Projectors and Idempotent Matrices

[hard sequoia]

I need help with these two questions. The solution to the first one (13) is so odd. A=I or rank(A)= trace(A) < 2

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[narrow knot]

Suppose A = {{a, b}, {c, d}}. We have:
A^2 = {{a^2 + bc, b(a + d)}, {c(a + d), bc + d^2}}
We need A = A^2. So, we get a system:
a^2 + bc = a
b(a + d) = b
c(a + d) = c
bc + d^2 = d
First, let's look at the second equation.
b(a + d) = b
(b - 1)(a + d) = 0
So, either b = 1 or a + d = 0.
Same for the third equation: c = 1 or a + d = 0.
So, now I think you can just look at different cases.

[tall shore]

a projector is necessarily idempotent

so that narrows down your search immediately

@hard sequoia

[narrow knot]

Yup.

[regal belfry]

That should be b(a+d-1) so b=0 or a+d=1

[narrow knot]

Ah, right! My bad.