2
$\begingroup$

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.

  • 0
    Really, why is (1) "clearly" true? Be careful about commutativity.2012-12-13
  • 3
    Hint for (3): Look at $(AB)^{n+1}$.2012-12-13
  • 0
    can you explain 3 and 4 please. i could not follow you.2012-12-13
  • 0
    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 something2012-12-13
  • 1
    Your reasoning seems to be wrong, as $(A+B)^{-1}\neq A^{-1}+B^{-1}$ in general. Statement 1 is true nonetheless.2012-12-13

2 Answers 2

1

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 $\det(\lambda I - AB)$.

  • 0
    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.2012-12-13
  • 0
    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 $k$.2012-12-13
  • 0
    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.2012-12-13
  • 0
    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.2012-12-13
  • 0
    still i didn't understand it?please clarify it in more detail.thanks for ur time..2012-12-13
  • 0
    It's already clear to an extent that it almost becomes a spoiler. You could try to spend more time on it. Good luck! ;-D2012-12-13
0

$BA$ is nilpotent only when $AB$ and $BA$ are commutatative. Not in general.. therefore 3 is not true..

  • 1
    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.2012-12-16