In , problem 5 the author shows that the differential of the determinant map $d(\det)_A$ for an invertible matrix $A$ is nonsingular by only showing that $d(\det)_A(A) \ne 0$. I don't really see why this shows that $d(\det)_A$ is nonsingular. I would like some further clarification on this point.
Need help understanding proof about critical values of the determinant map.
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multivariable-calculus
differential-topology
1 Answers
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$d(\det)_A$ is a map $M_n(\mathbb R) \to \mathbb R$. It is non-singular, if its rank equals $\min\{\dim M_n(\mathbb R), \dim \mathbb R\} = \min\{n^2, 1\} = 1$, which means that $d(\det)_A \ne 0$.
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0Oh I see. Since the differential is a linear map, the image is either $\{0\}$ or $\mathbb{R}$, so it suffices to show that it's not the zero map. – 2012-10-15