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Math Help - Invertible Matrices

  1. #1
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    Invertible Matrices

    So there is a preposition that says;
    If AB is invertible, then A is invertible and B is invertible too.

    I uploaded the proof that I made, however I think it's not enough.

    The problem with my proof, in my opinion, is that I showed that B and A is invertible from one side (either from left or right), so I think that in order to show that they are invertible I have to show that they have inverses from both side and they are the same, am I mistaken?

    I would appreciate if you can verify, and tell me if my proof is correct, however if it's not don't tell me what's the right proof, I want to work on it. Thank you!
    Attached Thumbnails Attached Thumbnails Invertible Matrices-invertible-matrix.png  
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  2. #2
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    Re: Invertible Matrices

    seems to me you can just say

    Because AB is given to be invertible

    $$(AB)^{-1}=B^{-1}A^{-1}$$

    and because A is given to be invertible

    $$(AB)^{-1}A=B^{-1}A^{-1}A$$

    $$B^{-1}=(AB)^{-1}A$$
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  3. #3
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    Re: Invertible Matrices

    You can simplify it a great deal:

    If AB is invertible, there is a matrix C such that C(AB) = (AB)C = I.

    By associativity, we have: (CA)B = I, and A(BC) = I.

    You are correct in that you have only shown B is left-invertible, and A is right-invertible.

    We have 4 cases to consider:

    1) A is invertible, B is not,

    2) B is invertible, A is not,

    3) Neither are invertible,

    4) Both are invertible.

    Let's look at case (1):

    We know there is a matrix C such that (AB)C = I. Multiplying by A-1, we get:

    A-1[(AB)C] = A-1I = A-1.

    (A-1A)(BC) = A-1

    BC = A-1

    (BC)A = A-1A = I

    B(CA) = I, showing B is right-invertible, with the same right-inverse as left-inverse, that is: B is invertible, contradicting our assumption. So case 1 cannot occur.

    I think you can easily do case (2).

    For case (3) show that B singular implies AB singular, which is impossible. What's left?
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  4. #4
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    Re: Invertible Matrices

    This is my the new proof. The credit goes to Deveno.
    Though I am curious , is it possible to improve my first proof and get to the conclusion that A and B is invertible?
    Attached Thumbnails Attached Thumbnails Invertible Matrices-proof2.png  
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  5. #5
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    Re: Invertible Matrices

    There is a problem with your proof for case (3), can you see what it is? (Hint: suppose A was the 0-matrix...does it follow from D ≠ I that AD ≠ A?).

    You could modify your original proof by using column reduction instead of row reduction, this would give you the inverses on "the other side".
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  6. #6
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    Re: Invertible Matrices

    Quote Originally Posted by Deveno View Post
    There is a problem with your proof for case (3), can you see what it is? (Hint: suppose A was the 0-matrix...does it follow from D ≠ I that AD ≠ A?).
    But that is the case when A is the zero matrix, and we know for sure that A can't be the zero matrix, otherwise AB wouldn't be invertible, am I mistaken?
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