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=x⟺x2−x+bc=0 which is a quadratic equation having solutions

x=1±√1−4bc2=0.5±√0.25−bc

But this is only possible if bc≤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∈{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 x∈L={0.5−√0.25−bc,0.5+√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 toCASE 3, and the point (0.0) corresponding toCASE 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 √0.25−bc=2 thus rendering x=0.5±2=−1.5 and 2.5 and form the matrix

A=[−1.53.75−12.5]

which will then be idempotent, as an example ofCASE 3.

QUESTIONs:

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

**Answer**

By the kernel decomposition theorem, you have

R3=Ker(A(A−I))=Ker(A)⊕Ker(A−I), so that

rank(A)+rank(A−I)=3. If one of rank(A) or

rank(A−I) is zero, we have the trivial cases A=0 or A=I. Otherwise,

one of rank(A) or rank(A−I) 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=(1−xa2−ya3x(1−xa2−ya3)y(1−xa2−ya3)a2xa2ya2a3xa3ya3),A2=(0a1xa101−xa3x(1−xa3)0a3xa3),A3=(00a00b001)

Similarly, when rank(A−I)=1, using the technical remark and reinjecting into

A×A=A, we see that A is of one of the following forms :

A4=(−xa2−ya3x(−1−xa2−ya3)y(−1−xa2−ya3)a21+xa2ya2a3xa31+ya3),A5=(1a1xa10−xa3x(−1−xa3)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*