in the book matrix theory but Fuzhen Zhang it is said on page 94, that if $J$ is a Jordan block and if $J^2=J$ then, J must be of size 1, that is, J is a number.
I do not get it, take the matrix with first row $(1,1)$ and second row $(0,1)$. This could serve as a Jordan block right?- But since this matrix is idempotent and not a number, this somehow contradicts this statement.
On
Any idempotent linear operator is diagonalisable, with eigenvalues in $\{0,1\}$ only. This is because the polynomial equation $x^2=x$ that characterises idempotent operators has simple roots $0,1$. In particular the non-diagonalisable Jordan block you describe cannot be idempotent, and indeed you can check that it isn't.
It is easy to see that the only diagonalisable Jordan blocks are those of size $1\times1$.
The matrix you cite is not idempotent.
A nice way to prove the claim you mention is to write $\left[ \begin{matrix} \lambda & 1 \\ 0 & \lambda \end{matrix} \right] = \left[ \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix} \right] + \left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] = \Lambda + N$, where $N$ is nilpotent. For a general Jordan block of size $k$, we have that the nilpotent part $N$ has $N^{k-1} \neq 0$ and $N^{k}= 0$. (The 1s on the superdiagonal of $N$ bump up to the next diagonal with each multiplication.)
Now $J^2 = \Lambda^2 + 2\Lambda N + N^2$. If $J^2 = J$ then $\Lambda^2 + 2\Lambda N + N^2 = \Lambda + N$. Assume $N \neq 0$ (that is, assume this is not a trivial Jordan block). If the diagonals of the two matrices are to be equal, then $\Lambda$ must be a $k \times k$ scalar matrix of 1s, or of zeroes. But then the superdiagonals cannot be equal because $2\Lambda N$ is either a diagonal of 2s or a diagonal of zeroes (not 1s, as in $J$).