Proving AB-BA is singular from given conditions

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Proving AB-BA is singular from given conditions

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Source: Romania NMO 2025 11.4

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#1 h Wednesday at 2:04 PM • 1 Y

Y by RobertRogo

Let $A,B \in \mathcal{M}_n(\mathbb{C})$ be two matrices such that $A+B=AB+BA$ . Prove that:
a) if $n$ is odd, then $\det(AB-BA)=0$ ;
b) if $\text{tr}(A)\neq \text{tr}(B)$ , then $\det(AB-BA)=0$ .

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#2 h Yesterday at 9:19 AM • 2 Y

Y by MS_asdfgzxcvb, RobertRogo

I hope the contest is over.
for a) Putting $A=X+1/2I_n,B=Y+1/2I_n$ we obtain $XY+YX=1/2I$ .
Since $XY,YX$ commute, they are simultaneously triangularizable and we may assume that they are triangular.
Since $XY,YX$ have same spectrum, $diag(XY)=[a_1,\cdots,a_n],diag(YX)=[b_1,\cdots,b_n])$ , where $(b_i)$ is a permutation of $(a_i)$ .
Moreover, for every $i$ , $b_i=1/2-a_i$ ; if $a_i\not= 1/4$ , then there is $j\not= i$ s.t. $b_i=a_j$ and therefore $b_j=a_i$ .
Finally, the eigenvalues are associated $2$ by $2$ via the indices $i,j$ .
If $n$ is odd, then there is at least one index $k$ s.t. $a_k$ is an isolated eigenvalue and $a_k=b_k=1/4$ .
Thus $\det(XY-YX)=\det(AB-BA)=0$ . $\square$

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#3 h Yesterday at 9:03 PM

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a) We have $(2A-I)(2B-I)-I=2(AB-BA)$ and $(2B-I)(2A-I)-I=2(BA-AB).$

The matrices $(2A-I)(2B-I)$ and $(2B-I)(2A-I)$ have the same characteristic polynomial, so $\det((2A-I)(2B-I)-I)=\det((2B-I)(2A-I)-I),$ yielding $2^n \det(AB-BA)=(-2)^n \det(AB-BA).$

If $n$ is odd, the last relation implies $\det(AB-BA)=-\det(AB-BA),$ so $\det(AB-BA)=0.$

b) Let $X=2A-I$ and $Y=2B-I.$ We have $XY+YX = \frac{1}{2} I.$ This implies $XYX+YXX= \frac{1}{2} X$ and $XXY+XYX = \frac{1}{2} X,$ so $XXY=YXX.$

Therefore $X(XY-YX)+(XY-YX)X=O.$ If $XY-YX$ is invertible, then $(XY-YX)^{-1}X(XY-YX)+X=O,$ and using the trace we find $\mathrm{tr}(X)=0.$ We similarly get $\mathrm{tr}(Y)=0,$ contradiction.

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