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Math Help - Problem of Idempotent matrix

  1. #1
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    Unhappy Problem of Idempotent matrix

    Assume A is idempotent matrix of order n,in other words: A^2 = E_{n},and det(A) > 0.show that: det(A + E_{n}) \not=0
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  2. #2
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    Quote Originally Posted by Xingyuan View Post
    Assume A is idempotent matrix of order n,in other words: A^2 = E_{n},and det(A) > 0.show that: det(A + E_{n}) \not=0
    wait a second! idempotent means A^2=A not A^2=I. are you sure you wrote the question correctly?
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  3. #3
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    I am not sure. I am reading a book of Linear Algebra. this problem is in that book. if n is odd ,and A^2 = E_{n},I have proved det(A + E_{n}) \not=0. when n is even ,i have no idea.
    If A^2 = A,I am not sure whether the conclusion was right?

    thanks very much.
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  4. #4
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    Quote Originally Posted by Xingyuan View Post
    I am not sure. I am reading a book of Linear Algebra. this problem is in that book. if n is odd ,and A^2 = E_{n},I have proved det(A + E_{n}) \not=0. when n is even ,i have no idea.
    If A^2 = A,I am not sure whether the conclusion was right?

    thanks very much.
    well, this is not very helpful! post the question exactly as you've seen it. do not add anything to it! then i'll help you to solve it.
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  5. #5
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    OK, Assume A^2 = E_{n},and det(A) > 0,show that : det(A + E_{n}) \not= 0

    thanks very much !
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  6. #6
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    Quote Originally Posted by Xingyuan View Post
    OK, Assume A^2 = E_{n},and det(A) > 0,show that : det(A + E_{n}) \not= 0

    thanks very much !
    assuming that E_n is the identity matrix (this is an unusual notation for the identity matrix and you should've mentioned what it means!) this still can't be the question because it's false!

    a trivial counter-example: n any even number and A = -E_n.
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  7. #7
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    Oh,yes .If n is odd .the conclusion is correct.thanks very much.

    If A^2 = A,and det(A) > 0. can we get det(A + E_{n})\not=0?
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  8. #8
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    Quote Originally Posted by Xingyuan View Post

    Oh,yes .If n is odd .the conclusion is correct.thanks very much.
    for odd values of n the problem is still false. a counter example is A=\begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. we have A^2=I, \ \det A= 1 > 0 but \det(A + I)=0.


    If A^2 = A,and det(A) > 0. can we get det(A + E_{n})\not=0?
    yes, and we only need \det A \neq 0. the proof is very simple: A(A+I)=A^2 + A=2A. thus: (\det A) \det (A+I)=2^n \det A and hence \det(A+I)=2^n, because \det A \neq 0.
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  9. #9
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    Thanks Very Much!!
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