0
$\begingroup$

I am having a hard time understanding the definitions of dimensions. Dimension of a finite-dimensional vector space is defined as the length of any basis of the vector space. The definition seems to be easy to understand. But I need examples to have a better understanding of dimensions so if anyone can explain me more about dimensions, it would much appreciated!

Thanks!

$\endgroup$
2
$\begingroup$

Dimension is actually quite a difficult notion to make rigorous -- we want to have a down-and-dirty definition, but it usually turns out that in a sufficiently general context,mathematical objects behave unexpectedly.

When dealing with real numbers, it seems natural to define the dimension of the unit interval to be 1, and the image under any continuous map should somehow "shrink" the dimension. (For instance, if the interval was mapped to a point.) However, Peano showed that there exist "space-filling-curves" that map $[0,1]$ continuously onto the unit box $[0,1]^2$. Weirder still, there are spaces (eg fractals) who only seem to have a correctly defined dimension (called the Hausdorff dimension) when we consider arbitrary real numbers (as opposed to integer values). The key fact here is that it's not the size of the set that matters, but it depends on how it sits in Euclidean space.

In the world of algebra, functions aren't quite as free as in analysis, so you actually can define dimension in a way that aligns with intuition quite nicely, although the definition is weird. There's the Krull dimension, defined for any ring (and doesn't make sense until you see the geometry).

So all in all, it's a difficult notion to define and this answer (originally a comment) is just to give you a few words that are google-able.

$\endgroup$
1
$\begingroup$

From a geometry perspective, we can look at $\mathbb{R}, \mathbb{R}^2, \mathbb{R}^3$. A basis is any minimal set of vectors in which "directions" on how to arrive at any point in the space is possible.

$$\mathbb{R} = \text{span}\{1\}$$

A one-dimensional space is a line. Think of this as being able to move only forward and backward. The notions of left-right, up-down are meaningless here. Any point on this line is a multiple of anything else that is nonzero, it's simplest to just use 1. To get to $5$, we can go along the line a total of $5$ $1's$. Having any additional vectors in our span is redundant, any fewer and we couldn't go anywhere.

$$\mathbb{R}^2 = \text{span}\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right\}$$

A two dimensional space is a plane. The notions of forward and backward or left and right are needed to get anywhere in this space. To get to any point in this space, we need two vectors. For example, to get to the point $ \begin{pmatrix}7\\-4\end{pmatrix}$ we need to go along the first vector $7$ times, and along the second vector, backwards 4 times. Having only one vector in our set we can only move along the direction of that vector. Having any more would again be redundant, since these two are enough.

$$\mathbb{R}^3 = \text{span}\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}$$

This is usually called $3-$space. We can think of this space as actually being the space we live in. We have up and down, left and right, forward and backward. So three vectors will be needed to describe coordinates here. To get to any point in this space, we need three vectors. To get to the point $ \begin{pmatrix}1\\-2\\1\end{pmatrix}$ we need to go along the direction of the first vector $1$ time, and along the second vector backwards 2 times, and along the third vector once. Having only one or two vectors in our set we can only move along the plane spanned by them. For example, I couldn't get to $\begin{pmatrix}0\\0\\2\end{pmatrix}$ if the third vector was deleted from our set.

The bases I gave above are called the Standard Bases for the respective vector spaces. Most you encounter will have a defined 'standard' basis. It's just the simplest to use since, for example in $\mathbb{R}^2,$

$$\begin{pmatrix}a\\b\end{pmatrix} = a\begin{pmatrix}1\\0\end{pmatrix}+b\begin{pmatrix}0\\1\end{pmatrix}$$

Other bases are possible, such as $$\mathbb{R}^2 = \text{span}\left\{\begin{pmatrix}3\\0\end{pmatrix},\begin{pmatrix}0\\2\end{pmatrix}\right\} = \text{span}\left\{\begin{pmatrix}1\\1\end{pmatrix},\begin{pmatrix}1\\-1\end{pmatrix}\right\}$$

Although 'direction giving' can be a little more complicated here, its still possible. As long as the set can describe (using linear combinations) any coordinate in the space and there are no redundant vectors included, the set forms a basis. Although

$$\mathbb{R}^2 =\text{span}\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}, \begin{pmatrix}1\\1\end{pmatrix}\right\}$$

is true, we have more vectors than necessary. The third is redundant, since the directions to get to (1,1) are possible with just the first two. This is not a basis. For vector spaces, bases have a unique size, and that is the definition of dimension.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.