How to establish that an anti-idempotent matrix is singular? In linear algebra, idempotent matrices are defined by
$$
A^2 = A \tag{1}
$$
for a square matrix $A$. Obviously, the identity matrix $I$ is an idempotent matrix. It can be also shown that if $M$ is idempotent, then $I - M$ is idempotent by a trivial calculation.
$$
(I - M) (I - M) = I - M - M + M^2 = I - M - M + M = I - M
$$
In a similar manner, we can define an anti-idempotent matrix $A$ by the condition
$$
A^2 = - A \tag{2}
$$
(A trivial example is the zero matrix.)
To find non-trivial examples of an anti-idempotent matrix $A$, I considered the case of $(2 \times 2)$ matrices:
$$
A = \left[ \matrix{ a & b \cr
                    c & d \cr} \right]
$$
If $A$ is anti-idempotent, then it must satisfy: $A^2 = -A$.
This leads to a set of $4$ equations:
$$
a^2 + b c = - a, \ a b + b d = - b, \ c a + c d = -c, \ \ b c + d^2 = -d 
$$
A simple manipulation results in the equations
$$
b (a + d + 1) = 0, c (a + d + 1) = 0, a^2 + b c = -a, b c + d^2 = -d.
$$
Taking $a = 2$, we see that $a + d + 1 = 0$ or $d = -3$.
We can choose $b$ and $c$ from $b c = -6$. One choice is $b = 2, c = -3$.
Thus, an anti-idempotent matrix is:
$$
A = \left[ \begin{array}{cc}
            2 & 2 \\
            -3 & -3 \\
        \end{array} \right]
$$
(It is easy to check that $A^2 = -A$.)
It can be easily shown : If $M$ is an anti-idempotent matrix, then $I + M$ is also anti-idempotent. Indeed,
$$
(I + M) (I + M) = I + M + M + M^2 = I + M + M - M = I + M.
$$
The examples I considered for anti-idempotent matrices yield singular matrices.
I like to know if it is generally true that anti-idempotent matrices are singular matrices. How to establish this result? Your comments are welcome.
 A: Since anti-idempotent matrices $A$ of any size $n\ge2$ verify $A^2+A=0$, then their minimal polynomial is given by $p(x)=x^2+x=x(x+1)$, meaning that their only eigenvalues can be either 0 or -1. Since the minimal polynomial is $p(x)$, then this means that the matrices are also all diagonalizable.
Therefore a matrix $A$ is anti-idempotent if and only if $A$ is diagonalizable and its eigenvalues are 0 and/or 1.
A: $det(A)^2=det(A^2)= (-1)^n det(A)$
so that
$det(A)(det(A)+(-1)^{n+1})=0$
Hence $det(A)$ is zero or $det(A)=(-1)^n$
However $A^2=-A$ means that $-1$ is an eigenvalue for $A$, if $A$ is not the zero matrix.
Moreover if $A$ is non-singular, then the dimension of the image of $A$ has to be equal to $n$. However the image of $A$ is contained (really is exactly equal) in the eigenspace of $A$ of eigenvalue $-1$, so that the geometric multiplicity of the eigenvalue $-1$ is exactly $n$.
(Another way is simply to multiply $A^2=-A$ by $A^{-1}$ to get directly $A=-I$)
Thus $A$ is diagonalizable and $A=B(-I)B^{-1}=-I$
This means that if $A$ is an idempotent non-singular matrix, then $A$ is equal to $-I$.
Now consider the general case, $A$ could be singular. Then by nullity-rank theorem
$V=ker(A)\oplus Im(A) $
But here $Im(A)=V_{-1}$ and in general $ker(A)=V_{0}$, where $V_\lambda$ is the eigenspace of $A$ of eigenvalue $\lambda$. Thus
$V=V_{0} \oplus V_{-1}$
That means $A$ is diagonalizable with eigenvalues $0$ and $-1$ and
$A=B\begin{pmatrix}-I_r & 0 \\ 0 & 0 \end{pmatrix} B^{-1}$
Thus $A$ is an anti-idempotent matrix if and only if $A=B\begin{pmatrix} -I_r & 0\\ 0 & 0 \end{pmatrix}B^{-1}$, $0\leq r\leq n$
Please observe that you can use the same argument to prove that if $A^2=A$ then $A=B\begin{pmatrix} I_r & 0\\ 0 & 0 \end{pmatrix}B^{-1}$ (or consider simply the anti-idempotent matrix  $-A$)
A: They are just matrices that satisfy $p(A) = 0$ where $p(z) = z^2 + z$. Note that $p$ is square-free, and hence any such $A$ is diagonalisable. The eigenvalues of $A$ are restricted to the roots of $p$, i.e. consisting of $0$ or $-1$, and nothing else.
So, the invertible anti-idempotent matrix are diagonalisable and have only $-1$ as an eigenvalue. It's not hard to see that $-I$ is the only possibility. Every other anti-idempotent matrix will not be invertible.
Another thing to note: $A$ is anti-idempotent if and only if $-A$ is idempotent!
