The dimension of a vector space is the number of elements in the basis for that vector space.

If we look at $\mathbb R^n$, then we say that the dimension of $\mathbb R^n$ is $n$. So every element in $\mathbb R^n$ can be expressed as a linear combination of these $n$ elements in the basis.

We further say that the elements of $\mathbb R^n$ are $n$-tuples.

My question:

I am slightly confused as to what is the relation between the $n$ in the dimension of $\mathbb R^n$ and the $n$ in the $n$-tuple.

The $n$ dimension gaurantees that the elements of $\mathbb R^n$ will be $n$-tuples, right. However if we have $n$-tuples as elements of some arbitrary vector space then we cannot say that the dimension of that vector space is $n$.

Am i correct in this regard.

  • $\begingroup$ I think you might avoid some confusion if you at least mentally stated first that $\mathbb{R}^n$ consists of $n$-tuples of real numbers. By definition $\mathbb{R}^n$ is the set of all $n$-tuples of real numbers. To say this you would not need to know what a vectorspace even is. Then you realize it is a vector space (with the usual laws) and then you observe this space has dimension $n$. $\endgroup$ – quid Sep 3 '15 at 16:31

If $\mathcal V$ is some field, then $\mathcal V^n$ is a vector field over $\mathcal V$ and it will always have dimension $n$, because $$\{(1,0,\dots,0), (0,1,\dots,0), \dots, (0,0\dots, 1)\}$$ is always a basis for $\mathcal V^n$.

However, you are right that just because a vector space consists of $n$-tuples, it may not be a $n$-dimensional vector space. For example, the space

$$X = \{(x, 0)| x\in\mathcal R\}$$ is a one dimensional vector space over $\mathbb R$.

  • $\begingroup$ Can we say If a vector space consists of $n$-tuples then it might be a proper subspace of an $n$ dimensional vector space, if not $n$ dimensional itself . Actually from your example , this occurred to me. $\endgroup$ – user118494 Sep 3 '15 at 16:26
  • 1
    $\begingroup$ @user118494 you need to say $n$-tuples of what? If $n$-tuples of real number (and you keep the usual operation of addition and multiplication) then yes a vectorspace whose elements are $n$-tuples will be a supspace of $\mathbb{R}^n$ and thus have dimension at most $n$. $\endgroup$ – quid Sep 3 '15 at 16:29

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