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Let $A$ and $B$ be complex $n×n$ matrices. Which of the following statements are true?

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

I am stuck on this problem and don't know where to begin. Can anyone help me please?

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Are you the same user as prakash? –  user1551 Dec 17 '12 at 12:48
    
we r friend....... –  pankaj Dec 17 '12 at 12:54
    
please guide us sir....... –  pankaj Dec 17 '12 at 12:55
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There was a good edit to this post that got rejected, I don't know why. –  Ben Millwood Dec 17 '12 at 12:55
    
I made the edit and I also don't know why it was rejected. –  Michael Albanese Dec 17 '12 at 12:57
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3 Answers

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For (1.) and (2.), consider how you would do it for real numbers (i.e. $1×1$ matrices). Then see if your argument continues to work for matrices (recall, or prove, that a product of invertible matrices is automatically invertible).

For (3.), just write out the definition of nilpotent, write out powers as multiplication, and recall that multiplication is associative.

For (4.), use the following facts (not in this order):

  • $\det(AB)=\det A \det B$
  • $AB + AC = A(B + C)$
  • $I = AA^{-1}$ for any invertible $A$.
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still not getting c and d –  pankaj Dec 17 '12 at 13:23
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Hints:
(a) If $A,B,A+B$ are invertible, then so is $A^{-1}(A+B)B^{-1}$.
(b) Take $A=B=I$ as @macydanim suggested.
(c) Use that $(BA)^k=B(AB)^{k-1}A$.
(d) $P_{AB}(x)=\det(xI-AB)=\det(A(xA^{-1}-B))=\det((xA^{-1}-B)A)$. Justify all the equalities.

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Your hints are kind of "here are all the hard and interesting parts of the answer, now do the tedious unenlightening busywork". I don't think that's the best approach. –  Ben Millwood Dec 17 '12 at 13:36
    
I do agree with you - from mathematical point of view. On the other hand, if you never worked with those 'tricks', it is hard to find them on your own. –  Dennis Gulko Dec 17 '12 at 13:39
    
Of course it is hard – if homework isn't hard, it often isn't useful! The key is that people discover the tricks by chance after spending a lot of time with the problem, thinking about it, considering it from different angles. It's really the latter that you want to happen, so you set them up with the tools they need and you let them go and sit with it for a bit. –  Ben Millwood Dec 17 '12 at 13:43
    
e.g. IMO the real purpose of (d) is not to prove the result but to make the student aware of the kind of manipulations one can apply with the determinant and multiplication, to make them "see through" the homomorphism. If you give them the manipulations, they don't have to gain that awareness. –  Ben Millwood Dec 17 '12 at 13:45
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