Dot product of vector and symmetric linear map?

[Combinatus]

Homework Statement

My book states as follows:

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

\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)

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I'm a bit confused about the \Gamma(u) \cdot v = (AX)^t Y part. Why isn't \Gamma(u) \cdot v = (AX) Y, 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 A = A^t, not that AX = (AX)^t.

X^t (AY) = u \cdot \Gamma(v) 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.

 

[ystael]

X^t (AY) = u \cdot \Gamma(v) 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.

No, in fact, that's exactly it. If, in your notation, w_1, w_2 are represented in coordinates by the column vectors Z_1, Z_2, then w_1 \cdot w_2 is represented in coordinates by Z_1^t Z_2 -- that is, the transpose-product is exactly the coordinate representation of dot product (in an orthonormal basis, anyway).

 

[Combinatus]

Ah, you're right. I thought in terms of matrix multiplication of two n x 1 matrices rather than a standard dot product in R^n for some reason. Thank you!