Let $A_n$ be square matrix where $n \geq 2$ and $A^2 +2A=0$. Then
- A is singular
- A is nonsingular
- 0 and -2 are eigenvalues of A
- either 0, or -2 is not an eigenvalue of A
- (1)-(4) are false.
My attempt: thanks to Cayley-Hamilton theorem, we know that characteristic polynomial of A equal zero. Replacing A with $\lambda$ gives $\lambda^2 + 2\lambda = 0$. Hence, 0 and -2 are eigenvalues of A, i.e. (3).
There are 2 problems: 1st, the answer should be (5), and 2nd problem is that assuming that (3) is true, (1) is also must be true due to 0 eigenvalue. So, I made mistake somewhere.