Let $n\in \mathbb{N}$. Show that the determinant map $\det: M_n(\mathbb{R})\rightarrow \mathbb{R}$ is infinitely diffeentiable and compute the total derivative $d(\det)$ at every point of $A\in M_n(\mathbb{R})$. Find a necessary and sufficient condition on the rank of $A$ for $d(\det)=0$ on $A$..
I really do not know how to find derivative of determinant map..
Let $n=2$ for simplicity...
Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ then $df:\mathbb{R}^2\rightarrow \{\rm{maps~} \mathbb{R}^2\rightarrow \mathbb{R}\}$ with $df(a)=\left(\frac{\partial f}{\partial x}(a)~\frac{\partial f}{\partial y}(a)\right)$ and $df(x)(x_1,x_2)=x_1\frac{\partial f}{\partial x}(a)+x_2\frac{\partial f}{\partial y}(a)$
Consider $\det:M_2(\mathbb{R})\rightarrow \mathbb{R}$ with $\begin{bmatrix}x&y\\z&w\end{bmatrix}\mapsto xw-yz$ then,
$$d(\det)\begin{bmatrix}a_1&a_2\\a_3&a_4\end{bmatrix}= \begin{bmatrix}a_3&-a_4\\-a_2&a_1\end{bmatrix}$$
Now i am confused... Do we have
$$d(\det)\begin{bmatrix}a_1&a_2\\a_3&a_4\end{bmatrix}\begin{bmatrix}b_1&b_2\\b_3&b_4\end{bmatrix}= \begin{bmatrix}a_3&-a_4\\-a_2&a_1\end{bmatrix} \begin{bmatrix}b_1&b_2\\b_3&b_4\end{bmatrix}=a_3b_1-a_4b_2-a_2b_3+a_1b_4$$
I do not know where to go from here...
Example (3x3): You have $$ \det (A) = \left|\begin{matrix} a & b & c\\d & e & f\\g & h & i\end{matrix}\right| = a\left|\begin{matrix}e & f\\h & i\end{matrix}\right| - b\left|\begin{matrix}d & f\\g & i\end{matrix}\right| + c\left|\begin{matrix}d & e\\g & h\end{matrix}\right|. $$ That is, $$ \frac{\partial\det}{\partial a}(A) = \left|\begin{matrix}e & f\\h & i\end{matrix}\right|,\quad \frac{\partial\det}{\partial b}(A) = -\left|\begin{matrix}d & f\\g & i\end{matrix}\right|,\quad \frac{\partial\det}{\partial c}(A) = \left|\begin{matrix}d & e\\g & h\end{matrix}\right|. $$ Hence, all these determinants have to be zero. Now, do the same stuff for d, e, and f, where you apply Laplace to the second row of the matrix. Thereafter, do it again for g, h, and i. You will see that the condition $D(det)(A) = 0$ means that all 2x2 subdeterminants have to vanish. This again is equivalent to the matrix A having rank at most 1 (why?). Try to generalize this approach to the case of nxn matrices.