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Proof of exchangeability in matrix multiplications with Identity matrix I

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Let A be a square matrix such that I-A is nonsingular.

    Prove that A * (I-a)^-1 = (I-A)^-1 * A

    2. Relevant equations

    Now I think that

    A^-1 * A = A*A^-1 = I
    and
    I*A = A*I
    are relevant for this.

    3. The attempt at a solution

    I tried to express (I-a)^-1 in respect to I without having an inversion in it. but somehow I can't get it to work.

    Any ideas?
     
  2. jcsd
  3. Mar 15, 2012 #2

    lanedance

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    this may be a place start... you'll need to assume A is non-singular as well...
    [tex] (A (I-A)^{-1})^{-1} = .. [/tex]
     
  4. Mar 15, 2012 #3
    thank you it looks like that might be the way to go.
    however, i dont really know where to go from there:

    1. Is taking the inverse an equivalent tranformation?
    in other words after using

    (A*B)^-1 = B^-1 * A^-1

    is this correct: (I−A) * A^-1 = A^-1 * (I−A)

    2. Where should I go from here?
    I'm guessing that I have to transform this further so I can show they are equal by using this formula: A^-1 * A = A * A^-1

    But how do I get there?
     
  5. Mar 15, 2012 #4

    lanedance

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    try multplying and see if it behaves an inverse should

    [tex] (B^{-1} A^{-1})(AB) [/tex]

    Ok so you get
    [tex] (A (I-A)^{-1})^{-1} = (I−A)A^-1 [/tex]

    continuing to multiply through and as you suggest using A^-1 * A = A * A^-1 = I
    [tex] (A (I-A)^{-1})^{-1} = (I−A)A^{-1} = A^{-1}−I = A^{-1}I −A^{-1}A = A^{-1}(I-A)[/tex]

    so you have
    [tex] (A (I-A)^{-1})^{-1} = A^{-1}(I-A)[/tex]

    Any ideas where to go from here, think about gettingt back to the LHS in the original question?
     
  6. Mar 15, 2012 #5

    Dick

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    You don't need to assume A is nonsingular. I would start from A=IA=(I-A)^(-1)(I-A)A. Now show (I-A)A=A(I-A). Continue from there.
     
  7. Mar 16, 2012 #6

    lanedance

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    starting from
    [tex](A (I-A)^{-1})^{-1} = A^{-1}(I-A) [/tex]

    then I was thinking take the inverse of both sides
    [tex]((A (I-A)^{-1})^{-1})^{-1} = (A^{-1}(I-A))^{-1} [/tex]
    [tex]A (I-A)^{-1} = (I-A)^{-1}A [/tex]

    Though as mentioned this assumes A is non-singular, which may be a leap we don't require. With this in mind I would try Dick's method.. also because he's usually always right about these things.
     
  8. Mar 19, 2012 #7
    Thank you both for your help!

    So if I get this right then Dick suggested to do this:

    Since: A = I*A = (I-A)^-1*(I-A)*A

    I can rewrite:


    A*(I-A)^-1 = (I-A)^-1*A

    (I-A)^-1*(I-A)*A*(I-A)^-1 = (I-A)^-1*(I-A)^-1*(I-A)*A

    question is, am i allowed to cancel in the following way:

    (I-A)^-1*______A_______ = (I-A)^-1*______________A

    leaving me with

    (I-A)^-1*A = (I-A)^-1*A

    That obviously would be great!
     
  9. Mar 19, 2012 #8

    lanedance

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    I can't quite follow your working, isn't your second expression what you are trying to prove?
     
  10. Mar 19, 2012 #9

    lanedance

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    As a start though, why not expand the expression below to show its true
    (I-A)A=A(I-A)
     
  11. Mar 19, 2012 #10

    lanedance

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    this is similar to noting that the identity matrix always commutes...
     
  12. Mar 19, 2012 #11
    thats what i did:

    (I-A)*A=A*(I-A)

    (I-A)*(I-A)^-1*(I-A)*A = (I-A)^-1*(I-A)*A*(I-A)

    What I was trying to ask is: can i do this

    (I-A)*(I-A)^-1*(I-A)*A*I = (I-A)^-1*(I-A)*A*(I-A)*I
     
  13. Mar 19, 2012 #12
    because that would mean I could do this:



    A*(I-A)^-1 = (I-A)^-1*A

    (I-A)^-1*(I-A)*A*(I-A)^-1 = (I-A)^-1*(I-A)^-1*(I-A)*A

    (I-A)^-1*(I-A)*A*(I-A)^-1*I = (I-A)^-1*(I-A)^-1*(I-A)*A*I

    leaving

    (I-A)^-1*A = (I-A)^-1*A

    0 = 0
     
  14. Mar 19, 2012 #13

    lanedance

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    still not sure I follow, are you starting by assuming the answer and working back from it? and i can't see how you get from the red and green to the last line, or follow the whole idea...

    how about this, you know
    [tex] (I-A)A = A(I-A)[/tex]

    now substitute it on the RHS for (I-A)A in the below expression
    [tex] A = (I-A)^{-1}(I-A)A[/tex]

    can you take it from here?
     
    Last edited: Mar 19, 2012
  15. Mar 19, 2012 #14
    oooh ok, yeah nevermind. thanks for tppinting that out ^^
    i just didnt get what you where going for
     
  16. Mar 19, 2012 #15
    man, now i thought i finally had it, but as i'm trying do write it down I cant seem to figure out how to proof (I-A)A=A(I-A) anymore.

    I expanded it to (I-A)*(I-A)^-1*(I-A)*A = (I-A)^-1*(I-A)*A*(I-A). But now what?

    Ok so overall I will now do this:


    A=I*A=(I-A)^-1*(I-A)*A

    subsitute in

    A*(I-A)^-1

    = (I-A)^-1*(I-A)*A*(I-A)^-1

    because: (I-A)*A=A*(I-A) i can rewrite as

    (I-A)^-1*A*(I-A)*(I-A)^-1

    which is again: (I-A)^-1*A*I

    = (I-A)^-1*A
     
  17. Mar 19, 2012 #16

    lanedance

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    yeah ok, so that look good

    for (I-A)A=A(I-A) just try multying the expressions out
     
  18. Mar 19, 2012 #17

    lanedance

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    (I-A)A = A.I-A.A

    then use the fact that the identity commutes
     
  19. Mar 19, 2012 #18
    thank you!
    wow I feel stupid.
    well

    so (I-A)A=A(I-A)

    I*A - A2 = A*I - A2
    A-A2 = A-A2

    alright
    thanks for all your help!
     
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