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Math Help - conjugates

  1. #1
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    conjugates

    is it true that A multiplied by its conjugate A* is always greater than zero?

    and if so how do i prove/show this?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathisthebestpuzzle View Post
    is it true that A multiplied by its conjugate A* is always greater than zero?

    and if so how do i prove/show this?
    depends on what A is, but in general, no.

    counter-example: take A = \sqrt{2} - 2. Then A^* = \sqrt{2} + 2. So that A \cdot A^* = 2 - 4 = -2 < 0
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    depends on what A is, but in general, no.

    counter-example: take A = \sqrt{2} - 2. Then A^* = \sqrt{2} + 2. So that A \cdot A^* = 2 - 4 = -2 < 0
    That's a strange sort of conjugation. I would have taken the conjugate of \sqrt{2} - 2 to be -\sqrt{2} - 2, in which case AA* is positive.

    Quote Originally Posted by mathisthebestpuzzle View Post
    is it true that A multiplied by its conjugate A* is always greater than zero?

    and if so how do i prove/show this?
    If A is meant to be a matrix and A* is its hermitian adjoint then AA* is always positive. Reason: \langle AA^*x,x\rangle = \langle A^*x,A^*x\rangle = \|A^*x\|^2\geqslant0.
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    another example is through the usage of complex numbers:

    A=(a+j), A* = (a-j)
    A.A*=(a^2 - aj + aj- j^2)
    = a^2-(-1^(1/2))^2
    =a^2 - (-1)
    = a^2 + 1 which is always larger than 0

    Complex Numbers
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Opalg View Post
    That's a strange sort of conjugation. I would have taken the conjugate of \sqrt{2} - 2 to be -\sqrt{2} - 2, in which case AA* is positive.
    indeed. what i did is how we conjugate stuff on my planet. apparently on earth they do it differently. good stuff! (excuse: i was drunk -- that's always my excuse)

    could we not use something like A = 2 + \sqrt{145} then?
    Last edited by Jhevon; April 27th 2008 at 12:55 PM.
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  6. #6
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    The word conjugate is one of those words in mathematics which has many different meanings. There is a standard algebraic/number theoretic definition. Let F be an extension field over a smaller field K. Given two algebraic elements \alpha,\beta\in F we say \alpha,\beta are conjugates iff they has the same minimal polynomial. So a+bi and a-bi ( b\not = 0) are conjugates because they have the same minimal polynomials.

    Though that definition above is applied to fields we can extend it a little and apply it to \mathbb{Z} (which is not a field) as well. Given two algebraic number \alpha,\beta we say \alpha and \beta are conjugates if they have the same minimal polynomial over \mathbb{Z}. In your case \alpha = \sqrt{2} - 2 and \beta = -\sqrt{2} - 2 have x^2+4x+2 as their minimal polynomial.
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  7. #7
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    these conjugates are over a complex field.
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