Help Solve a Mystery: Lawden's "An Introduction to Tensor Calculus

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Hi,

Anyone out there have Lawden's book, "An Introduction to Tensor Calculus, Relativity and Cosmology"? Expression 39.13 (on p 109) and the sentence that precedes it have me stumped. My understanding of what he's saying is that a symmetric quadratic form can always be diagonalized so that the coefficients are 1. I was not aware of this. It seems to me that's the same thing as saying all symmetric matrices are similar to the identity--that is, its eigenvalues are all 1. If someone could clear this up for me it would be awesome. By the way, I know this is more of a linear algebra question, but I mentioned it in here because I figured people in this forum might be familiar with the book.

thank you :)
 
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Okay I figured out some of that typesetting stuff so I can actually write what Lawden says.

We have the quadratic form

<br /> d s^2 = g_{ij} d x^i d x^j<br />

where g_{ij} is the (symmetric) metric tensor at point A, x^i = a^i. Define the y-coordinates by the relation

<br /> x^i = b^i{}_{j} y^j<br />,

which is to say

<br /> d x^i = b^i{}_{j} d y^j<br />.

Lawden says, "...it is a well known result from algebra that the quadratic form g_{ij} d x^i d x^j can be reduced to the diagonal form

(d y^1)^2 \ + \ (d y^2)^2 \ + \ ... \ + \ (d y ^N)^2

at the point A by proper choice of the b^i{}_j (some of these coefficients may have to be given imaginary values)." This is important because he is showing that at points where the metric tensor is stationary, a coordinate frame can be chosen that is locally Cartesian. At any rate, I'm not familiar with this well known result from algebra. If anyone is, please explain it to me.

thanks!
 
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