So say we have two ($n \times n$)-matrices $C$ and $D$ and also that $CD+C+D=0$
Prove that $CD=DC$.
I'm kind of curious about this one and don't know where to start.
Asked
Active
Viewed 107 times
1 Answers
8
Hint: Add $I$ to both sides of the original equation, and factor the left-hand side to get $$ (C+I)(D+I)=I $$ Now remember that inverses are always two-sided.
hmakholm left over Monica
- 286,031
-
Are you guaranteed that any matrix here but $I$ has an inverse? – yo' Mar 16 '15 at 17:51
-
$(C+I)(D+I)=CD+CI+ID+I=(CD+C+D)+I=0+I$. And as Henning says in real matrices having one sided inverse implies two sided inverse. So the point is $C+I$ and $D+I$ are inverses for themselves. – Asinomás Mar 16 '15 at 17:51
-
ah damn, stupid me. – yo' Mar 16 '15 at 17:54
-
Is it possible that then both C and D are zero-matrices? – Matrix1 Mar 16 '15 at 19:21
-
@Matrix1: Sure. $C=D=0$ satisfies both your assumption and the conclusion. – hmakholm left over Monica Mar 16 '15 at 19:22
-
Thanks a lot gentlemen for your guick feedback! I think I now get the idea behind all of it. – Matrix1 Mar 16 '15 at 19:30