Suppose that $A, B$ are idempotent matrices ($A^2=A$), such that $A + B$ is idempotent, prove that $AB = BA = 0$
Asked
Active
Viewed 6,848 times
10
-
1Are you sure you don't want to prove $AB+BA=0$ ? – Belgi Jan 26 '13 at 23:49
-
1No it is definitely AB=BA=0 – Chance Jan 26 '13 at 23:51
-
2Probably there is a typo on your book/list/etc... – Sigur Jan 26 '13 at 23:57
-
@Belgi Thanks.. – Git Gud Jan 26 '13 at 23:58
-
@GitGud - No problem, I was doing the same thing and I thought it would work – Belgi Jan 27 '13 at 00:00
-
It must be a typo in the question. – Chance Jan 27 '13 at 00:06
-
With square matrices over a characteristic $0$ field K, the following generalization ($n\geq 2$) is actually true. Assume $p_1,\ldots,p_n$ are idempotents. If $p_1+\ldots+p_n$ is idempotent, then $p_ip_j=0$ for all $i\neq j$. Note that the converse is trivial, so this is actually an equivalence. – Julien Jan 27 '13 at 00:41
2 Answers
16
This is true in every ring where you can divide by $2$. So it is true in particular for matrices.
We have $(A+B)^2=A^2+AB+BA+B^2=A+AB+BA+B$. Since $A+B$ is idempotent, it follows that $$AB+BA=0.$$ We call this equation (E).
Right-multiply (E) by $B$. We find $AB+BAB=0$, hence $AB=-BAB$.
Left-multiply (E) by $B$. This yields $BAB+BA=0$, hence $BA=-BAB$.
Equating the last two equations, we find $AB=BA$.
Now $AB+BA=0$ clearly yields $2AB=2BA=0$. Dividing by $2$, we get $$ AB=BA=0. $$
Julien
- 44,791
-
Very nice! I thought in this direction when I saw I coldn't get the result, but I figured it was a typo and I tried to find a trivial example so I tooked $B=A$ and then I saw that in $F_2$ I get such an example. Did you know this before, or just proved it ? (+1) – Belgi Jan 27 '13 at 00:17
-
9
This is incorrect.
For example take $F=\mathbb{F}_{2}$ the field with two elements and $A=B=I$ over $F$.
$A,B$ are clearly idempotent and $$A+B=I+I=2I=0$$ hence $$(A+B)^{2}=0^{2}=0=A+B$$ is also idempotent.
But $$AB=BA=I^{2}=I\neq0$$
From your assumption you can only get $AB+BA=0$.
Belgi
- 23,150
-
Yes I can only seem to get $AB+BA=0$ from this assumption. Could possibly be a typo. – Chance Jan 27 '13 at 00:02
-
1@Chance: It is true for square matrices over a field in characteristic distinct from $2$. See my answer. – Julien Jan 27 '13 at 00:13
-
-
-