Operator with given spectrum which is not projector.

Question

I'm stuck in making an example of such operator $A$, that spectrum of $A$ is $\{0,1\}$, but A is not a projector. Could you give me such example, please

Answer 1

You can take, for instance,$$\begin{array}{rccc}A\colon&\mathbb R^3&\longrightarrow&\mathbb R^3\\&(x,y,z)&\mapsto&(y,0,z).\end{array}$$Its spectrum is $\{0,1\}$, but$$A\bigl(A(x,y,z)\bigr)=(0,0,z)\neq A(x,y,z),$$unless $y=0$. So, $A$ is not a projection.

Answer 2

The desired counterexample is supplied by any real square matrix $A$ whose minimal polynomial is $p(\lambda)=\lambda^n(\lambda-1)^k$ for $n \ge 1, k\ge 1$ and $n > 1$ and/or $k > 1$. For example, $$ A=\pmatrix{1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0} $$ The minimal polynomial is $p(\lambda)=\lambda(\lambda-1)^2$. So the spectrum of $A$ is $\{0,1\}$, but $A$ is not a projection because $A(A-I)\ne 0$.