Complement of a subspace

1. Hyperplane. Write $V=\ker\varphi$ with $\varphi$ a non-zero linear form. Then $E\setminus V=\{\varphi>0\}\sqcup\{\varphi<0\}$. Both pieces are open, disjoint, non-empty (replace $x$ by $-x$), and convex (preimages of intervals under a linear map), hence path connected. So there are exactly two components.

2. Codimension $\ge2$. Let $a,b\in E\setminus V$.

*Which points $c$ are bad for $a$?* Suppose $(1-t)a+tc\in V$ for some $t\in[0,1]$. Then $t\ne0$ (because $a\notin V$), and so $c=\frac1t\big(v-(1-t)a\big)$ with $v\in V$, i.e.
$$c\in V+\mathbb{R}a.$$
$V+\mathbb{R}a$ is a linear subspace of dimension $\dim V+1$ (since $a\notin V$), which is $\le N-1<N$. Likewise define $W_b=V+\mathbb{R}b$. If $c\notin W_a$ the whole segment $[a,c]$ avoids $V$ and similarly for $[c,b]$ and $W_b$.

*Choice of $c$.* $E$ is not the union of two proper subspaces $W_a$, $W_b$ (a vector space over an infinite field is never a union of finitely many proper subspaces). Pick $c\notin W_a\cup W_b$. Then $c\notin V$ and the path $[a,c]\cup[c,b]$ (two segments) lies in $E\setminus V$.

Conclusion. $E\setminus V$ is path connected.

Remark. In $M_n(\mathbb{R})$, the complement of the hyperplane $\{\operatorname{tr}=0\}$ has two components (Problem 5), whereas the complement of the subspace of scalar matrices $\mathbb{R}I$ (codimension $n^2-1\ge2$ for $n\ge2$) is connected. The result is false for *non-linear* sets of codimension $1$ and may fail even for codimension $2$ non-linear sets that are not "nice".