Are there non-identity matrices that fit the definition of $A^2 = A$?

Question

Consider the 'pseudo' definitions below:

$A$ is an $n \times n$ matrix

$A$ is $m_1$ if $A^2= I$ (the identity matrix)

$A$ is $m_2$ if $A^2 = A$

Currently, the only type of matrix that is $m_2$ that I can think of is the identity matrix. This would then mean that any $m_2$ matrix is also an $m_1$ matrix, also the definition also implies that an $m_1$ matrix is also $m_2$. Are there other matrices that are $m_2$ but not m1?

Answer 1

Yes, there are many such matrices. These are known as "self-inverse" matrices. I can provide an example:

A rotation matrix $R \in \mathbb{R}^{2x2}$ rotates vectors by an angle $\theta$. The rotation matrix that rotates vectors by 180 degrees is not an identity matrix. It is actually $-I$. However, if you are looking for something more interesting, any 2x2 matrix of the form:

a b

c -a

And where $a^2 + bc - 1 = 0$ is self inverse and not identity. More information can be found on the wikipedia page

Answer 2

As CalvinLin pointed out in the comments, such matrices are called projection matrices, or idempotent matrices. One obvious property is that if $x\in R(A)$ then $Ax = A^2\xi = A\xi = x$. Such matrices also have the property that they decompose $\mathbb{F}^n$ into their range and nullspace, in the sense that $\mathbb{F}^n = R(A)\oplus N(A)$ where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$. This is easily proven by noticing that if $x\in R(A)\cap N(A)$ then $$ 0 = Ax = x\implies R(A)\cap N(A)=\{0\}, $$ and $$\DeclareMathOperator{\rank}{rank}\DeclareMathOperator{\dim}{dim} \dim(R(A)+N(A))=\operatorname{rank}(A)+\dim N(A) - \dim R(A)\cap N(A)=\rank(A)+\dim N(A)=n. $$ This has the immediate consequence that any such matrix is similar to a "partial" identity matrix: If $S = \begin{pmatrix}X & Y\end{pmatrix}$ is full rank, where $R(X)=R(A)$ and $R(Y)=N(A)$, then $S$ is nonsingular, and $$ AS = \begin{pmatrix}AX & 0\end{pmatrix}= \begin{pmatrix}X&Y\end{pmatrix}\begin{pmatrix}I_r & 0 \\0 & 0_{n-r}\end{pmatrix}\implies S^{-1}AS = \begin{pmatrix}I_r & 0 \\0 & 0_{n-r}\end{pmatrix}, $$ where $r=\rank{A}$.

Moreover $I-A$ is also idempotent and $N(A) = R(I-A)$.

For your second question, the matrices that satisfy $A^2 = I$ are also called involutory. If $A$ is involutory, it must be nonsingular since $A = A^{-1}$, so $R(A)=\mathbb{F}^n$. But if it's also idempotent, then for all $x\in\mathbb{F}^n$ we have that $Ax = x\implies A = I$. In other words, the only matrix that satisfies $m_1$ and $m_2$ is the identity. Another nice result, which you can attempt to prove yourself, is that $A$ is idempotent iff $I-2A$ is involutory.