Operator $T$ with $T^2=T$ which is not a projection

Question

Can someone help me with this exercise? I know that if $T$ is a bounded self-adjoin operator with $T^2=T$ then it is a projection. I need an example of an operator with $T^2=T$ which is not a projection (not self-adjoint, of course)

Answer 1

I suppose that your projections are orthogonal projections. You can take, for instance,$$\begin{array}{rccc}T\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&(x+y,0).\end{array}$$It is a projection from $\mathbb{R}^2$ onto $\{(x,0)\,|\,x\in\mathbb{R}\}$. But it is not orthogonal with respect to the usual inner product.

Answer 2

This example may fit the case. Let P be an operator in $\mathcal{B}(\mathcal{H})$ written in the form \begin{pmatrix} I&X\\ 0&0\\ \end{pmatrix} where $I$ is the orthogonal projection onto $\mbox{ran}\, P$, the range of $P$, and $X$ can be any bounded linear operator from $(\mbox{ran}\, P)^{\perp}$ into $\mbox{ran}\, P$.

In general, given an orthogonal projection $E$, and an invertible operator $X$, not unitary, in $\mathcal{B}(\mathcal{H})$, then $XEX^{-1}$ is what you want.