Find two matrices A,B$\in {{M}_{n\times n}}$ with the propertie that there not exists such a continuous function $f:[0,1]\to M$ with $f(0)=A,f(1)=B$

Question

If $M_{n\times n}$ is the set of invertible matrices with real entries. Find two matrices $A,B\in M_{n \times n}$ with the propriety that there not exists such a continuous function

$$f:[0,1]\to M, \quad f(0)=A, f(1)=B $$

the only way i was thinking was is the inverse function such as $f^{-1}(A)=0, \quad f^{-1}(B)=1,$ but this doesnt seem to get me anywhere.

Answer 1

If $\det(A)>0>\det(B)$, then there is no such function, because otherwise the range of the map $\det\circ f$ would contain $\det(A)$ and $\det(B)$, but not $0$.

Answer 2

Let A be any matrix with positive determinant, B any matrix with negative determinant. If there is such a $f$ , then $det(f(t))$ is a continuous function of $t\in[0,1]$ (it is a polinomial), with $det(f(0))=det(A)>0$ and $det(f(1))=det(B)<0$. Then, by the intermediate value theorem there must be a $t\in[0,1]$ such that $det(f(t))=0$ giving a contradiction.

Answer 3

Hint.

The existence of the function $f$ means that there exists a path between the two matrices $A$ and $B$, which implies that $A$ and $B$ are in the same connected component of $M:=M_{n\times n}$.

On the contrary, the continuous map $\det:M\to\mathbb{R}$ tells you that $M$ is not connected (since $\det(M)$ is not).

Now, find two elements of $M$ that are not in the same component of $M$.