Operator theory: $T^2=T$, but $T^{*}\neq T$ Give an example of operator $T:\mathbb{R}^2\to\mathbb{R}^2$ with $T^2=T$, but $T^{*}\neq T$. 
What could I consider ? 
thanks :) 
 A: *

*Restrict yourself to linear operators (matrices) and make the general ansatz


$$T(a,b,c,x\ne c^*-c)=\begin{pmatrix}a&c\\{}\\c+x&b\end{pmatrix}.$$


*

*Then square it (if you're lazy plug in {{a,c},{c+x,b}}^2//FullSimplify in Wolfram Alpha) and solve one linear and one quadratic equation which stem from the condition $T^2-T=0.$


You find the family of solutions
$$T(c,x\ne c^*-c)=\begin{pmatrix}\tfrac{1}{2}\pm \tfrac{1}{2}\sqrt{1-4\ c\ (c+x)}&c\\{}\\c+x&\tfrac{1}{2}\mp \tfrac{1}{2}\sqrt{1-4\ c\ (c+x)})\end{pmatrix}.$$
For $c=0$ and $c=-x$ you'd get the most simple examples
$$\begin{pmatrix}1&0\\{}\\x&0\end{pmatrix},\ \ \  \begin{pmatrix}1&-x\\{}\\0&0\end{pmatrix},$$
with $x\ne 0$.
A: You can take $$T = \begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}$$
A: An oblique projection would do. The equation $T^2=T$ is a defining property of projections, and $T^*=T$ (for a projection) is connected with orthogonality.
A: Edit: the following is a result of comments by Harald:
Let
$$S\binom{x}{y}:=\binom{x+y}{-y}\,\,,\,\,\text{or in matrix form:}\;\;[S]=\begin{pmatrix}1&1\\0&\!\!\!-1\end{pmatrix}$$
Since $\,S^2=I\,$ we get that
$$T:=\frac{1}{2}\left(S+I\right)\;\;\;\text{fulfills}\;\;\;T\binom{x}{y}=\binom{x+\frac{1}{2}y}{0}\,\,,\,\,\text{or as matrix:}\;\;[T]=\begin{pmatrix}1&\frac{1}{2}\\{}\\0&0\end{pmatrix}$$
yet
$$[T^*]=\begin{pmatrix}1&0\\{}\\\frac{1}{2}&0\end{pmatrix}\neq[T]$$
