1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Now:
    [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...)

    Thanks.
     
  2. jcsd
  3. Jun 24, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's the same as (y*)x, yes. Which is the same as your expression.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Proof x.(My) = (M*x).y
  1. X^n-y^n proof (Replies: 16)

  2. Y' = f'(x) proof (Replies: 2)

  3. Proof of x^n - y^n = (Replies: 6)

Loading...