Families of Idempotent $3\times 3$ Matrices

Solution 1:

By the kernel decomposition theorem, you have ${\mathbb R}^3={\sf Ker}(A(A-I))={\sf Ker}(A)\oplus{\sf Ker}(A-I)$, so that ${\sf rank}(A)+{\sf rank}(A-I)=3$. If one of ${\sf rank}(A)$ or ${\sf rank}(A-I)$ is zero, we have the trivial cases $A=0$ or $A=I$. Otherwise, one of ${\sf rank}(A)$ or ${\sf rank}(A-I)$ is $1$, and the other is $2$. Remember also that when $A$ is idempotent, ${\sf rank}(A)$ coincides with ${\sf trace}(A)$.

Technical remark. A $3\times 3$ matrix has rank $1$ iff it has one of the three forms $[C,xC,yC],[0,C,xC],[0,0,C]$ where $x$ and $y$ are constants and $C$ is a column with at least one nonzero entry.

When ${\sf rank}(A)=1$, using the technical remark and reinjecting into $A\times A=A$, we see that $A$ is of one of the following three forms :

$$ \begin{array}{lcl} A_1&=&\left(\begin{array}{ccc} 1-xa_2-ya_3 & x(1-xa_2-ya_3) & y(1-xa_2-ya_3) \\ a_2 & xa_2 & ya_2 \\ a_3 & xa_3 & ya_3 \\ \end{array}\right), \\ A_2&=&\left(\begin{array}{ccc} 0 & a_1 & xa_1 \\ 0 & 1-xa_3 & x(1-xa_3) \\ 0 & a_3 & xa_3 \\ \end{array}\right), A_3=\left(\begin{array}{ccc} 0 & 0 & a \\ 0 & 0 & b \\ 0 & 0 & 1 \\ \end{array}\right) \\ \end{array} $$

Similarly, when ${\sf rank}(A-I)=1$, using the technical remark and reinjecting into $A\times A=A$, we see that $A$ is of one of the following forms :

$$ \begin{array}{lcl} A_4&=&\left(\begin{array}{ccc} -xa_2-ya_3 & x(-1-xa_2-ya_3) & y(-1-xa_2-ya_3) \\ a_2 & 1+xa_2 & ya_2 \\ a_3 & xa_3 & 1+ya_3 \\ \end{array}\right), \\ A_5&=&\left(\begin{array}{ccc} 1 & a_1 & xa_1 \\ 0 & -xa_3 & x(-1-xa_3) \\ 0 & a_3 & 1+xa_3 \\ \end{array}\right), A_6=\left(\begin{array}{ccc} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 0 \\ \end{array}\right) \\ \end{array} $$

All the $A_i$ are idempotent. We have thus obtained a necessary and sufficient condition, made up of six cases (eight if you include the degenerate cases $A=0$ and $A=I$).

Solution 2:

This is a partial answer, but something that you may find useful. Let $A$ be a $3\times3$ matrix and denote $t=\text{tr}(A)$, $d=\det(A)$ and $a=a_1+a_2+a_3$, where $a_k=\det(A_{\hat k\hat k})$ is the subdeterminant corresponding to the $k$th diagonal element of $A$. The characteristic polynomial of $A$ is $$ p(x) = \det(A-xI) = -x^3+tx^2-ax+d. $$ By the Cayley-Hamilton theorem $p(A)=0$, when the polynomial is naturally interpreted for matrices. That is, $$ -A^3+tA^2-aA+dI=0. $$ We can factor this polynomial: $$ -(A^2-A)(A-(t-1))+(t-a-1)A+dI=0. $$ Suppose that $A$ is idempotent: $A^2=A$. The above equation gives now $(t-a-1)A+dI=0$. If $t-a-1\neq0$, this means that $A$ is a multiple of the identity. It is easy to see that the only such solutions are $A=I$ and $A=0$ (and these are of course idempotent matrices).

All other idempotent matrices must therefore satisfy $t-a-1=0$ and $d=0$. That is, $\det(A)=0$ and $$ \text{tr}(A) = 1+\det(A_{\hat1\hat1})+\det(A_{\hat2\hat2})+\det(A_{\hat3\hat3}). $$ The second condition can be alternatively expressed as $\text{tr}(A)=1+\text{tr}(\text{cof}(A))$, where $\text{cof}(A)$ is the cofactor matrix of $A$. Trying to find explicit solutions to these equations seems messy.

Let me stress that these conditions are necessary for all idempotent matrices other than $0$ and $I$, but not sufficient. The characteristic polynomial does not retain all information about the original matrix. The following matrix satisfies both of my conditions but is not idempotent: $$ \begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&0&0 \end{pmatrix}. $$

As a remark, the same approach works better for a $2\times2$ matrix $A$. Now $p(x)=x^2-tx+d$, so $A^2-tA+dI=0$. Using $A^2=A$ gives $(1-t)A+dI=0$. Again, either $A$ is a multiple of the identity (these two cases we know), or $1-t=0$ and $d=0$.

The equations are much easier so solve now. Suppose $ A = \begin{pmatrix} a&b\\ c&d \end{pmatrix}. $ We have $ad-bc=0$ and $a+d=1$. If $c=0$, we get $a\in\{0,1\}$ and $d=1-a$, and $b$ can be anything. If $c\neq0$, we get $a=1-d$ and $b=(d-d^2)/c$, and $d$ can be anything. It is easy to check which of these solutions are actually idempotent.