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Let $\text{det}: M_n(\mathbb{R})\rightarrow \mathbb{R}$ be the determinant map from $n\times n$ matrices to $\mathbb{R}$, then how can I show that 1 is a regular value of this map? I am specifically having trouble to calculate the derivative of this map on matrices.

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    $\begingroup$ By definition of a regular value, $1$ is a regular value of the determinant if for every matrix $M$ with determinant $1$, $\mathrm{d}_M(\det) : M_n \to \mathbb{R}$ is a non zero linear map. So you just have to compute this linear map. You can look for "differential of the determinant" to do so. $\endgroup$
    – Didier
    Jan 16, 2021 at 19:15

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Since $M \mapsto \det M$ is a function of more than one variable, finding a formula for its derivative is best done via directional derivatives. Let's denote and calculate the directional derivative of the determinant function at the matrix $M$ in the direction of the matrix $N$ by \begin{align*} \partial_N\det(M) &= \left.\frac{d}{dt}\right|_{t=0}\det(M+tN)\\ &= \left.\frac{d}{dt}\right|_{t=0}(\det M)\det(I+tM^{-1}N)\\ &= (\det M)\left.\frac{d}{dt}\right|_{t=0}\det(I+tM^{-1}N). \end{align*} On the other hand, \begin{align*} \det(I + tA) &= t^{n}\det(t^{-1}I + A)\\ &= t^{n}(t^{-n} + (\operatorname{tr} A)t^{-n+1} + \cdots)\\ &= 1 + t(\operatorname{tr}A) + t^2(\cdots) \end{align*} and therefore, $$ \left.\frac{d}{dt}\right|_{t=0}\det(I + tA) = \operatorname{tr} A. $$ Substituting this into the equation above, we get $$ \partial_N\det(M) = (\det M)\operatorname{tr}(M^{-1}N). $$ From here, it is straightforward to verify that $1$ is a regular value of the determinant function.

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  • $\begingroup$ Hai Prof, I don't quite know how to see the equality $\det(t^{-1}I + A) = t^{-n} + \text{tr} A\, t^{-n+1} + \cdots$ Would you elaborate please? $\endgroup$ Jan 17, 2021 at 13:39
  • $\begingroup$ Recall that the characteristic polynomial of $A$ is $\det (\lambda I + A) = \lambda^n + \lambda^{n-1} (\operatorname{tr} A) + \cdots + \det A$. Set $\lambda = t^{-1}$. $\endgroup$
    – Deane
    Jan 17, 2021 at 16:36
  • $\begingroup$ Thank you. ${}{}$ $\endgroup$ Jan 17, 2021 at 18:28
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Let $M$ such that $det(M)=1$, write $M=(m_1,...,m_n)$ where $m_i\in\mathbb{R}^n$. consider $f(t)$ the matrix whose first column is $m_1+tm_1$ and whose colum $i$ is $m_i,i\neq 1$. We have $det(f(t))=1+t$, this implies that $det'_M(f'(t))=1$, we deduce that $det$ is not degerated at $M$.

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