Operator theory: $T^2=T$, but $T^{*}\neq T$

Question

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 :)

Answer 1

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

$$[T^*]=\begin{pmatrix}1&0\\{}\\\frac{1}{2}&0\end{pmatrix}\neq[T]$$

Answer 2

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.

Answer 3

You can take $$T = \begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}$$

Answer 4

• 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 $x\ne 0$.