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Problem: We are given $n\times n$ square matrices $A$ and $B$ with $AB+BA=0$ and $A^2+B^2=I$. Show $tr(A)=tr(B)=0$.

Thoughts: We have $tr(BA)=tr(AB)=-tr(BA)=0$. We also have the factorizations $(A+B)^2=I$ and $(A-B)^2=I$ by combining the two relations above.

Let $\alpha_i$ denote the eigenvalues of $A$, and $\beta_i$ the eigenvalues of $B$. We have, by basic properties of trace,

$\sum \alpha_i^2 +\sum \beta_i^2=n$

from $A^2+B^2=I$.

I'm not sure where to go from here.

I would prefer a small hint to a complete answer.

  • 12
    Something seems a bit off. Let $A=I$ and $B=0$, so that $AB+BA=0$ and $A^2+B^2=I$, giving $\mbox{tr}(A)=n$. Perhaps there are a few more conditions?2012-06-14
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    Well, that is certainly a counterexample. But the problem is stated exactly as it is in Golan. I'm not sure how to fix it, or if it can be fixed.2012-06-14
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    if we say that they are invertible?2012-06-14
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    Welp, you may have found the [outright error](http://www.ams.org/mathscinet-getitem?mr=2140296) Watkins wasn't able to find.2012-06-14
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    yes, if invertible, then $AB + BA = 0$ is enough2012-06-14
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    Just to note that a small modification of @Sam's example gives both traces non-vanishing: just chose $A$ and $B$ to be orthogonal projections to two subspaces which are orthogonal complements. Perhaps to fix the question requires a condition on $AB$ (something like $AB \neq 0$ maybe).2012-06-14
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    This is exactly a property of Pauli matrix http://en.wikipedia.org/wiki/Pauli_matrices that are also the generators of SU(2) group. The condition $A^2+B^2=cI$ being $c$ a constant identify an invariant of the group (a Casimir invariant). So, turning the attention to group theory maybe things can become clearer.2012-06-14
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    @WillieWong: I think the condition should be $det A,\ det B\ne 0$.2012-06-14
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    @Jon: see the comments of [Serkan](http://math.stackexchange.com/questions/158190/show-trace-is-zero#comment364883_158190) and [haohaolee](http://math.stackexchange.com/questions/158190/show-trace-is-zero#comment364906_158190). Invertible means we don't need $A^2 + B^2 = I$ at all!2012-06-14
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    @WillieWong Switching to $AB+BA = I$ seems promising.2012-06-14
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    @WillieWong or the correct problem could be $A^2 + B^2 = 0$ and $AB + BA = 0.$2012-06-14

2 Answers 2

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It appears from the context in the book that the correct problem is $$ A^2 + B^2 = A B + B A = 0. $$ The middle step is that $(B-A)^2 = 0,$ so we name the nilpotent matrix $N=B-A.$ Wait, I think that is enough. Because it is also true that $(A+B)^2 = 0.$ So $A+B$ and $B-A$ both have trace $0.$ So $tr \; \; 2B = 0.$ That finishes characteristic other than 2. We don't need full Jordan form for nilpotent matrices, just a quick proof that $N^2 = 0$ implies that the trace of $N$ is zero. Hmmm. This certainly does follow from the fact that a nilpotent matrix over any field has a Jordan form, but I cannot say that I have seen a proof of that.

Alright, in characteristic 2 this does not work, in any dimension take $$ A = B, $$ $$ A = B \; \; \; \mbox{then} \; \; A^2 + B^2 = 2 A^2 = 0, \; AB + BA = 2 A^2 = 0. $$

In comparison, the alternate problem $$ A^2 + B^2 = A B + B A = I $$ has the same thing about nilpotence, however in fields where $2 \neq 0$ and $2$ is a square we get a counterexample with $$ A \; = \; \left( \begin{array}{rr} \frac{1}{\sqrt 2} & \frac{-1}{2} \\ 0 & \frac{1}{\sqrt 2} \end{array} \right) $$ and $$ B \; = \; \left( \begin{array}{rr} \frac{1}{\sqrt 2} & \frac{1}{2} \\ 0 & \frac{1}{\sqrt 2} \end{array} \right) $$

