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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.

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

3 Answers 3

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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!

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  • $\begingroup$ 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. $\endgroup$
    – Dr. Sundar
    Mar 28, 2022 at 8:52
  • $\begingroup$ @Dr.Sundar No problem! $\endgroup$ Mar 28, 2022 at 8:53
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$det(A)^2=det(A^2)= (-1)^n det(A)$

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 $-A$)

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    $\begingroup$ 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! $\endgroup$
    – Dr. Sundar
    Mar 28, 2022 at 9:19
  • $\begingroup$ @Dr.Sundar uh thank you very much, I missed to write $\endgroup$ Mar 28, 2022 at 9:21
  • $\begingroup$ @Dr.Sundar okay now I wrote better I hope. Please tell me if you agree with me when you have time :) $\endgroup$ Mar 28, 2022 at 10:32
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    $\begingroup$ Your reply not only answered my query, but also enlightened me with some deep thinking. I thank you once again for great help! 🙏🙏 $\endgroup$
    – Dr. Sundar
    Mar 28, 2022 at 23:01
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    $\begingroup$ @Dr.Sundar no problem, we are here to think and to learn together 💪 $\endgroup$ Mar 28, 2022 at 23:02
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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.

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