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Proof x.(My) = (M*x).y

  1. Jun 24, 2008 #1
    1. The problem statement, all variables and given/known data
    This was a question in a recent exam and I would like to know if the answer I gave is correct since I not 100% sure...

    If [itex]\mathbf{x}[/itex] and [itex]\mathbf{y}[/itex] are complex vectors in C^n (complex) and [itex]M[/itex] is a square (n x n)-matrix (also in C^n), prove that:
    [tex]\mathbf{x} \cdot (M \mathbf{y}) = (M^* \mathbf{x}) \cdot \mathbf{y}[/tex]
    (where M* denotes the complex conjugate + the transpose of M: [itex]M^* = \overline{M}^T[/itex]
    (The dot denotes the standard complex dot-product)

    3. The attempt at a solution
    I did the following:

    [tex]\mathbf{x} \cdot \mathbf{y} = \mathbf{x}^T \mathbf{\overline{y}}[/tex]
    [tex]\mathbf{x} \cdot (M \mathbf{y}) = \mathbf{x}^T \overline{M \mathbf{y}} = \mathbf{x}^T \overline{M} \overline{\mathbf{y}}[/tex]

    [tex]\mathbf{x}^T \overline{M} = (M^* \mathbf{x})^T[/tex] because [tex](M^* \mathbf{x})^T = \mathbf{x}^T (M^*)^T = \mathbf{x}^T \overline{M}[/tex]

    So, now we have:
    [tex]\mathbf{x} \cdot (M \mathbf{y}) = (M^* \mathbf{x})^T \overline{\mathbf{y}} = (M^* \mathbf{x}) \cdot \mathbf{y}[/tex]

    Is this solution correct? Or did I make an error somewhere? (I'm not entirely sure of the very first statement for example...)

  2. jcsd
  3. Jun 24, 2008 #2


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    Yes, that's basically correct. But if you aren't sure of the first statement, look up how you defined the inner product. You could also have written your definition x.y=(y*)x. An equally valid definition of inner product is x.y=(x*)y. The first inner product is antilinear in y and the second one is antilinear in x. The choice of one or the other is a matter of convention.
  4. Jun 24, 2008 #3
    Well that was really the problem. The book we are using first defines inner products on the real numbers, it defines the standard dot product as:
    [tex]\vec{x} \cdot \vec{y} = \sum_{i=1}^n x_i y_i[/tex]
    and it is then shown that x.y = x^T y.

    Then it goes on later about the complex numbers. It does define the standard dot product as:
    [tex]\vec{x} \cdot \vec{y} = \sum_{i=1}^n x_i \overline{y_i}[/tex]
    but it does NOT explicitly show that the dotproduct can also be written as [tex]\mathbf{x}^T \overline{\mathbf{y}}[/tex].
    I can easily verify that this works but I wasn't sure whether it is true for every condition...

    So it is correct? Yay :)
  5. Jun 24, 2008 #4


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    That's the same as (y*)x, yes. Which is the same as your expression.
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