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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?2011-02-28
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    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.2011-02-28
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    @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. :]2011-02-28

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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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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...2011-02-28
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    @Zev: Well, one might say I'm cheating, since I'm teaching a linear algebra course and we just finished the rational canonical form. (-:2011-02-28
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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?2011-02-28
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    @Mehrdad: 0 matrix is not invertible.2011-02-28
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    Oh... hm, what if we removed the invertibility constraint? Then how would you solve this?2011-02-28
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    @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).2011-02-28
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    @Merhdad: see my answer for the case of non-invertibility2011-02-28
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    @Arturo: Oh shoot... we haven't learned about characteristic polynomials yet! (This is the chapter before we learn about eigenvalues.) Is it possible to do this without those techniques?2011-02-28
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    @Arturo: I see that any diagonal matrix with 0 or -1 on the diagonal works, but how do you *get* to that? (Solution is easy to verify, but solving the problem isn't as easy.)2011-02-28
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    @Mehrdad: I just added the explanation, which uses the minimal polynomial.2011-02-28
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    @Arturo: I wish I knew what a minimal polynomial was... :(2011-02-28
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    @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.2011-02-28
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    @Arturo: Ah, that makes sense. Thanks! +12011-02-28
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    @Mehrdad: The minimal polynomial $\mu(t)$ is the monic polynomial of smallest degree such that $\mu(A)=0$.2011-02-28
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    @Arturo: Haha I think I'll wait until I learn what monic and minimal polynomials are, but thanks for your explanation nonetheless. :)2011-02-28
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    @Mehrdad: "Monic" means the leading coefficient is $1$. And I just *told* you what the "minimal polynomial" is.2011-02-28
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    @Arturo: Oh wow, that was simple... thanks! (Yeah, I understood that you'd explained what a minimal polynomial is, but I didn't get it since I didn't know what monic meant.)2011-02-28