Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 6 down vote accepted

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 $$ B \; = \; \left( \begin{array}{rr} \frac{1}{\sqrt 2} & \frac{1}{2} \\ 0 & \frac{1}{\sqrt 2} \end{array} \right) $$

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.