# Show trace is zero

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.

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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? –  Alex R. Jun 14 '12 at 11:01
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. –  Potato Jun 14 '12 at 11:02
Welp, you may have found the outright error Watkins wasn't able to find. –  Willie Wong Jun 14 '12 at 11:42
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). –  Willie Wong Jun 14 '12 at 12:00
@Jon: see the comments of Serkan and haohaolee. Invertible means we don't need $A^2 + B^2 = I$ at all! –  Willie Wong Jun 14 '12 at 13:55

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.

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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 $A^2+B^2=I$. –  Jyrki Lahtonen Jun 14 '12 at 13:49
@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. –  Jon Jun 14 '12 at 14:00
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. –  Jon Jun 14 '12 at 14:08
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 :-) –  Jyrki Lahtonen Jun 14 '12 at 18:55
@JyrkiLahtonen: My comment was not for you. Your comments were really helpful. Rather I thank you a lot. –  Jon Jun 14 '12 at 19:12