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I will prove this just for the case $2\times 2$ being the approach similar for $n\times n$. Any matrices can be written through the generators of U(2) group as $$ A=a_0I+a_1\sigma_1+a_2\sigma_2+a_3\sigma_3=a_0I+{\bf a}\cdot{\bf \sigma} $$

$$ B=b_0I+b_1\sigma_1+b_2\sigma_2+b_3\sigma_3=b_0I+{\bf b}\cdot{\bf \sigma} $$ having $tr(\sigma_i)=0$, $\sigma_i\sigma_j+\sigma_j\sigma_i=2\delta_{ij}I$, $\sigma_i\sigma_j-\sigma_j\sigma_i=2i\epsilon_{ijk}\sigma_k$ and $\sigma_i^2=I$. Besides, we assume $det A, det B\ne 0$. Now we have $$ AB+BA=2a_0b_0+2(a_0{\bf b}\cdot{\bf \sigma}+b_0{\bf a}\cdot{\bf \sigma})+2{\bf a}\cdot{\bf b}=0. $$ and so, it must be $a_0=b_0=0$ and ${\bf a}\cdot{\bf b}=0$. Similarly, one has $$ A^2+B^2=(|{\bf a}|^2+|{\bf b}|^2)I $$ and then $|{\bf a}|^2+|{\bf b}|^2=1$. This implies $tr(A)=tr(B)=0$. These matrices are orthogonal in some sense.

  • 4
    Why does this not contradict Willie's example $$A=\pmatrix{1&0\cr0&0\cr},\qquad B=\pmatrix{0&0\cr0&1\cr}?$$ Here $AB=BA=0$ and $A^2+B^2=I$.2012-06-14
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    @JyrkiLahtonen: Nice argument. The reason is that there is another condition to get both equations satisfied. In your case is $a_0=a_3=\frac{1}{2}$ and $b_0=-b_3=\frac{1}{2}$. I will fix it. Thanks a lot.2012-06-14
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    Of course I have forgotten the condition I have pointed out in my comment above. You must have both determinants of $A$ and $B$ different from zero.2012-06-14
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    But I don't understand, how you can conclude $a_0=b_0=0$? Let's try: I get $a_0{\mathbf b}=-b_0{\mathbf a}$ and $a_0b_0=-{\mathbf a}\cdot {\mathbf b}$.2012-06-14
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    @JyrkiLahtonen: I solve the equation by setting $a_0=b_0=0$ and ${\bf a}\cdot {\bf b}=0$. As your example is there to show, other solutions are singular.2012-06-14
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    But how can we tell that your suggested $a_0=b_0=0$, ${\mathbf a}\cdot{\mathbf b}=0$ is the only solution leading to non-singular $A,B$? That's at the heart of my question. How did you use non-singularity to deduce that we must have $a_0=b_0=0$?2012-06-14
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    @JyrkiLahtonen: Your solution implies the the determinant is zero. Select the values I gave in my preceding answer to you. These correspond exactly to $A$ and $B$ you gave to me initially and to your alternative solution. Both matrices have a null determinant.2012-06-14
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    But that was just a single example. Referring to it doesn't explain why adding the assumption $\det A\neq0, \det B\neq0$ makes your step $$AB+BA=0\implies a_0=b_0=0$$ valid. That example only showed that the assumption is needed, but exactly how do you use it to arrive at your conclusion?2012-06-14
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    But all that is moot. Serkan and haohaolee already settled that in the case where both $A$ and $B$ are invertible the anticommutator relation implies that $tr(A)=tr(B)=0$ unless we are working over a field of characteristic two. All because $B=-A^{-1}BA$ implies $tr(B)=-tr(B)$ et cetera.2012-06-14
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    My Pauli matrices are rusty. Is there a formula for $\det A$ in terms of $a_i$:s? That may be the part I'm missing. It is not the quaternionic norm $a_0^2+a_1^2+a_2^2+a_3^2$ for sure, because physics uses different normalization.2012-06-14
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    @JyrkiLahtonen: You hit the point. The determinant for $A$ is $a_0^2-a_1^2-a_2^2-a_3^2$ and the same for $B$. So, your solution implies immediately that both determinants are zero, in agreement with your first example and as I pointed out from the start.2012-06-14
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    This downvote is nonsense and unfair. The argument is plainly correct. Please, remove it. Thanks.2012-06-14
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    I didn't downvote. I am finally making progress with your argument. The matrices $I$ and $\sigma_i, i=1,2,3$ are linearly independent. So your equation $AB+BA=0$ implies that $a_0b_0+a\cdot b=0$ (coefficient of $I$) and $b_0a-a_0b=0$ (coefficients of the $\sigma_j$:s). Therefore $$b_0\det A=b_0a_0^2-b_0a\cdot a=b_0a_0^2+a_0b\cdot a=a_0(a_0b_0+a\cdot b)=0,$$ implying that $b_0=0$ and a symmetric argument shows that $a_0=0$. The trace condition follows, because the $\sigma_j$ are traceless. +1 (at long last :-)2012-06-14
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    @JyrkiLahtonen: My comment was not for you. Your comments were really helpful. Rather I thank you a lot.2012-06-14
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    The other equation in my preceding comments should have been $b_0a0=-a_0b$, i.e. I committed a sign error. But it was used correctly in explaining Jon's argument.2012-06-15