I-P)A(I-P)=(I-P)A=A-PA. $$ Since $A^2-I=0$, for any $x\in H$ $$\tag4 0=(A^2-I)x=(A-I)(A+I)x. $$ That is, $(A+I)x\in\ker(A-I)=PH$. Thus $P(A+I)=A+I$, and so $$\tag5 A-PA=-(I-P).$$ Now, using $(3)$ and $(5)$, $$\tag6 (I-P)A(I-P)=(I-P)A=-(I-P). $$
So the matrix representation of $A$ becomes $$\tag7 A=\begin{bmatrix} P&PA(I-P)\\ 0&-(I-P)\end{bmatrix} . $$ As $P$ acts as the identity on $\ker(A-I)$ and $I-P$ acts as the identity on $\ker(A-I)^\perp$, we can write $(7)$ as $$ A=\begin{bmatrix}I&PA(I-P)\\0&-I\end{bmatrix} . $$