# Why can the metric tensor always be diagonalized?

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?

• Would Mathematics be a better home for this question? Commented Feb 7, 2022 at 19:54
– Gold
Commented Feb 7, 2022 at 19:55

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 $$\Lambda^{-1} = \Lambda^T$$.