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For every square matrix A over $\mathbb{R}$, we'll assign: $t(A)=\sum_{i=1}^{n}a_{ii}$. And for every matrix $A,B \in M_n$ exists: $t(AB)=t(BA)$. Prove that if $AB-BA=A$ then $\det(A)=0$ and $t(A)=0$.

Since we know that $t(AB)=t(BA)$ its obvious that $t(AB)-t(BA)=0$. And from $AB-BA=A$ we also know that $t(AB-BA)=t(A)$.

I was thinking on continue it like so: $t(AB-BA)=A \Rightarrow t(AB)-t(BA)=t(A)$ and from $t(AB)-t(BA)=0$ so $t(A)=0$. (Same logic for determinants)

But I'm 100% sure that this is incorrect, can I get any direction?

Gary
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  • $A$ has to be nilpotent, so we have $t(A)=\det(A)=0$. – Dietrich Burde Apr 03 '23 at 13:04
  • @DietrichBurde: How does one see that $A$ has to be nilpotent? – Jason DeVito - on hiatus Apr 03 '23 at 14:07
  • @JasonDeVito From the duplicate: For any $k$, you have $A^{k+1}=AA^k=(AB-BA)A^k=ABA^k-BA^{k+1}$. So $$ \text{tr}(A^{k+1})=0,\ \ k=0,1,2,\ldots $$ We deduce that $\text{tr}(p(A))=0$ for every polynomial $p$ with $p(0)=0$. This implies that all eigenvalues of $A$ are zero (because otherwise we can get a polynomial $p$ that is 1 at all nonzero eigenvalues, and $p(A) $ would not have zero trace). So $A$ is nilpotent. – Dietrich Burde Apr 03 '23 at 15:04
  • @DietrichBurde: Oh, I had completely missed that it had been closed as a duplicate. Sorry for wasting your time - I could have checked out the duplicate myself. But your comment makes perfect sense, thanks! – Jason DeVito - on hiatus Apr 03 '23 at 15:19

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Your logic is correct for $t(A) = 0$. but determinant does not work that way. Here is a derivation : if $det(A) \neq 0$ then $A^{-1}$ exists. Hence,$$AB-BA = A \implies ABA^{-1} = I+B \implies det(ABA^{-1}) = det(I+B)$$ $$\implies det(A) \times det(B) \times det(A)^{-1} = det(I+B) \implies det(B) = det(I+B)$$.

This is not true for every $B$. Hence $A^{-1}$ does not exist. Hence $det(A) = 0$.

Note that i am assuming your expression $AB-BA=A$ is required to hold for every matrix $B$. Let me know if you have some specific matrix $B$ in mind.

Balaji sb
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  • If the eigenvalues of $B$ are in conjugate pairs and their real parts are all $-1/2$ then $\det B=\det(I+B)$ – Empy2 Apr 03 '23 at 13:47
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    Thats why i said i assumed $AB-BA = A$ holds for every matrix $B$. Hence this implies $det(B) = det(I+B)$ for every matrix $B$ which is clearly not true. – Balaji sb Apr 03 '23 at 13:49
  • Sorry, I missed that. On the other hand, if $B=0$ then $A=0$. – Empy2 Apr 03 '23 at 14:24