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Dot product of vector and symmetric linear map?

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data

    My book states as follows:


    If u and v have the coordinate vectors [tex]X[/tex] and [tex]Y[/tex] respectively in a given orthonormal basis, and the symmetric, linear map [tex]\Gamma[/tex] has the matrix [tex]A[/tex] in the same basis, then [tex]\Gamma(u)[/tex] and [tex]\Gamma(v)[/tex] have the coordinates [tex]AX[/tex] and [tex]AY[/tex], respectively. This gives:

    [tex]\Gamma(u) \cdot v = (AX)^t Y = (X^t A^t) Y = X^t A^t Y = X^t AY = X^t (AY) = u \cdot \Gamma(v)[/tex]


    I'm a bit confused about the [tex]\Gamma(u) \cdot v = (AX)^t Y[/tex] part. Why isn't [tex]\Gamma(u) \cdot v = (AX) Y[/tex], thus rendering the operation undefined (assuming that X and Y are row vectors with at least two rows)? After all, as far as I could figure, a symmetric, linear map would only yield that [tex]A = A^t[/tex], not that [tex]AX = (AX)^t[/tex].

    [tex]X^t (AY) = u \cdot \Gamma(v)[/tex] bestows similar confusion upon me as well. It seems to me as if the vectors are just casually transposed for the dot product to "work out", although that probably isn't it.

    I'm probably missing something very trivial. I've looked around for alternative proofs to no avail.
    Last edited: Feb 19, 2010
  2. jcsd
  3. Feb 19, 2010 #2
    No, in fact, that's exactly it. If, in your notation, [tex]w_1, w_2[/tex] are represented in coordinates by the column vectors [tex]Z_1, Z_2[/tex], then [tex]w_1 \cdot w_2[/tex] is represented in coordinates by [tex]Z_1^t Z_2[/tex] -- that is, the transpose-product is exactly the coordinate representation of dot product (in an orthonormal basis, anyway).
  4. Feb 19, 2010 #3
    Ah, you're right. I thought in terms of matrix multiplication of two n x 1 matrices rather than a standard dot product in [tex]R^n[/tex] for some reason. Thank you!
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