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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?

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@Eric: Thanks for the edit. Just curious, though -- why shouldn't we use the characters I'd used? – Mehrdad Feb 28 '11 at 3:29
Well Latex is prettier, and I think has more characters available. Also you can center equations and such. Most mathematical documents are written with Latex so it is familiar to people, and good to learn for the future. Maybe someone else can provide a more detailed answer. – Eric Naslund Feb 28 '11 at 3:32
@Eric: Huh ok. I was going to use LaTeX at first, but since it was possible not to, then I just didn't (so as to avoid loading images each time). Thanks anyway. :] – Mehrdad Feb 28 '11 at 3:33
up vote 10 down vote accepted

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 $A=-I$.

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Hm... by the way, isn't A = 0 also a solution? – Mehrdad Feb 28 '11 at 3:18
@Mehrdad: 0 matrix is not invertible. – InterestedGuest Feb 28 '11 at 3:19
@Mehrdad: Every projection satisfies $P^2=P$. For your original problem, any matrix that is similar to a matrix that is diagonal and all diagonal entries are either $-1$ or $0$ satisfies the equation $A^2+A=0$ (though any $0$ will mean the matrix is not invertible). – Arturo Magidin Feb 28 '11 at 3:20
@Merhdad: see my answer for the case of non-invertibility – Arturo Magidin Feb 28 '11 at 3:21
@Mehrdad: I don't know if you can solve the general case (where $A$ is not invertible) without knowing more. Perhaps that is why you were given only the easy invertible case to solve. – Arturo Magidin Feb 28 '11 at 3:25

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, $k=0,\ldots,n$.

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+1 for your more general answer, Arturo. It also reminds me how rusty my linear algebra has gotten... – Zev Chonoles Feb 28 '11 at 3:32
@Zev: Well, one might say I'm cheating, since I'm teaching a linear algebra course and we just finished the rational canonical form. (-: – Arturo Magidin Feb 28 '11 at 3:34

It is given that $A$ is invertible and hence multiply by $A^{-1}$ to get $$A^{-1}A^2 + A^{-1}A = 0 \Rightarrow A + I =0 \Rightarrow A = -I$$

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