Let $H_1, \dots , H_k$ be a collection of hyperplanes in $\Bbb R^{k+1}$ with normal vectors $n_1, \dots, n_k$. Show that the intersection $H_1 \cap \dots \cap H_k$ forms a line if and only if the vectors $n_1, \dots, n_k$ are linearly independent.
For the implication $\Longleftarrow$ : The proof I was given says that we consider the matrix $N=[n_1, \dots, n_k] \in \Bbb R^{(k+1) \times k}$. Then $\bigcap_{i=1}^k H_i=\{x \in \Bbb R^{k+1} \lvert x \cdot N=c\}$ for a fixed $c \in \Bbb R^{k}$. If the vectors $n_1, \dots, n_k$ are linearly independent, then $rank(N) = k$. Furthermore, there exists $x_∗$ such that $x_∗ \cdot N = c$. This is true, for example, because N can be completed to an invertible square matrix. This means that $\bigcap_{i=1}^k H_i= x_* + \ker(N)$.
However, I agree with the fact that $x_* + \ker(N) \subseteq \bigcap_{i=1}^k H_i$ but I don't see why $\bigcap_{i=1}^k H_i \subseteq x_* + \ker(N)$.
Maybe I can answer my own question by saying that for $x \in \bigcap_{i=1}^k H_i$, we have that $(x-x_*)\cdot N=0 \implies x-x_* \in \ker(N) \implies x \in x_* + \ker(N)$