Let $n\geq 3$ be an odd integer and $K$ be a field with characteristic $2$.

Let $A,B\in M_n(K)$ s.t. $A^2+B^2=I_n$; is it true that $AB+BA$ is a singular matrix?

Remark. i) It is not difficult to see that the result is false when $n$ is even.

ii) The proposition is true for $n=3$; but I do not know the answer if $n=5,7,\cdots$.

EDIT. answer to @ i707107 . i) user1551 gave a counterexample for even $n$: $A=\begin{pmatrix}0&1\\1&0\end{pmatrix},B=\begin{pmatrix}0&1\\0&0\end{pmatrix}$.

ii) To obtain the reported result for $n=3$, I did a PC computation using Grobner basis theory (the basis contains $234$ elements). The case $n=5$ has too large complexity.

Remark. You are right; the problem is equivalent to show that $A+B$ admits $1$ as eigenvalue.

  • 3
    $\begingroup$ Would you please include the proofs for your remarks? I found an equivalent formulation: Since $(A+B)^2= I+AB+BA$, we have $\det(I-(A+B)^2) = \det(I-(A+B))\det(I+(A+B)) = (\det (I+A+B))^2$, so the same question to yours is that "is it true that $A+B+I$ is a singular matrix?" $\endgroup$ – Sungjin Kim Dec 22 '17 at 16:51

Edit. In hindsight, the original proof is overly complicated. It can be made somewhat simpler:

As $A^2+B^2=I$, the matrix square $A^2$ commutes with all polynomials in $B$. Hence $A$ commutes with $C=AB-BA$. Now, if $C$ is nonsingular, then $C^{-1}$ commutes with $A$ too and in turn $$ I=(AB-BA)C^{-1}=ABC^{-1}-BAC^{-1}=ABC^{-1}-BC^{-1}A. $$ Taking traces on both sides, we get $n=0$, which is a contradiction because $n$ is odd. $\square$

(The original proof goes here.)

The techniques for proving the classical result that "$A$ commutes with $[A,B]$ implies that $[A,B]$ is nilpotent" (over a field of characteristic zero) can be reused here.

First, as $A^2+B^2=I$, the matrix square $A^2$ commutes with all polynomials in $B$. Therefore $A$ commutes with $AB-BA$ when $1=-1$.

Next, if we define an iterate $X_1 = B$ and $X_{k+1}=X_k(AB-BA)$ for every $k\ge1$, one can prove by mathematical induction that $(AB-BA)^{k+1}=AX_k-X_kA$ and hence all positive powers of $C=AB-BA$ are traceless.

Now, let $\lambda_1,\ldots,\lambda_r$ be those distinct and nonzero eigenvalues of $C$ over the algebraically closure of the underlying field and $m_1,\ldots,m_r$ be their algebraic multiplicities respectively, so that $m_1+\ldots+m_r\le n$ and equality holds iff $C$ is nonsingular. Then the condition $\operatorname{tr}(C^k)=0$ for every $k\ge1$ results in a Vandermonde system of equations: $$ \pmatrix{ \lambda_1&\lambda_2&\cdots&\lambda_r\\ \lambda_1^2&\lambda_2^2&\cdots&\lambda_r^2\\ \vdots&\vdots&&\vdots\\ \lambda_1^r&\lambda_2^r&\cdots&\lambda_r^r\\ } \pmatrix{m_1\\ m_2\\ \vdots\\ m_r}=0. $$ If this were a field of characteristic zero, the above would imply that all $m_i$s are zero and $C$ is nilpotent. However, as we are working over a field of characteristic $2$ here, the above only implies that every $m_i$ is congruent to zero modulo $2$, i.e. every $m_i$ is even. Thus $m_1+\ldots+m_r<n$ because $n$ is odd, meaning that $C$ has a zero eigenvalue.

  • $\begingroup$ It's better to put $X_0=B$. You wrote a pretty proof; in fact the great idea (of biblical simplicity) is in lines $3,4$. That $A^2$ commute with $AB-BA$ is clear; yet, I never thought $A,AB-BA$ could commute! How did you think about it ? On the other hand, you showed the following -valid for every $n$-: if $A^2$ is a polynomial in $B$, then the characteristic polynomial of $AB-BA$ is in the form $x^{n-2k}P^2(x)$, where $P\in K[x]$ is a polynomial of degree $k\leq n/2$ s.t. $P(0)\not= 0$. $\endgroup$ – loup blanc Dec 30 '17 at 18:20
  • $\begingroup$ @loupblanc It's just accidental. I had no idea how to solve the problem. Although what we want to show is that $AB-BA$ is singular (rather than nilpotent), and although $AB-BA$ isn't nilpotent in general even if it commutes with $A$, I just wanted to know why the usual proof failed in characteristic 2. But it turned out that most of the proof can be salvaged. $\endgroup$ – user1551 Dec 31 '17 at 5:41

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