Is the set $S = \{\vec{x} \in \mathbb{R}^n : \|\vec{x}\| = |x_1|\}$ a subspace of $\mathbb{R}^n$?

Question

I'm pretty sure it's not closed under vector addition but verification would be appreciated.

Answer 1

Notice $$\sqrt{x_1^2+x_2^2+\cdots+x_n^2}=|x_1|\implies x_1^2+x_2^2+x_3^2+\cdots+x_n^2=|x_1|^2.$$ But $x_1^2=|x_1|^2$, so subtracting gives $x_2^2+x_3^2+\cdots+x_n^2=0$. This is only possible if all but the first component of the vector is zero (can you see why?). Also, $r \vec{e}_1\in S$ for any real $r\in\mathbb{R}$ and $\vec{e}_1$ the unit vector on the $x_1$-axis (check this!), so we may conclude $$S=\{(r,0,0,\dots,0)\in\mathbb{R}^n:r\in\mathbb{R}\}.$$ Can you tell whether or not this is a vector space in its own right?