# 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.

• FYI, you showed that $I + M$ is idempotent, not anti-idempotent. Mar 28, 2022 at 9:02

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!

• Thank you for a quick reply! The last line in your reply gave a precise reply to my query. I know that among the set of idempotent matrices, only $I$ is non-singular and all the rest are singular matrices. By your characterization of anti-idempotent matrices, we can deduce that $-I$ is the only non-singular anti-idempotent matrix and other anti-idempotent matrices are singular matrices. Thanks a lot for a quick proof. Mar 28, 2022 at 8:52
• @Dr.Sundar No problem! Mar 28, 2022 at 8:53

$$det(A)^2=det(A^2)= (-1)^n det(A)$$

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 $$-A$$)

• Can you throw some light on your answer, "Moreover if A is non-singular, then the image of A has to be equal to n.." // If $A$ is non-singular, then the rank of $A$ is $n$ and its range is equal to $R^n$. . Some typo? I think you meant the image (range) of $A$ is equal to $R^n$, the \$n-space.. Kindly correct the typo, thanks! Mar 28, 2022 at 9:19
• @Dr.Sundar uh thank you very much, I missed to write Mar 28, 2022 at 9:21
• @Dr.Sundar okay now I wrote better I hope. Please tell me if you agree with me when you have time :) Mar 28, 2022 at 10:32
• Your reply not only answered my query, but also enlightened me with some deep thinking. I thank you once again for great help! 🙏🙏 Mar 28, 2022 at 23:01
• @Dr.Sundar no problem, we are here to think and to learn together 💪 Mar 28, 2022 at 23:02

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.