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let $A_{2\times 2}$ matrix, and The matrix $B$ is order square,such $$AB-BA=A$$ show that $$A^2=0$$

My idea: since $$Tr(AB)=Tr(BA)$$ so $$Tr(A)=Tr(AB-BA)=Tr(AB)-Tr(BA)=0$$

Question:2

if $A_{n\times n}$ matrix,and the matrix $B$ is order square,such $$AB-BA=A$$ then we also have $$A^2=0?$$ and then I can't Continue .Thank you

math110
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4 Answers4

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You say that $tr(A) = tr(AB)-tr(BA)=0$. Therefore Cayley Hamilton equation tells us that $A^2 = \det(A) I_2$.

On the other hand we have $A^2= A^2B-ABA=ABA-BA^2$. Therefore $2A^2 = A^2B-BA^2=0$ since $A^2$ is a multiple of the identity.

Beni Bogosel
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  • It's very very Nice!Thank you – math110 Jul 17 '14 at 09:37
  • so if $A_{n\times n},B_{n\times n}$ this reslut is also true?and you methods is also works? – math110 Jul 17 '14 at 09:43
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    @chinamath : We use the fact that we are in the $2 \times 2$ case because $A$ satisfies its characteristic polynomial which has the form $X^2 - \mathrm{tr}(A) X + \det(A)$. In the $n \times n$ case you would have to deal with the other exterior powers (which are the other coefficients of the characteristic polynomial), and this is not obvious that it would work. – Patrick Da Silva Jul 17 '14 at 09:57
  • Is there a counter example to the question in characteristic two? – Patrick Da Silva Jul 17 '14 at 10:02
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    @PatrickDaSilva In characteristic two $A = \binom{0,1}{1,0}$, $B = \binom{0,0}{0,1}$ form a counterexample. – martini Jul 17 '14 at 10:09
  • @martini,you example is not counterexample – math110 Jul 17 '14 at 10:15
  • @math110 Why? $AB = \binom{0,1}{0,0}$, $BA = \binom{0,0}{1,0}$, so $AB - BA = A$ but $A^2 = I_2 \ne 0$. – martini Jul 17 '14 at 10:19
  • Great proof. But I would say that $A^2 = -\det(A) I_2$ – the_candyman Jul 17 '14 at 10:22
  • @martini,your example $AB-BA=\begin{bmatrix}0&1\ -1&0\end{bmatrix}\neq A ?$ – math110 Jul 17 '14 at 10:29
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    @math110 In characteristic two, $-1 = 1$, so: yes. I do not aim to give a counterexample for other characteristic, as there the above proof thows no such examples exist. – martini Jul 17 '14 at 10:29
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    A remark: I think that the general result (in characteristic $0$) in dimension $n$ is that $A$ is nilpotent. To see this, one prove by induction that $kA^k=A^k B-B A^k$, and let $M(x)=x^s+\cdots+a_k x^k+\cdots+a_0$ be the minimal polynomial of $A$. Then $\sum_{k=0}^s ka_k A^k=0$, and $\sum ka_k x^k=sM(x)$, hence $a_k=0$ for $k<s$, $M(x)=x^s$, and as $s\leq n$, $A^n=0$. – Kelenner Jul 17 '14 at 10:45
  • @Kelenner: I think it is $2^kA^k=A^kB-BA^k$. To prove that $A$ is nilpotent it is enough to prove that $tr(A^k)=0$ for each $k$, which you did. After that you can deduce that all eigenvalues are zero using a Vandermonde system. – Beni Bogosel Jul 17 '14 at 15:06
  • @Beni Bogosel: In fact, you have $2A^2=A^2B-BA^2$, so you get multipling by $A$ $2A^3=A^3B-ABA^2$ and $2A^3=A^2BA-BA^3$, summing you get $$4A^3=A^3B-BA^3+A(AB-BA)A=A^3B-BA^3+A^3$$ so $3A^3=A^3B-BA^3$. – Kelenner Jul 17 '14 at 15:58
  • @Kelenner: I'm sorry. I didn't make all the computations. – Beni Bogosel Jul 17 '14 at 17:20
2

An alternative geometric approach:

  • We have $A\in \mathfrak{sl}_2$ and it can be assumed that $B\in \mathfrak{sl}_2$ as well. Hence $A$, $B$ can be seen as vectors $\vec{a},\vec{b}\in \mathbb{C}^3$. In this picture, $[A,B]\sim \vec{a}\wedge \vec{b}$ and $\operatorname{Tr}AB\sim \vec{a}\cdot\vec{b}.$

  • Now since $\vec{a}\wedge \vec{b}\sim \vec{a}$, taking the scalar product with $\vec{a}$, we get $\operatorname{Tr}A^2\sim\vec{a}\cdot \vec{a}=0 $. As for $A\in\mathfrak{sl}_2$ one has $A^2=\frac{\operatorname{Tr}A^2}{2} I$, the result follows.

Start wearing purple
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Hint:

  1. $\operatorname{tr}{A}=0$

  2. If $\det{A}=0$, then $A^2=0$ (By Cayley-Hamilton theorem)

Suppose $\det{A}\neq 0$ then $A$ is invertible. $$\begin{align}I&=A^{-1}A\\&=A^{-1}(AB-BA)\\&=B-A^{-1}BA\end{align}$$ Implies $\operatorname{tr}{I}=0$ (!)

Sourav Ghosh
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$AB - BA = A \Rightarrow A^2B - ABA = A^2 \Rightarrow$ $$-ABA = A^2(I-B)$$ $AB - BA = A \Rightarrow ABA - BA^2 = A^2 \Rightarrow$ $$ABA = (I+B)A^2$$ Adding this identities: $0 = A^2(I-B) + (I+B)A^2$

etothepitimesi
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Darth Geek
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