$A^2=cA$ for some $c \neq 0$ Let $A \in \mathbb{C}^{n \times n}$ and $0 \neq c \in \mathbb{C}$ a given constant. Suppose that $A$ has the following property: 
$$A^2 = cA.$$
Questions. 
1) Is there a matrix class for matrices those have this special property? If there is what is the name of it?
2) It is true or not, that $A$ has rank $1$. If it is prove it, if not give a counterexample.
3) Anything interesting about $A$?
 A: The zero matrix satisfies $A^2 = cA$. As mentioned above, if $A$ is invertible then $A = cI$. Thus $A$ need not have rank $1$.
Otherwise, since $c \ne 0$, $(\frac{1}{c}A)^2 = \frac{1}{c}A$, so $\frac{1}{c}A$ is idempotent, hence is a projection onto $\text{im}(\frac{1}{c}A) = \text{im}(A)$, a proper nonzero subspace of $\mathbb{C}^n$. Thus such matrices $A$ are precisely scalar multiples of a projection matrix.
A: Since $A^2=cA,\; c\ne0$ then the polynomial with simple roots $x^2-cx=x(x-c)$ annihilates $A$ so it's a diagonalizable matrix and it's similar to
$$\operatorname{diag}(\underbrace{c,\cdots,c}_{r\;\text{times}},0,\cdots,0)=c\operatorname{diag}(\underbrace{1,\cdots,1}_{r\;\text{times}},0,\cdots,0)=cJ_r$$
so $A$ is similar to $cJ_r$ which's the matrix of $cP_V$: composition of an homothetie with a projection onto the image of $A$.
A: The identity matrix $I$ works with $c=1$.  And the matrix $aI$ works with $c=a$.  These are examples with rank $n$.  You could get a matrix of any rank $k$ with $1\le k<n$ by taking a matrix with $1$s in the first $k$ diagonal entries and $0$s everywhere else (both lower in the diagonal and in the off-diagonal).
