Let $G=GL(n, \mathbb{C})$ and $M=\{Q \in GL(n, \mathbb{C}) | Q^t = Q \}$. Suppose $G$ acts on $M$ through the following: $\forall g \in G, \forall m \in M$, $g\cdot m = gmg^t$. Question: Is the action transitive or not?
Is the action transitive?
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linear-algebra
group-theory
1 Answers
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Every symmetric matrix $Q\in M$ is the matrix of a quadratic form of maximum rank over $\mathbb{C}^n$. The action of $G$ represents base change for the quadratic form represented by $Q$. Every quadratic form has orthonormal bases. Thus the identity matrix is in every orbit, so yes, the action is transitive.
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0Thanks a lot! Unfortunately, I am not aware of Slyvester's law of inertia... – 2012-04-24