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I know that the Killing form is non-degenerate for semi-simple Lie Algebras. $$\Gamma(v,w) = \gamma_{ij} v^i w^j$$

I just wanted to double check that this was equivalent to saying that the Killing matrix $\gamma_{ij}$ was invertible for semi-simple Lie Algebras.

Dietrich Burde
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Alex Gower
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    The word "killing matrix" should not be used in the title. Better is "the representing matrix of the Killing form". Have you computed such a matrix for the Killing form of $\mathfrak{sl}_2(\Bbb C)$? This is very helpful. – Dietrich Burde Dec 22 '21 at 10:07
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    Note that the "Killing matrix" depends on a choice of basis (so "the Killing matrix" or "the matrix for the Killing form" doesn't make sense without reference to any basis). However change of basis changes the Killing matrix by $M\mapsto AMA^t$ for some invertible $A$, so invertibility is independent of the choice. (I now see this has partly been mentioned in the answer.) – YCor Dec 24 '21 at 09:45

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For any bilinear form (on a finite-dimensional vector space over a given field) it is true that the form is non-degenerate if and only if its representing matrix (w.r.t. any basis) is invertible. See e.g. Non-degenerate bilinear forms and invertible matrices.

By the way, I have not heard the nomenclature "the Killing matrix" before, and would like to point out that this representing matrix (although not its invertibility) depends on what basis we choose for the Lie algebra.

Torsten Schoeneberg
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