What matrix A with dimension n x n is always true given A - A^2 = O

Question

What matrix A with dimension n x n is always true given A - A^2 = O where O is the zero matrix with dimension n x n and I with dimension n x n is the identity matrix.

A) A is a diagonal matrix
B) A = A^2
C) A = I
D) A = O
E) I = A^2

Just got out of an exam with this question and I chose B. I thought about this question for a little bit longer after I handed in my exam and I came to the conclusion that A = I.

A - A^2 = O
A = A^2
AA^-1 = (A^-1)(A^2)
I = A

Can someone confirm if this is right?

Answer 1

It seems to me that the other answers have misinterpreted the question. My understanding is that the question asks:

Given that $A$ is a matrix satisfying $A - A^2 = 0$, which of the following is always true?

The only correct answer to this question is B. Certainly, if $A = A^2$, then we can say that $$ A = (A - A^2) + A^2 = 0 + A^2 = A^2 $$ As a counterexample to all of the other choices, consider the matrix $$ A = \pmatrix{1&1\\0&0} $$

Answer 2

It is right (answer C, the proof needs a little modification), but D) is also true. A counterexample to E): $$ A=\begin{bmatrix} 0&1\\ 1&0 \end{bmatrix} $$ And for example $2I$ is a counterexample to A).

Edit: And corected version of your proof: $A - A^2 = O$, $A = A^2$, and now, instead of

$AA^{-1} = (A^{-1})(A^2)$

either $AA^{-1} = A^2(A^{-1})$ or $A^{-1}A = (A^{-1})(A^2)$

and finally $I = A$.

Answer 3

I think $B$, $C$ and $D$ are all correct.

Infact $B$ is the question rewritten in another way. And as already established $C$ and $D$ are correct.