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Let $A$ be an $n \times n$ matrix with real entries. Prove that no matrix $A^2 + A + I$ is nilpotent.

I tried to approach by contradiction and suppose that such an $A$ did exist so $(A^2 + A + I)^n = [0]$. I then need to figure $A^2 + A + I$, but the function $x^2 + x + 1$ doesn't factor over $\mathbb{R}$, and the factorization over $\mathbb{C}$ isn't very helpful. Is the fact that it doesn't factor over $\mathbb{R}$ enough?

Ben Grossmann
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Tom Baker
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    This seems to be false as stated. For instance, the matrix $$ A = \pmatrix{0&-1\1&-1} $$ satisfies $A^2 + A + I = 0$, and $0$ is nilpotent – Ben Grossmann Nov 25 '20 at 22:45
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    This is false as written; if $A = \left[ \begin{array}{cc} 0 & -1 \ 1 & -1 \end{array} \right]$ you can check that $A^2 + A + I = 0$, which is nilpotent. Does "nilpotent" here mean "nilpotent and nonzero"? – Qiaochu Yuan Nov 25 '20 at 22:46

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This statement is false. We find that the companion matrix associated with the polynomial $(x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$ is given by $$ A = \pmatrix{0&0&0&-1\\1&0&0&-2\\0&1&0&-3\\0&0&1&-2}. $$ This matrix has real entries, but it satisfies $(A^2 + A + I)^2 = 0$. In fact, we find that $A^2 + A + I \neq 0$, which means that $A^2 + A + I$ is a non-zero nilpotent matrix.

For posterity: as discussed in the comments to this answer, you can get an answer in "real Jordan form", namely $$ A = \pmatrix{C&I\\0 & C}, \quad C = \pmatrix{-1/2 & -\sqrt{3}/2\\\sqrt{3}/2& -1/2}. $$ We could also take $$ C = \pmatrix{0&-1\\1&-1}, $$ which is an interesting blend of approaches.

If you use another version of the real Jordan form, you might instead end up with the example $$ A = \left[\begin{array}{cc|cc} -1/2 & 1 & -\sqrt{3}/2 & 0\\ 0 & -1/2 & 0 & -\sqrt{3}/2\\ \hline \sqrt{3}/2 & 0 & -1/2 & 1\\ 0 & \sqrt{3}/2 & 0 & -1/2 \end{array}\right]. $$

Ben Grossmann
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    Nice! I had just constructed a $4 \times 4$ example using a generalized Jordan block but this works and is a bit easier. – Qiaochu Yuan Nov 25 '20 at 22:50
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    @QiaochuYuan That works too. Another idea that occurred to me was to build a complex example and use the standard representation of $\Bbb C^2$ in $M_2(\Bbb R)$ to replace the blocks and get a $4 \times 4$ example, but I guess that really amounts to the same thing as using the "real Jordan form", which I guess is what you mean by a generalized Jordan block. – Ben Grossmann Nov 25 '20 at 22:54
  • @lonzaleggiera You're right, thanks for the correction – Ben Grossmann Mar 27 '24 at 15:51