Families of Idempotent 3×33\times 3 Matrices

I did the following analysis for 2×2 real idempotent (i.e. A2=A) matrices:
[abcd]2=[a2+bc(a+d)b(a+d)cbc+d2]=[abcd]
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 x2+bc=xx2x+bc=0 which is a quadratic equation having solutions
x=1±14bc2=0.5±0.25bc
But this is only possible if bc0.25 for otherwise the above expression is not real. This gives us the following cases:

CASE 1: If b,c=0 we have x{0,1} and since a+d=1 is unnecessary we have four possibilities: (a,d){(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)(0,0) we have xL={0.50.25bc,0.5+0.25bc} 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:

enter image description here

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 0.25bc=2 thus rendering x=0.5±2=1.5 and 2.5 and form the matrix
A=[1.53.7512.5]
which will then be idempotent, as an example of CASE 3.

QUESTIONs:

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

Answer

By the kernel decomposition theorem, you have
R3=Ker(A(AI))=Ker(A)Ker(AI), so that
rank(A)+rank(AI)=3. If one of rank(A) or
rank(AI) is zero, we have the trivial cases A=0 or A=I. Otherwise,
one of rank(A) or rank(AI) is 1, and the other is 2. Remember also
that when A is idempotent, rank(A) coincides with trace(A).

Technical remark. A 3×3 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 rank(A)=1, using the technical remark and reinjecting into
A×A=A, we see that A is of one of the following three forms :

A1=(1xa2ya3x(1xa2ya3)y(1xa2ya3)a2xa2ya2a3xa3ya3),A2=(0a1xa101xa3x(1xa3)0a3xa3),A3=(00a00b001)

Similarly, when rank(AI)=1, using the technical remark and reinjecting into
A×A=A, we see that A is of one of the following forms :

A4=(xa2ya3x(1xa2ya3)y(1xa2ya3)a21+xa2ya2a3xa31+ya3),A5=(1a1xa10xa3x(1xa3)0a31+xa3),A6=(10a01b000)

All the Ai 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).

Attribution
Source : Link , Question Author : String , Answer Author : Ewan Delanoy

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