1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Matrix proofs

  1. Apr 20, 2013 #1
    1. The problem statement, all variables and given/known data

    First Question: Be A and B square matrix. Show that if A and B are invertible matrix, then: (A + B)^(-1) = A^(-1) * [I + B*A^(-1)]^(-1)

    3. The attempt at a solution

    First Queston: (A+B)^(-1) = [A^(-1) + A^(-1)*B*A^(-1)]^(-1) (I distributed the a^(-1) outside the []). Then I don't know what else I can do


    1. The problem statement, all variables and given/known data

    Second Question: Be A and P matrix of nxn dimension with P being invertible. So det (P^(-1)*A*P) = det(A)

    3. The attempt at a solution

    I would like to transform the (P^(-1)*A*P) into (I*A), but since matrix multiplication ain't commutative, I don't know how to get there
     
  2. jcsd
  3. Apr 21, 2013 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can verify this by showing that
    $$(A+B)(A^{-1}[I + BA^{-1}]^{-1}) = I$$
    What properties of determinants do you know about?
     
  4. Apr 21, 2013 #3
    I was only able to develop it until here:

    [itex]A(A^{-1}[I + BA^{-1}]^{-1}) + B(A^{-1}[I + BA^{-1}]^{-1}) =[/itex]

    [itex]AA^{-1}[I + BA^{-1}]^{-}) + BA^{-1}[I + BA^{-1}]^{-1} = [/itex]

    [itex]I[I + BA^{-1}]^{-1} + BA^{-1}[I + BA^{-1}]^{-1} = [/itex]

    [itex][I + BA^{-1}]^{-1} + B[A^{-1} + A^{-1}BA^{-1}]^{-1} = [/itex]

    I know that de determinant of A^(-1) = 1/ det A, that det A = det A transpose, that the det A with a constant K that multiplies one of its lines is equal to K that multiplies det A and so that if this constant multiplies all lines. then det A = K^n * det A
     
  5. Apr 21, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think the problem is asking you to assume that the wrong things are invertible. A and B invertible doesn't show A+B is invertible. You need to assume that everything is invertible. Then for the first use that C^(-1)D^(-1)=(DC)^(-1). Figure out what to put D and C equal to.
     
  6. Apr 21, 2013 #5

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Now try using the distributive property.
     
  7. Apr 21, 2013 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Another important property: ##\det(AB) = \det(A)\det(B)##. Do you know this one? If so, the second problem should be easy.
     
  8. Apr 22, 2013 #7
    How? The identity is irrelevant, so only the [itex]a^{-1}[/itex] is left, and I already distributed it without seeing any meaningful result


    Ah yes, had forgotten this one.

    Didn't really understand. Can you try to explain it once more?
     
  9. Apr 22, 2013 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What I mean is take the inverse of both sides of (A + B)^(-1) = A^(-1) * [I + B*A^(-1)]^(-1). What you first did, distributing the A^(-1) inside of the [] is quite wrong. There's a ^(-1) on the bracket.
     
  10. Apr 22, 2013 #9

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Both terms have a common factor on the right, namely ##(I + BA^{-1})^{-1}##:

    $$I(I+BA^{-1})^{-1} + BA^{-1}(I+BA^{-1})^{-1}$$

    So factor out that common term and see what you get.
     
  11. Apr 24, 2013 #10
    Yes, I have to admit that I was suspicious about this distributive. Thanks for pointing it out.

    I see it now. Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Matrix proofs
  1. Matrix Proof (Replies: 9)

  2. Matrix Proof (Replies: 1)

  3. Matrix Proof (Replies: 6)

  4. Matrix proof (Replies: 3)

  5. Matrix proof (Replies: 8)

Loading...