How does existence of complemented subspace imply extending automorphisms?

Question

Suppose we have an inner space $X$ with a complemented finite-dimensional subspace $M$. Denote the complement by $N$.

My questions are:

1. If $X$ is infinite-dimensional and $M$ is finite-dimensional, does that mean $N$ is infinite-dimensional?

2. If $X$ has a complemented subspace, does that imply that any autohomeomorphism of a finite-dimensional subspace extends to an autohomeomorphism of the whole $X$?

I have seen the "2." in tomasz´s comments under one of my previous questions, that´s why I am interested whether this holds generally and why.

Thank you for any insights.

Answer 1

1. If $X$ is infinite-dimensional and $M$ is finite-dimensional, does that mean $N$ is infinite-dimensional?

Yes. In general, if $X=M\oplus N$ then $\dim X=\dim M+\dim N$ and so $X$ being infinite dimensional implies that one of $M,N$ has to be infinite dimensional as well.

2. If $X$ has a complemented subspace, does that imply that any autohomeomorphism of a finite-dimensional subspace extends to an autohomeomorphism of the whole $X$?

Yes. So first of all, a linear subspace $M\subseteq X$ has a complement $N$, if $N\subseteq X$ is a linear subspace such that $X\simeq M\oplus N$ and the isomorphism is both linear and a homeomorphism. WLOG we can assume that $X=M\oplus N$. Let $f:M\to M$ be an autohomeomorphism. Then it extends to $F:X\to X$ via $F(a,b):=(f(a),b)$ which is an autohomeomorphism as well. Note that dimensions here don't matter.