Find all invertible $n\times n$ matrices $A$ such that $A^2 + A = 0$ This was a question on one of our practice midterms:
Find all invertible $n \times n$ matrices $A$ such that $$A^2 + A = 0.$$
I was told to expand $A^2$ and then solve, but that seems like a really ugly (and hard-to-generalize) solution... are there any better ones?
 A: It is given that $A$ is invertible and hence multiply by $A^{-1}$ to get $$A^{-1}A^2 + A^{-1}A = 0 \Rightarrow A + I =0 \Rightarrow A = -I$$
A: If $A^2 + A = 0$, then the minimal polynomial of $A$ divides $t^2+t = t(t+1)$. That means that the characteristic polynomial of $A$ must be of the form $(-1)^nt^k(t+1)^r$, where $k+r=n$. $\lambda=0$ cannot be a root, though (since you specify that $A$ is invertible, so $0$ is not an eigenvalue), so the characteristic polynomial is necessarily $(-1)^n(t+1)^n$.
But that means that the minimal polynomial of $A$ must be $t+1$ (since it must divide the characteristic polynomial, and also $t(t+1)$). This implies that $A+I=0$, hence $A=-I$.
Added. If we drop the requirement that $A$ be invertible, then the invertible case proceeds as above. In the noninvertible case, the minimal polynomial is $t(t+1)$ or $t$. If it is $t$, then $A=0$. If the minimal polynomial is $t(t+1)$, then the matrix is diagonalizable (since the minimal polynomial splits and is square free), and the only eigenvalues are $0$ and $-1$. So the matrix $A$ is similar to a diagonal matrix in which every diagonal entry is $0$ or $-1$.
In summary, if we drop the requirement that $A$ be invertible, then there are $n+1$ similarity classes for possible $A$'s, each similarity class corresponding to a diagonal matrix that has $k$ diagonal entries equal to  $0$, followed by $n-k$ entries equal to $-1$s in the diagonal, $k=0,\ldots,n$. 
A: Note that
$$A^2+A=0\iff A^2=-A\iff A=-I$$
where $I$ is the identity matrix, and where we have used that $A$ is invertible in going from $A^2=-A$ to $A=-I$ (we multiplied by $A^{-1}$ on both sides). So the only solution is $A=-I$.
