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Thread: Easy matrix help?

  1. #1 Easy matrix help? 
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    Question: Show that the equation Ax = x can be rewritten as (A - I)x = 0
    where A is a n by n matrix and x is a n by 1 matrix.

    My (probably incorrect) answer:

    Ax = x
    Ax(x^-1) = x(x^-1)
    A(xx^-1) = xx^-1
    AI = I
    A = I
    A - I = 0
    (A - I)x = 0x
    (A - I)x = 0

    Might anyone tell me how I should be going about solving this?


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  3. #2 Re: Easy matrix help? 
    . DrRocket's Avatar
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    Quote Originally Posted by m84uily
    Question: Show that the equation Ax = x can be rewritten as (A - I)x = 0
    where A is a n by n matrix and x is a n by 1 matrix.

    My (probably incorrect) answer:

    Ax = x
    Ax(x^-1) = x(x^-1)
    A(xx^-1) = xx^-1
    AI = I
    A = I
    A - I = 0
    (A - I)x = 0x
    (A - I)x = 0

    Might anyone tell me how I should be going about solving this?
    You are right. That is not correct. It is not even close.

    What are you doing in a class on matrix algebra if your skills in elementary algebra are so poor ?

    This question is quite simple. You need to figure this one out for yourself in order to learn the subject. To just show you how to do it would be counter-productive.

    Just think about it a bit.


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  4. #3  
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    Thanks for the reply DrRocket!

    Okay, I'll give it another go; this time I'll explain my reasoning as I go through the steps. I'd appreciate it if someone were to tell me if I've done something wrong.

    Ax = x

    Ax - x = 0
    (x was moved from the right to the left side making it negative on that side and resulting in 0 on the other)

    (A - I)x = 0
    (x was factored out from Ax resulting in A and was factored out from x which results in I, the matrix equivalent to 1)

    Thanks for the replies in advance.
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  5. #4  
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    The second post looks a lot better.

    The most important feature of the Identity Matrix I is that it works as the "one" for matrix multiplications.

    Thus:

    AX = X
    <=>
    AX = IX
    <=>
    AX - IX = 0
    <=>
    (A - I)X = 0

    Which is pretty much what you said.

    (It seems what you showed in your previous post was that if X was invertible then AX = X is equivalent with A = I.)
    "Complexity is stupidity disguised as intellect."
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  6. #5  
    . DrRocket's Avatar
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    Quote Originally Posted by m84uily
    Thanks for the reply DrRocket!

    Okay, I'll give it another go; this time I'll explain my reasoning as I go through the steps. I'd appreciate it if someone were to tell me if I've done something wrong.

    Ax = x

    Ax - x = 0
    (x was moved from the right to the left side making it negative on that side and resulting in 0 on the other)

    (A - I)x = 0
    (x was factored out from Ax resulting in A and was factored out from x which results in I, the matrix equivalent to 1)

    Thanks for the replies in advance.
    That works. And yes, it really is that easy.
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