$A,B \in M_2(\mathbb C)$ be such that $AB-BA=B^2$ ; then is it true that $AB=BA$?

Question

Let $A,B \in M_2(\mathbb C)$ be such that $AB-BA=B^2$. Then is it true that $AB=BA$ ?

If we can show $\mathrm{tr}(B)=\det (B)=0$, then we are done by using $B^2-(\mathrm{tr}(B)) B+\det (B)=0$; but I don't know how to show even that. Please help. Thanks in advance.

I don't know if it helps, but, $(A-B)(A+B)=A^2-BA+AB-B^2=A^2+B^2-B^2=A^2$. — Michael Burr, Dec 23 '15 at 13:27
@Thomas So I guess you could say the problem is to show whether $B^2 = 0$. It doesn't really matter. — Arthur, Dec 23 '15 at 13:31
@Arthur well, yes. But that's a rather complicated way of stating this. — Thomas, Dec 23 '15 at 13:33
@Thomas It's rather common problem type to give some condition on two matrices and then ask whether they commute. Asking whether $B^2 = 0$ would seem off in that context. — Arthur, Dec 23 '15 at 13:34

Sangchul Lee · Accepted Answer · 2015-12-23T15:47:29.263

We prove the following generalized claim:

Claim. Assume that $AB - BA = B^p$ for some $p \geq 2$. Then $AB = BA$.

This is equivalent to proving that $B^p = 0$. To this end, we notice that

$\operatorname{tr}(B^p) = \operatorname{tr}(AB - BA) = 0$ since $\operatorname{tr}(AB) = \operatorname{tr}(BA)$.
We have $\det(B^p) = 0$. Indeed, if $\det(B^p) \neq 0$, then $B$ is invertible. Multiplying $B^{-p}$ to the left, we have $$B^{-p}AB - B^{1-p}A = I_2.$$ But this is impossible since $$2 = \operatorname{tr}(I_2) \neq \operatorname{tr}((B^{-p}A)B - B(B^{-p}A)) = 0.$$

Applying Cayley-Hamilton theorem, we find that $B$ is nilpotent:

$$B^{2p} = \operatorname{tr}(B^p)I - \det(B^p)I_2 = 0.$$

Therefore we must have $B^2 = 0$ and hence $B^p = 0$. ////

Addendum: Counter-example for $p = 1$. For any $a_1, a_2, b \in \Bbb{C}$ with $b \neq 0$, put

$$ A = \begin{pmatrix} a_1 & a_2 \\ 0 & a_1 - 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & b \\ 0 & 0 \end{pmatrix}. \tag{*} $$

Then $AB - BA = B$.

Conversely, if $A, B \in M_2(\Bbb{C})$ is such that $AB - BA = B$ and $B \neq 0$, then $A$ and $B$ can be written as $\text{(*)}$ up to unitary change of basis.

(Remark. Using the Jordan normal form, it is straightforward that this is true up to change of basis.)

Indeed, since $\det(B) = 0$, $B$ has rank 1 and we may write

$$ B = b \begin{pmatrix} u_1 \bar{v}_1 & u_1 \bar{v}_2 \\ u_2 \bar{v}_1 & u_2 \bar{v}_2 \end{pmatrix} $$

for some $b \neq 0$ and for some unit vectors $u = (u_1, u_2), v = (v_1, v_2)$. Then $\operatorname{tr}(B) = 0$ is equivalent to $\langle u, v \rangle = 0$ and hence $u, v$ are orthogonal. In particular, $B$ corresponds to the linear map

$$ Bx = b\langle v, x \rangle u. $$

(Here, the 2nd argument of the Hermitian inner product $\langle \cdot, \cdot \rangle$ is chosen to be linear.) Then the claim follows by rewriting everything w.r.t. the orthonormal basis $\{u, v\}$ and solving the equation $AB - BA = B$.

@SangchulLee : From $B^4=O$ , we don't actually even need any minimal polynomial or so ; just that then $B$ is nilpotent and $B$ is $2 \times 2$ matrix ! so $B^2=O$ — , Dec 23 '15 at 15:03
@user228169, But isn't it essential to consider minimal polynomial when proving that fact? I may be wrong, but it seems to me that some non-trivial observation is required to push the exponent $m$ of $A^m = 0$ all the way down to the ambient dimension. — Sangchul Lee, Dec 23 '15 at 15:14
@SangchulLee : No no , not necessarily essential ; if say $k$ be the degree of nilpotency , then there is a non-zero vector $v$ such that $A^{k-1}v \ne 0$ and then you can show that ${v,Av,...,A^{k-1}v }$ is linearly independent ; hence $ k \le n$ — , Dec 23 '15 at 15:17

score 6 · Answer 2 · edited Apr 13 '17 at 12:21

6

It is easy to show that a matrix $B\in M_n(K)$ satisfying, for some $p\ge 1$ $$ AB-BA=B^p $$ is nilpotent and hence satisfies $B^n=0$, see Proposition $2.2$ here. The proof uses the Vandermonde determinant. In your case we obtain that $B$ is a nilpotent matrix of size $2$, so that $B^2=0$, and hence $AB=BA$.

More generally, $AB-BA$ is nilpotent, if it commutes with $A$ (or $B$), see here.

edited Apr 13 '17 at 12:21

Community

1

answered Dec 23 '15 at 13:45

Dietrich Burde

130,978

Dear @Dietrich,Don't you need to assume $p >1$ ? – Arpit Kansal Dec 23 '15 at 13:52
Oh, in your proof of Prop 2.2 you jumped twice to get that trace $0$,, with no warning :-) – orangeskid Dec 23 '15 at 13:55
Dear @ArpitKansal: No, it is also true for $p=1$. This has been discussed on MSE very often i see, e.g., here, or here. Also here., or here. – Dietrich Burde Dec 23 '15 at 14:00
Dear @DietrichBurde I see.Thank you.Regards, – Arpit Kansal Dec 23 '15 at 14:10
Dear @ArpitKansal I should be more precise. For $p=1$ it is still true that $B$ is nilpotent. The counterexample above states that $AB=BA$ need not be true then. – Dietrich Burde Dec 26 '15 at 20:16
Dear @DietrichBurde I see your point.Thank you,again! – Arpit Kansal Dec 27 '15 at 04:05

score 0 · Answer 3 · answered Dec 23 '15 at 13:47

0

We have $B^3= BAB-AB^2=AB^2-BAB$, hence $AB^2=B^2A$.

Hence $A$ commute with $B^2$, but we know that $B^2= tr(B)B-det(B)Id$ Therefore, if $tr(B)\not =0$, $A$ commmute with $B$.

If $tr (B)=0$, $B^2=det(B) Id$. If $det(B) =0, B^2=0$ and we are done.

If not the two eigenvalues of $B$ are opposite (tr=0) and non zero, then $tr(B^2)=\lambda^2\not =0$, contradiction as $tr B^2=tr(AB-BA)=0$

answered Dec 23 '15 at 13:47

Thomas

7,470

Actually $B^3=BAB-B^2A=AB^2-BAB$, and it hard to see how this leads to $AB^2=B^2A$. A wrong answer in my opinion. (-1) – user26857 Sep 02 '16 at 14:44

$A,B \in M_2(\mathbb C)$ be such that $AB-BA=B^2$ ; then is it true that $AB=BA$?

3 Answers3