# Regular value of the determinant map

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.

• 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. Jan 16, 2021 at 19:15

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.
• 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? Jan 17, 2021 at 13:39
• Recall that the characteristic polynomial of $A$ is $\det (\lambda I + A) = \lambda^n + \lambda^{n-1} (\operatorname{tr} A) + \cdots + \det A$. Set $\lambda = t^{-1}$. Jan 17, 2021 at 16:36
• Thank you. ${}{}$ Jan 17, 2021 at 18:28
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 $$M$$.