# Families of Idempotent 3×33\times 3 Matrices

I did the following analysis for $2\times2$ real idempotent (i.e. $A^2=A$) matrices:

So in particular we have $(a+d)c=c$ and $(a+d)b=b$ so if either $b$ or $c$ is nonzero we have $a+d=1$. We also see that $a$ and $d$ both satisfy the equation $x^2+bc=x\iff x^2-x+bc=0$ which is a quadratic equation having solutions

But this is only possible if $bc\leq 0.25$ for otherwise the above expression is not real. This gives us the following cases:

CASE 1: If $b,c=0$ we have $x\in\{0,1\}$ and since $a+d=1$ is unnecessary we have four possibilities: $(a,d)\in\{(0,0),(1,0),(0,1),(1,1)\}$.

CASE 2: If $bc=0.25$ we have $x=0.5$ so $a=d=0.5$.

CASE 3: If $bc<0.25$ yet $(b,c)\neq(0,0)$ we have $x\in L=\{0.5-\sqrt{0.25-bc},0.5+\sqrt{0.25-bc}\}$ and to have $a+d=1$ we must have $\{a,d\}=L$ so that if $a$ is one solution, then $d$ is forced to be the other solution. Or the other way around.

The cases can be illustrated via the following diagram graphing the hyperbola $xy=0.25$ corresponding to CASE 2, the area $xy<0.25$ corresponding to CASE 3, and the point $(0.0)$ corresponding to CASE 1:

The blue bands show the graphs of $xy=k$ for $k=0.05$ to $0.20$ and the cyan bands show $xy=k$ for $k=-0.05,-0.10,...$ For instance one could choose $(b,c)=(3.75,-1)$ so that $\sqrt{0.25-bc}=2$ thus rendering $x=0.5\pm 2=-1.5$ and $2.5$ and form the matrix

which will then be idempotent, as an example of CASE 3.

QUESTIONs:

1. Can similar descriptions be derived for $3\times 3$ matrices?
2. Is this a well known description of idempotent $2\times 2$ matrices?

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 :

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 :

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 $A=I$).