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Thread: 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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    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 $\displaystyle A$ is, but in general, no.

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

    counter-example: take $\displaystyle A = \sqrt{2} - 2$. Then $\displaystyle A^* = \sqrt{2} + 2$. So that $\displaystyle A \cdot A^* = 2 - 4 = -2 < 0$
    That's a strange sort of conjugation. I would have taken the conjugate of $\displaystyle \sqrt{2} - 2$ to be $\displaystyle -\sqrt{2} - 2$, 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: $\displaystyle \langle AA^*x,x\rangle = \langle A^*x,A^*x\rangle = \|A^*x\|^2\geqslant0$.
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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 $\displaystyle \sqrt{2} - 2$ to be $\displaystyle -\sqrt{2} - 2$, 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 $\displaystyle A = 2 + \sqrt{145}$ then?
    Last edited by Jhevon; Apr 27th 2008 at 11:55 AM.
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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 $\displaystyle F$ be an extension field over a smaller field $\displaystyle K$. Given two algebraic elements $\displaystyle \alpha,\beta\in F$ we say $\displaystyle \alpha,\beta$ are conjugates iff they has the same minimal polynomial. So $\displaystyle a+bi$ and $\displaystyle a-bi$ ($\displaystyle b\not = 0$) 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 $\displaystyle \mathbb{Z}$ (which is not a field) as well. Given two algebraic number $\displaystyle \alpha,\beta$ we say $\displaystyle \alpha$ and $\displaystyle \beta$ are conjugates if they have the same minimal polynomial over $\displaystyle \mathbb{Z}$. In your case $\displaystyle \alpha = \sqrt{2} - 2$ and $\displaystyle \beta = -\sqrt{2} - 2$ have $\displaystyle x^2+4x+2$ 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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