4

Let $A,B$ be real symmetric matrices, and for any $n\in \Bbb N^{+}$, and for all $x,y\in \Bbb R$, we have $$tr(xA+yB)^n=x^ntr(A^n)+y^ntr(B^n).$$

Show that $AB=0$.

My try:since $$(xA+yB)^n=x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n$$ so $$tr\left(x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n\right)=x^ntr(A^n)+y^ntr(B^n)$$

then $$tr\left(\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}\right)=0$$ so $$\binom{n}{1}x^{n-1}ytr(A^{n-1}B)+\cdots+\binom{n}{n-1}xy^{n-1}tr(AB^{n-1})=0$$ then I can't

I found this is a nice result, maybe this is an odd problem, and I can't solve it. Thank you.

math110
  • 93,304

3 Answers3

6

This answer is an amalgamation of ideas from two deleted posts by @julien and @user1551 together with some of my thoughts.

Following @julien's insight, let us look at the polynomial condition at $n = 4$. If one compare the coefficient of the $x^2 y^2$ terms on both sides, we get

$$2\text{tr}(A^2B^2) + \text{tr}( (AB)^2 ) = 0$$

Combine this with the fact

$$\text{tr}((AB+BA)^2) = \text{tr}((AB+BA)^T(AB+BA)) \ge 0 \\\implies \text{tr}(A^2B^2) + \text{tr}( (AB)^2 ) \ge 0$$

We get $\text{tr}(A^2B^2) \le 0$. However

$$\text{tr}(A^2B^2) = \text{tr}(ABBA) = \text{tr}((BA)^\top BA ) \ge 0$$

This implies $\text{tr}(A^2B^2) = 0$.

Following @user1551's idea. Since $A$ is real symmetric, we can choose a basis such that $A$ is a diagonal matrix and its first $k=\text{rank}(A)$ diagonal entries are the only non-zero entries. $\text{tr}(A^2B^2)$ then become a positive linear combination of the first $k$ diagonal entries of $B^2 = B^\top B$ which are non-negative themselves.

As a result, $\text{tr}(A^2 B^2) = 0$ implies the first k diagonal entries of $B^2$ vanish. This means the first $k$ rows/columns of $B$ are zero and hence $AB=0$.

Update

It turns out there is a much simpler argument. The expression $ \displaystyle \| C \|_F = \sqrt{ \text{tr}( C^\top C ) }$ is the famous Frobenius norm over the ring of real matrices! The vanishing of the coefficient of the $x^2 y^2$ terms $$\text{tr}((AB+BA)^2) + 2\text{tr}(BAAB) = 0$$ is equivalent to $$\quad \|AB+BA\|_F^2 + 2\|AB\|_F^2 = 0$$ This implies $\|AB\|_F = 0$ and hence $AB = 0$.

achille hui
  • 122,701
  • +1: so we could indeed conclude that $\mathrm{tr}(C^2)=0$ from $n=4$ in my deleted argument. But @user1551's idea reaches the conclusion a bit faster. Nice team work :-) – Julien Nov 27 '13 at 21:20
  • +1. The observation that the $x^2y^2$ term in $tr((xA+yB)^4)$ is equal to $tr((AB+BA)^2)+2tr(A^2B^2)$ is crucial. Thanks for the answer. – user1551 Nov 27 '13 at 21:27
-1

[Ignore this: Pick $x = 1, y = 0, n = 2$ to get $tr(A^2) = (tr A)^2$; you can do the same for $B$. {I've left it in only so that the comments still make sense.}]

Now pick $x = y = 1, n = 2$ to get $tr(A+B)^2 = tr(A^2) + tr(B^2)$.

But $tr( (A+B)^2 ) = tr(A^2) + tr(AB + BA) + tr(B^2)$.

So $tr(AB + BA) = 0$. So $2 tr(AB) = 0$.

That should get you well on your way.

John Hughes
  • 93,729
  • Then How go this way?Thank you – math110 Nov 27 '13 at 15:22
  • Are you sure of your first line? – Avitus Nov 27 '13 at 15:24
  • I confess, I'm not sure of the first line -- the equation given in the question lacks parentheses. It might be clearer if you swapped things so they read $(tr A)^2 = tr(A^2)$, which corresponds to the order in the given equation.

    But I don't actually use the first line elsewhere in the argument, so it's moot.

     – John Hughes Nov 27 '13 at 15:31
  • I think that the first line gives $tr(A^2)=tr(A^2)$; what follows is independent of it, though. In your answer you arrive at $2tr(AB)=0$. – Avitus Nov 27 '13 at 16:05
-2

$(xA+yB)^n=x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n$

I don't understand, this is not true in general, but only if $AB=BA$ ! But this is not supposed, isn't it ?...

I know a more powerful result : it is enough in fact that the equality holds for $x=1$ and $y=1$ to ensure that $AB=0$. Therafter we have immediately $BA=0$ exchanging the role of $A$ and $B$.

And the converse is true, if $AB=0$ then $BA$=0 because of the symmetry (transpose) hence we have the equality of traces ... for all $x$ and $y$.

sergbelh
  • 1