Can we use the multiplication map to prove that invertible matrices form an open subset of $M_n(R)$ Define a map  $f : M_n(R) \times M_n(R) \to M_n(R)$, by $f((A,B)) = AB$. We know that this map is continuous. Can we try to prove that invertible matrices form an open subset of $M_n(R)$ by somehow using this map. (I already know how to prove it by using the map of determinant)
 A: Unfortunately we cannot. Invertible matrices would be represented by pairs (A,B) such that AB = I. Namely this set of pairs is $f^{-1}(I)$ which is a closed set in $M_n(R) \times M_n(R)$.
A: You can using the implicit function theorem.
We have ${\partial f(A,B) \over \partial A}(H) = HB$, which is invertible whenever $B$ is invertible. (The inverse map is $H \mapsto H B^{-1}$.)
Note that $f(A^{-1},A) = I$, hence the implicit function theorem tells us that there are open $U,V$ with $A \in U$, $A^{-1} \in V$ and a $C^1$ function $\phi: U \to V$ such that $f(\phi(X), X ) = I$ for $X \in U$ and
$\phi(A) = A^{-1}$.
Since $\phi(X) = X^{-1}$, and $U$ is open, we see that the set of invertible matrices is open.
(In fact, this shows that the derivative of the inverse map
is ${\partial \phi(A) \over \partial A}(H) = - A^{-1}H A^{-1}$.)
Alternative: A simpler alternative is to use the fact that $A \mapsto \det A$ is a continuous map.
A: Suppose  $A^{-1}$ exists . We have :  $A+C =A(1+A^{-1} C)$ is invertible $\iff 1+A^{-1} C$ is invertible. For brevity let $B=A^{-1} C.$ Suppose $C$ belongs to the open ball about $A$ with radius $1/(2\|A^{-1}\|).$ Then for $n\in N$ we have $\|B^n\|<2^{-n}.$ Now for $n\in N$ let $ D_n=1+\sum_1^n (-B)^n.$ Then $D_n$ converges in norm to a limit $D.$ The continuity of $f(X,Y)=X Y$ implies that $(1+B) D_n$  converges in norm to $(1+B) D.$ But $(1+B) D_n= 1-B^{-n-1}$ converges in norm to $1.$ Therefore $(1+B) D=1$ so $1+B=1+A^{-1} C$ is invertible .
A: Use $\text{det}^{-1}(\Bbb{R} \setminus {0})$ is open .
