Continuity of map from $M(n,\mathbb{R})$ to itself.

Question

I am self studying topology of metric spaces and i encountered an exercise which requires to prove the map going from $M(n,\mathbb{R})$ to $M(n,\mathbb{R})$ defined by $A \mapsto A^{k}$ where k is any positive integer and $M(n,\mathbb{R})$ is given with euclidean norm. I am able to do it with $k=2$ but don’t know how to show it for any natural number k. I also thought of induction on k but can’t process. Please give me a hint to show this.

Answer 1

Yes, it can be done by induction. If $k=1$, you have the identity map, which is continuous. And, if $k\in\Bbb N$ and if the map $A\mapsto A^k$ is continuous, then the map $A\mapsto A^{k+1}$ is continuous too, since it is the previous map times the identity map. And the product of two continuous maps is continuous.

Answer 2

Realizing, $M(n,\mathbb {R})\cong \mathbb {R}^{n^2}$. You have the multiplication map $*:\mathbb{R}^{n^2}\times \mathbb{R}^{n^2}\to \mathbb{R}^{n^2}$ is nothing but bunch of polynomials in each co-ordinates of the target space.( by the usual rule of multiplication of matrices). Hence not only the map is continuous, it is infact smooth $(C^{\infty})$. You can also write the Jacobian explicitly.