# With the standard basis, which coordinate system do we mean?

In linear algebra we use the standard basis $\{\hat e_1,\hat e_2,\dots ,\hat e_n\}$.

However, I have never grasped which coordinate system we mean when talking about the standard basis.

With the standard basis, do we actually mean the cartesian coordinate system?

## 4 Answers

Let's get this ironed out a bit. Vector spaces in general do not have standard bases. They have vectors, but one needs to choose a basis in the vector space to express these vectors numerically, just like one needs to choose units in order to express physical quantities numerically. And like for choosing units, it is a somewhat arbitrary affair initially; however once it is done everything can be expressed numerically in a unique manner.

What this means in case of a (finite dimensional real) vector space$~V$ is that after choosing a basis$~[b_1,\ldots,b_n]$, every vector of$~V$ corresponds to an $n$-tuple of numbers, its coordinates with respect to that basis, which $n$-tuple can be seen as an element of the set $\Bbb R^n$. So choosing a basis leads to a correspondence between vectors and elements of $\Bbb R^n$ (and of course choosing a different basis would have led to a different correspondence). The vector space operations (addition, scalar multiplication) of $V$ then correspond to similar operations on the set $\Bbb R^n$. It turns out that independently of what $V$ actually was (as long as it has dimension$~n$), and independently of which basis was chosen for it, one always gets the same corresponding operations in $\Bbb R^n$ (namely coordinate-by-coordinate addition, respectively multiplication of all coordinates by the same factor). This means that (maybe somewhat confusingly) $\Bbb R^n$ itself is a vector space.

Now many introductory courses in linear algebra skip this introduction, and introduce vectors as being elements of $\Bbb R^n$ to begin with. Or maybe they don't, and people just forget about how this was introduced, just like in primary school we learn to add and multiply numbers while forgetting the apples, pumpkins, dollars or whatever were used to introduce us to the world of numbers in the first place. Anyway, vectors correspond to elements of $\Bbb R^n$, and that is where computations usually take place.

Now at some point one needs to get more precise about what effect choosing a different basis has (for instance because one wants to consider choosing a basis well adapted to a particular problem); the notion of basis comes to haunt us again, although we thought we could forget about it once we used a basis to get our correspondence of $V$ with $\Bbb R^n$, and to do our computations there. So consider a basis$~[v_1,\ldots,v_n]$ possibly different from the one we used to make that correspondence of $V$ with $\Bbb R^n$; passing these vectors through our correspondence we get a list$~[c_1,\ldots,c_n]$ of $n$ elements of $\Bbb R^n$ that form a basis of$~\Bbb R^n$. The coordinates of a vector $v\in V$ with respect to$~[v_1,\ldots,v_n]$ are the same as the coordinates of the corresponding element$~c$ of $\Bbb R^n$ with respect to$~[c_1,\ldots,c_n]$. But by definition of our correspondence $c\in\Bbb R^n$ consists of the coordinates of $v$ with respect to out original basis $[b_1,\ldots,b_n]$. So the problem of converting the coordinates of $v$ with respect to$[b_1,\ldots,b_n]$ into its coordinates with respect to$~[v_1,\ldots,v_n]$ is the same as finding the coordinates of$~c$ with respect to the basis $[c_1,\ldots,c_n]$ of$~\Bbb R^n$.

This conversion is a somewhat involved computation that I will not detail here. However, one thing is obvious: in the case that $[v_1,\ldots,v_n]$ is identical to$~[b_1,\ldots,b_n]$ the coordinates are the same in both cases, so the process of converting the coordinates amounts to returning them unchanged. This means that in this particular case any $n$-tuple $c\in\Bbb R^n$ is equal to its own coordinates with respect to the basis $[c_1,\ldots,c_n]$ of$~\Bbb R^n$. So $[c_1,\ldots,c_n]$ is a very special basis of$~\Bbb R^n$ indeed, and indeed the unique basis of $\Bbb R^n$ with this property. It is called the standard basis of $\Bbb R^n$; its elements are usually called $e_j$ rather than $c_j$, and they are explicitly given as the columns of the identity matrix.

