Solving for Invertibility of (B+I) Given B=B^2

[transcendency]

Homework Statement

Given a matrix B, if B = B², is (B+I) invertible?

2. The attempt at a solution

det(B) = 0 or 1

rref(rref(B) + I) is I, so (rref(B) + I) is invertible

if det(B) = 1:
let E₁E₂...E_n = B
then E₁E₂...E_n(rref(B) + I) = B + E₁E₂...E_n

I'm not sure if what I did is even useful =(

 

[CompuChip]

What can you say about det(B + I)?

 

[transcendency]

If det(B) = 1, then B^-1B = B^-1B², B = I
so det(B+I) = det(2I) != 0, so B+I is invertible.

I'm still stuck on if det(B) = 0..

I'm quite sure that if B = B², then B must be a diagonal matrix with entries being either 1 or 0, but I don't know how to prove it.

 

[HallsofIvy]

If B= B² then B²- B= B(B- I)= 0.

 

[matt grime]

I'm quite sure that if B = B², then B must be a diagonal matrix with entries being either 1 or 0, but I don't know how to prove it.

That follows from the fact you know its minimal poly divides X^2-X, hence you know all the possible eigenvalues (0 and 1), asndyou know the geometric multiplicity is 0 or 1.

Alternatively remember that a definition of an eigenvalue is:


t is an eigenvalue of X if and only if X-tI is not invertible.