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.
Best Answer
The fact that $A^2 + 2A = 0$ implies that the minimal polynomial of $A$ divides $\lambda^2 + 2\lambda$. So the options for the minimal polynomial are $p_1(\lambda) = \lambda^2 + 2\lambda$, $p_2(\lambda) = \lambda$ and $p_3(\lambda) = \lambda + 2$. If the minimal polynomial of $A$ is $p_2$ then $A = 0$ (and so the only eigenvalue of $A$ is $0$) and if the minimal polynomial is $p_3$ then $A = -2I$ (and so the only eigenvalue of $A$ is $-2$).
If the minimal polynomial of $A$ is $p_1$ then since the minimal polynomial and the characteristic polynomial have the same roots it implies that the eigenvalues of $A$ are $0$ and $2$.
Using $A = -2I$, $A = 0$ and $A = \mathrm{diag}(0,-2)$, you see that the only possible answer is $5$.