# We know that the the norm of a vector is norm(vec(u))=sqrt(u_1^2+u_2^2+...+u_n^2), where vec(u) in RR^n. How does the d xx 1 vector look like using the definition of norm, and what does RR^n really means? Or what is Rn for the d xx 1 vector?

We know that the the norm of a vector is $‖\stackrel{\to }{u}‖=\sqrt{{u}_{1}^{2}+{u}_{2}^{2}+...+{u}_{n}^{2}}$, where $\stackrel{\to }{u}\in {\mathbb{R}}^{n}$. How does the d x 1 vector look like using the definition of norm, and what does ${\mathbb{R}}^{n}$ really means? Or what is ${\mathbb{R}}^{n}$ for the d x 1 vector?
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Jacey Humphrey
It is on the unit sphere.
It might help to work with lower specific number of n, say $n=1,2,3$
If $n=1$, then $u=1$ or $u=-1$
If $n=2$, then it lies on the unit circle, ${u}_{1}^{2}+{u}_{2}^{2}=1$
If $n=3$, then it lies on the unit circle, ${u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}=1$
${\mathbb{R}}^{2}$ means $\left\{\left(x,y{\right)}^{T}|x,y\in \mathbb{R}\right\}$, similary for general ${\mathbb{R}}^{d}$