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.
 A: 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.
A: 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 $M$.
