Skip to main content

[Community Question] Linear-algebra: Is $\mathbb{R}^n$ a vector space or a metric space?

One of our user asked:

In my various courses, for instance, linear algebra and vector calculus, I am somewhat confused with what precisely $\mathbb{R}^n$ is.

From the definition of the Cartesian product, I would conceptualise $\mathbb{R}^n$ as the metric space with some distance operator, where all the points are just $n$-tuples. This is surely a distinct notion from vectors as isn't the point $A = (1,2,3)$, for instance, different from the vector $\vec{a} =\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}$ ? But if we were to consider the points in $\mathbb{R}^n$ as vectors then clearly it is a vector space. However I don't know whether these two conceptions of $\mathbb{R}^n$ are actually equivalent. Surely the vectors do not correspond to a specific point in space, unlike the points in $\mathbb{R}^n$.

Forgive me if this is a silly question, or if my question seems garbled. Also please help me with tags if they are inappropriate.


Comments

Popular posts from this blog

[Community Question] Calculus: Manifold with boundary - finding the boundary

One of our user asked: I have the manifold with boundary $M:= \lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1\geq 0, x_1^2+x_2^2+x_3^2=1\rbrace \cup\lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2\leq1\rbrace$ and I need to find the boundary of this manifold. I think it is $\lbrace (x_1,x_2,x_3) \in \mathbb R^n : x_1= 0, x_2^2+x_3^2=1\rbrace$ , the other option is that the boundary is the empty set? I think the first is right? Am I wrong?

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).

Order of elements of the Prüfer groups $\mathbb{Z}(p^{\infty})$

Let $\mathbb{Z}(p^{\infty})$ be defined by $\mathbb{Z}(p^{\infty}) = \{ \overline{a/b} \in \mathbb{Q}/ \mathbb{Z} / a,b \in \mathbb{Z}, b=p^i$ $ with$ $ i \in \mathbb{N} \}$ , I wish show that any element in $\mathbb{Z}(p^{\infty})$ has order $p^n$ with $n \in \mathbb{N}$ . i try several ways but I have not been successful, some help ?? thank you