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Let $A$ , $B$ be complex $n \times n$ matrices. Which of the followings are true?

  1. If $ A$,$B$ and $A+B$ are invertible, then $A^{-1}+B^{-1}$ is invertible.
  2. If $ A$,$B$ and $A+B$ are invertible, then $A^{-1}-B^{-1}$ is invertible.
  3. If $AB$ is nilpotent, then $BA$ is nilpotent.
  4. Characteristic polynomial of $AB$ and $BA$ are equal if $A$ is invertible.

Clearly 1 is true and I found a example in 2 that 2 is not correct, but I have no idea on 3 and 4. Kindly help me.

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  • $\begingroup$ Really, why is (1) "clearly" true? Be careful about commutativity. $\endgroup$ Dec 13, 2012 at 14:02
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    $\begingroup$ Hint for (3): Look at $(AB)^{n+1}$. $\endgroup$ Dec 13, 2012 at 14:04
  • $\begingroup$ can you explain 3 and 4 please. i could not follow you. $\endgroup$
    – poton
    Dec 13, 2012 at 14:29
  • $\begingroup$ for 1 $($A+B$)($A+B$)^{-1}=I$ as $A+B$ is invertble. that is $($A+B$)$(A^{-1}+b^{-1})$ =I. am I right or missing something $\endgroup$
    – poton
    Dec 13, 2012 at 14:34
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    $\begingroup$ Your reasoning seems to be wrong, as $(A+B)^{-1}\neq A^{-1}+B^{-1}$ in general. Statement 1 is true nonetheless. $\endgroup$
    – user1551
    Dec 13, 2012 at 14:38

2 Answers 2

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As someone has given an excellent hint on 3, I will give you hints on 1 and 4:

1) What is $B^{-1}(A+B)A^{-1}$?

4) Note that $\det X\det Y = \det XY = \det Y\det X$. Now try to factor something out from $\det(\lambda I - AB)$.

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  • $\begingroup$ 3)since $AB$ is nilpotent so $(AB)^{n+1}=0$ and $(AB)^{n+1}=A(BA)^nB$.from which it follows that $BA$ is nilpotent.and then 3 is right am i right. $\endgroup$
    – poton
    Dec 13, 2012 at 14:58
  • $\begingroup$ No. That hint contains a typo. You should look at $(BA)^{n+1}$ instead of $(AB)^{n+1}$. That is, show that $(BA)^{n+1}=0$ using the fact that $(AB)^k=0$ for some positive integer $k$. $\endgroup$
    – user1551
    Dec 13, 2012 at 15:01
  • $\begingroup$ 1)by calculation $B^{-1}(A+B)A^{-1}$=$A^{-1}+B^{-1}$. since left hand side is invertible so the result follows.am i right now. 4) but i still could not follow the hints for 4.plesa elaborate it.thanks. $\endgroup$
    – poton
    Dec 13, 2012 at 15:02
  • $\begingroup$ Basic idea: $\det(\lambda I - AB)=\det XY = \det YX=\det(\lambda I - BA)$. What $X$ and $Y$ can be used? Don't forget the condition given in the question, i.e. $A$ is invertible. $\endgroup$
    – user1551
    Dec 13, 2012 at 15:06
  • $\begingroup$ still i didn't understand it?please clarify it in more detail.thanks for ur time.. $\endgroup$
    – poton
    Dec 13, 2012 at 16:40
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$BA$ is nilpotent only when $AB$ and $BA$ are commutatative. Not in general.. therefore 3 is not true..

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    $\begingroup$ You mean when $A$ and $B$ commute with each other, right? Also, you can consider adding an explicit example or a hint for such an example. $\endgroup$
    – levap
    Dec 16, 2012 at 16:07

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