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Math Help - idempotent matrix problem.. HELP :/

  1. #1
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    Cool idempotent matrix problem.. HELP :/

    Hi, I have the following problem with an idempotent matrix and I am stuck...

    If A is an idempotent (A^2 = A), is E - nA invertible /E is the identity matrix and n is a real number/ and why?

    I've tried with setting (E - nA)(E - nA) but it doesn't get me anywhere..
    My instinct is telling me there it should be invertible in a lot of the cases, but... :/
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  2. #2
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    Quote Originally Posted by chefobg57 View Post
    Hi, I have the following problem with an idempotent matrix and I am stuck...

    If A is an idempotent (A^2 = A), is E - nA invertible /E is the identity matrix and n is a real number/ and why?

    I've tried with setting (E - nA)(E - nA) but it doesn't get me anywhere..
    My instinct is telling me there it should be invertible in a lot of the cases, but... :/
    Take

    (1 0)
    (1 0) = A

    The above is an idempotent matrix, but I - A (i.e., n = 1) isn't invertible.
    In this same case, if n =/= 1 then I - nA IS invertible.

    In general, and if you meant to ask, by any chance, for what values of n is I - nA invertible? If so, remembering that
    1/(1-x) = 1 + x + x^2 +.... for |x| < 1 we can try to do the same for I/(I - nA):

    I/(I - nA) = I + nA + n^2A +... = I + A(n + n^2 +...) ==> it's clear that we must have |n| < 1 for the part in the parentheses to converge to n/(1-n) and then it's easy to check that (I - nA)^(-1) = I + [n/(1-n)]A, even if we cannot justify the 1/(I - nA) thing!

    Tonio
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  3. #3
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    Wait, isn't this true?

    We are told that P is idempotent. First, however, we are going to show that I P is idempotent too. In other words, we need to show that:
    (I P)2 = I P
    Expanding the left side of the equation, we get:
    (I P)2 = (I P)(I P) = II 2P + P2 (3)
    However, we are told that P is idempotent and we know that I is an idempotent matrix, which means P2 = P and I2=I. Thus, equation (3) can also be written as:
    (I P)2 = (I P)(I P) = II 2P + P2 = I 2P + P = I P
    or (I P)2 = I P
    This shows that I P is idempotent too. Now, in part 2, we proved that an idempotent that is invertible is equal to the identity matrix. We just showed that I P is an idempotent matrix, which suggests that it is invertible only if I P = I. This means that P must be equal to 0 for that to hold. Thus, since P=0 is an idempotent matrix, it can be concluded that I P will be invertible.
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  4. #4
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    Quote Originally Posted by chefobg57 View Post
    Wait, isn't this true?

    We are told that P is idempotent. First, however, we are going to show that I P is idempotent too. In other words, we need to show that:

    (I P)2 = I P

    Expanding the left side of the equation, we get:

    (I P)2 = (I P)(I P) = II 2P + P2 (3)

    However, we are told that P is idempotent and we know that I is an idempotent matrix, which means P2 = P and I2=I. Thus, equation (3) can also be written as:

    (I P)2 = (I P)(I P) = II 2P + P2 = I 2P + P = I P



    or (I P)2 = I P

    This shows that I P is idempotent too. Now, in part 2, we proved that an idempotent that is invertible is equal to the identity matrix. We just showed that I P is an idempotent matrix, which suggests that it is invertible only if I P = I. This means that P must be equal to 0 for that to hold. Thus, since P=0 is an idempotent matrix, it can be concluded that I P will be invertible.

    Sorry but I didn't read all the above: did you read AND CHECK my example with A? Is this right (because, of course, I could be wrong)? If the answer is yes then the claim is false. Period.

    Tonio
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  5. #5
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    Quote Originally Posted by chefobg57 View Post
    Wait, isn't this true?

    We are told that P is idempotent. First, however, we are going to show that I P is idempotent too. In other words, we need to show that:

    (I P)2 = I P

    Expanding the left side of the equation, we get:

    (I P)2 = (I P)(I P) = II 2P + P2 (3)

    However, we are told that P is idempotent and we know that I is an idempotent matrix, which means P2 = P and I2=I. Thus, equation (3) can also be written as:

    (I P)2 = (I P)(I P) = II 2P + P2 = I 2P + P = I P



    or (I P)2 = I P

    This shows that I P is idempotent too. Now, in part 2, we proved that an idempotent that is invertible is equal to the identity matrix. We just showed that I P is an idempotent matrix, which suggests that it is invertible only if I P = I. This means that P must be equal to 0 for that to hold. Thus, since P=0 is an idempotent matrix, it can be concluded that I P will be invertible.

    Now I read the argument above and I jsut can't understand how did you
    arrive to the conclusion that 1-P is invertible from the two lines above!!

    Tonio
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  6. #6
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    Well, I was just wondering if I-P is invertible or not to begin with.. The thing I posted was my claim it is invertible.. Sorry for overwhelming you...

    I am really only a beginner in Linear Algebra and I am not sure about your explanation..

    Thanks in advance for helping me out!
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