Representation of an involution operator with an operator matrix 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

 A: 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 
$$\tag3
(I-P)A(I-P)=(I-P)A=A-PA=A(I-P). 
$$
On $(I-P)H$ we have, for any $x\in H$,
$$\tag4
0=(A^2-I)(I-P)x=(A-I)(A+I)(I-P)x.
$$
Now, using $(3)$,  $$\tag5(A+I)(I-P)=A-AP+I-P=(I-P)(A+I)(I-P).$$ So $(A+I)(I-P)x\in \ker(A-I)^\perp$; using that $A-I$ is injective on $\ker(A-I)^\perp$, we obtain from $(4)$ that 
$$\tag6
(A+I)(I-P)=0. 
$$
Combining $(5)$ and $(6)$ we get 
$$
(I-P)A(I-P)=-(I-P). 
$$
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 
$$
A=\begin{bmatrix}I&PA(I-P)\\0&-I\end{bmatrix} . 
$$
