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Inner Product of 0 vector, & Complex numbers

  1. Jun 5, 2009 #1
    Hello,

    Can someone help me understand why the Inner Product of a Null vector with itself can be non zero if complex numbers are involved?

    And why using the complex conjugate resolved this?

    I may have understood this wrong. It could be that an Inner Product of any non-Null vector with itself can be zero if complex numbers are involved. Either way does someone have an example?
     
  2. jcsd
  3. Jun 5, 2009 #2

    dx

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    It cannot. The inner product of the null vector with itself is zero, and the inner product of a non-null vector cannot be zero. Both these facts follow from the following axiom for inner products:
    [tex] \langle v|v \rangle \geq 0 [/tex], and [tex]\langle v|v \rangle = 0[/tex] if and only if [tex]v = |0\rangle [/tex]​
     
  4. Jun 5, 2009 #3

    HallsofIvy

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    The inner product on a vector space over the complex numbers must satisfy
    [tex]<u, v>= \overline{<v, u>}[/tex] and, in particular, [itex]<u, av>= \overline{a}<u, v>[/itex] for any scalar a.

    For C2, ordered pairs of complex numbers, the inner product [itex]<(a+ bi),(a+ bi)> = (a+ bi)\overline(a+ bi)= (a+bi)(a-bi)= a^2+ b^2[/itex] which is 0 only if a= b= 0.

    If you mistakenly use the inner product for real numbers, that is, without the "complex conjugation" you could have [itex]<a+ bi, a+ bi>= a^- b^2[/itex] which can be 0. But, as I said, that is a mistake.
     
  5. Jun 5, 2009 #4
    Ahha! Thank you!
     
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