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I'm reading through some general relativity notes. I have reached a part that I don't understand, probably because my linear algebra is not good enough.

My questions relating to the image below are:

  1. Why can the metric tensor always be diagonalized? Is it because it is already symmetric?
  2. Is the proof shown mathematically in the co-ordinate transform below?
  3. the answer to (2) is "yes", could someone attempt to explain why this proves that it is diagonalizable?

enter image description here

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To briefly answer your 3 questions:

  1. The metric tensor can be diagonalized because it is symmetric.
  2. The coordinate transform shows how the metric tensor transforms under a linear transformation of the underlying space (which strictly speaking would be the tangent space to spacetime at a given point). It does not show that the metric tensor can always be diagonalized.
  3. The answer to 2 was no.

Let's be very precise to avoid confusion: when we say that a matrix $G$ is diagonalized by an invertible matrix $\Lambda$, that means that $G' = \Lambda^{-1}G\Lambda$ is diagonal. If we see $G$ as describing a linear transformation in a given basis, then $G'$ describes the same linear transformation in the basis consisting of the columns of $\Lambda^{-1}$.

When we say that a metric tensor is diagonalized by a matrix, we mean that the matrix representing it after the change of basis, which we saw is $G' = \Lambda^{T}G\Lambda$, is diagonal.

The result linked by @Gold asserts that a symmetric matrix can always be diagonalized (implicitly meaning: as the matrix of a linear transformation) by an orthogonal matrix. Using this you are done, since for an orthogonal matrix $\Lambda$ we have $\Lambda^{-1} = \Lambda^T$.

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