Consider $L:=\left\{x\in\mathbb{R}^k: cx=\delta\right\}$ with $c=(c_1,\ldots,c_k)$ and $\delta\in\mathbb{R}$. Show that $\text{dim}(L)=k-1$.
Do not know how to show that. Anyhow my first idea was to pick any $j\in\left\{1,...,k\right\}$ and then $$ x_j=\frac{1}{c_j}\sum_{i\neq j}c_ix_i, $$ so that for any $x=(x_1,\ldots,x_k)^t\in L$ it is $$ x=x_1\cdot\begin{pmatrix}1\\\vdots\\c_j^{-1}c_1\\0\\\vdots\\ 0\end{pmatrix}+\ldots+x_{j-1}\begin{pmatrix}0\\\vdots\\1\\c_j^{-1}c_{j-1}\\0\\\vdots\\0\end{pmatrix}+x_{j+1}\begin{pmatrix}0\\\vdots\\c_j^{-1}c_{j+1}\\1\\\vdots\\0\end{pmatrix}+\ldots+x_k\cdot\begin{pmatrix}0\\\vdots\\c_j^{-1}c_k\\0\\\vdots\\1\end{pmatrix}. $$
So it is $\text{dim}(L)=k-1$.
But I am not sure if this is correct, because from what do I know that $c_j\neq 0$?