So this is what a standard basis is: it only exists for the very particular vector spaces $\Bbb R^n$, and it allows us to pretend that the components of an arbitrary $n$-tuple of numbers are coordinates of that $n$-tuple itself with respect to some basis (namely the standard basis). It is a bit like pretending that to measure numbers in $\Bbb R$ we need a unit of measure, and $1$ is the unique unit of measure that will make any number come out to be itself after measuring (which is less confusing than having it come out as a different number).

The coordinate system comes from the basis. What it means for a list of vectors $(v_1, \ldots v_n)$ to be a basis for $V$ over $\mathbb{F}$, is that each vector $v \in V$ can be expressed as $$v = \alpha_1 v_1 + \ldots + \alpha_n v_n$$ for unique $\alpha_1, \ldots, \alpha_n \in \mathbb{F}$. This means that, for each vector, there's a bijective mapping between the vectors in $V$ and a corresponding vector $(\alpha_1, \ldots, \alpha_n) \in \mathbb{F}^n$. This bijective mapping is the coordinate system, and naturally, it depends entirely on the choice of basis $(v_1, \ldots, v_n)$.

For example, the vector space $P_2(\mathbb{C})$, has a basis $(1, x, x^2)$. The coordinate mapping takes an arbitrary polynomial of degree $2$ or less $ax^2 + bx + c$, and expresses it as $(c, b, a) \in \mathbb{C}^3$, since $$ax^2 + bx + c = c(1) + b(x) + a(x^2)$$ is the unique linear combination of $(1, x, x^2)$ that produces $ax^2 + bx + c$.

In the case where the vector space is already $\mathbb{F}^n$, the vectors we're dealing with are already in $\mathbb{F}^n$, so this opens up the possibility that the coordinate mapping could be the identity function. That is, given the right basis, every vector could be its own coordinate vector. The basis that does this is the standard basis. It's this property that makes the standard basis so... standard.

As with all the bases, the mapping from a vector to its coordinate vector will be linear, in the sense that linear combinations of vectors will produce the corresponding linear combination of coordinate vectors. You can't find a basis that will, say, produce polar coordinates! So, if you like, it'll always be akin to Cartesian coordinates, in a sense.

It depends on the vector space you're working on. For example, if the vector space is the Euclidean Space $\mathbb{R^n}$, then the standard basis would be $$e_i = (0,0,...1,...,0)$$ where all positions except the $i$th one are zero.

Similarly, if you're working with $n$ degree polynomials, the standard basis would be $1$, $x$, $x^2$,..., $x^n$.

Basically yes, we do mean the cartesian coordinate system. However keep in mind that there is no canoncial basis of a vector space. There is always a choice involved. What this means is just, that you write your basis vectors as $$e_i=(0,\ldots, 1,\ldots,0)$$ where the single $1$ is of course at the $i$-th position. However without choosing a basis you would not be able to write a vector $$v=(v_1,\ldots,v_n)$$ because the $v_i$ depend on your choice of basis.

• No, the vector spaces $\Bbb R^n$ (and only such vector spaces) do have standard bases. And there you can write $v=(v_1,\ldots,v_n)$ without any choice, because that is what vectors in $\Bbb R^n$ are. – Marc van Leeuwen Jul 7 '17 at 18:29
• Polynomial rings $k[X]$ have a standard basis as well. But thanks for your clarification. My "basically yes" was supposed to answer the question about the cartesian coordinate system. I see that this might be understood differently. – Rene Recktenwald Jul 10 '17 at 11:00
• @MarcvanLeeuwen But $v=(v_1,...,v_n)$ means nothing more than $v=\sum v_ie_i$, and it depends on the choice of basis. – Tomasz Tarka Jul 10 '17 at 11:19
• @TomaszTarka: No, that is like saying that $10.5$ means nothing more that $10.5\times 1$; while true, if you take it as definition, you see that you wind up in a recursive knot. Try to explain what $e_2=(0,1,0)$ means (according to you), and I hope you can see what I mean. – Marc van Leeuwen Jul 10 '17 at 14:01
• writing $(1,0,0)\in \mathbb{R}^3$ still depends on the choice of the basis. If you don't mention anything it is understood to be relative to the standard basis. But it is not canonical. I think you are talking past each other – Rene Recktenwald Jul 11 '17 at 8:05