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Math Help - Matrix little demonstration

  1. #1
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    Matrix little demonstration

    Hello,

    I'm a bit bloqued on a little demontration.

    A^4=0 , (I-A)^-1 = I+A+A+A

    Here is what i've done:

    I=(I-A).(I+AI+AI+AI)
    => I=I+AI+AI+AI-AI-AI-AI-(A^4)I
    => I=I+A+A-A-A-A^4
    => I=I-A^4
    => A^4=0

    But it is prove in reverse...

    The right way:
    If you know that matrix A^4=0, how to prove that (I-A)^-1 = I+A+A+A

    But in ths case (harder I supose), I don't know how to do.

    Can you help me?
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  2. #2
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    Re: Matrix little demonstration

    Consider (I-A)[I+A+A^2+A^3]

    (what are the assumptions that you have forgotten to mention? And for good measure also consider [I+A+A^2+A^3](I-A))

    .
    Last edited by zzephod; May 25th 2013 at 06:43 AM.
    Thanks from topsquark
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  3. #3
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    Re: Matrix little demonstration

    Sorry but I have no particular assumptions.
    The exercice is presented as such
    A^4=0 , (I-A)^-1 = I+A+A+A
    It's already a problem for me at the base.

    Or what do you mean?

    If y consider (I-A).(I+A+A+A), I start in the same oposite direction that i've done no?

    I still have a misanderstanding with this.
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  4. #4
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    Re: Matrix little demonstration

    Quote Originally Posted by Adi0101 View Post
    Sorry but I have no particular assumptions.
    The exercice is presented as such
    A^4=0 , (I-A)^-1 = I+A+A+A
    It's already a problem for me at the base.

    Or what do you mean?

    If y consider (I-A).(I+A+A+A), I start in the same oposite direction that i've done no?

    I still have a misanderstanding with this.
    (I-A)[I+A+A^2+A^3]=[I+A+A^2+A^3] - [A+A^2+A^3+A^4]=I-A^4=I

    Hence if (I-A) is invertible and A^4=0 then [I+A+A^2+A^3] is its' inverse ..
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  5. #5
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    Re: Matrix little demonstration

    Okay, I got t now!

    Thanks a lot!
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  6. #6
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    Re: Matrix little demonstration

    Hello, Adi0101!

    I'm a bit bloqued on a little demontration.

    A^4\,=\,0 \quad\Longleftrightarrow\quad (I-A)^{-1} \:=\: I+A+A^2+A^3


    Here is what I've done:

    We have: . (I-A)^{-1} \:=\:I+A+A^2+A^3


    Multiply by (I-A):

    . . (I-A)(I-A)^{-1} \:=\: (I-A)(I+A+A^2+A^3)

    . . . . . . . . . . . . . . I \;=\;I^2 + IA + IA^2 + IA^3 - AI - A^2 - A^3 - A^4

    . . . . . . . . . . . . . . I \;=\;I + A + A^2 + A^3 - A - A^2 - A^3 - A^4

    . . . . . . . . . . . . . . I \;=\;I - A^4

    n . . . . . . . . . . . A^4 \;=\;0


    But to prove it in reverse . . .

    Just run the step in reverse . . .

    We have: . A^4 \;=\;0

    . . . . . . . . . 0 \;=\;-A^4

    Add I\!:\qquad\; I \;=\;I-A^4

    . . . . . . . . . I \;=\;I - A + A - A^2 + A^2 - A^3 + A^3 - A^4

    . . . . . . . . . I \;=\;(I-A) + A(I-A) + A^2(I-A) + A^3(I-A)

    . . . . . . . . . I \;=\;(I-A)(I+A+A^2+A^3)


    Multiply by (I-A)^{-1}

    . . (I-A)^{-1}I \;=\;(I-A)^{-1}(I-A)(I+A+A^2+A^3)

    . . . (I-A)^{-1} \;=\; I+A+A^2+A^3
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  7. #7
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    Re: Matrix little demonstration

    Great!
    It seems so simple now...

    Thanks a lot for this detailed demontration!
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