# What exactly is standard basis?

I am confused about the difference between coordinates and basis. My confusion is following:

Let $e_i$ denote the standard basis and $v_i$ denote a non-standard basis of a finite $n$-dimensional vector space $V$. Then $e_i = (\delta_{ij})$ (all entries zero except $i$-th). But: the coordinates of $v_i$ with respect to basis $v_i$ also are $(\delta_{ij})$. Now the definition of standard basis becomes circular: it is supposed to be the basis with vectors $(\delta_{ij})$ but every basis vector has these coordinates with respect to itself.

So what exactly is standard basis?

• The coordinates of $v_i$ w.r.t. $e_i$ are not necesserily equal to $\delta_{ij}$. Maybe w.r.t. $v_i$? Aug 14, 2013 at 8:31
• @MarkusMayr Yes! You are right! I corrected it! Thank you!
– blue
Aug 14, 2013 at 8:44
• Vector spaces admit many bases in general. But there are many specific families of vector spaces that come equipped with one $-$ a "standard" one. E.g. $K^n$ or $K[x]$.
– anon
Aug 14, 2013 at 8:44

As other answers have already pointed out, the notion of "standard basis" only applies to the very particular $K$-vector spaces $K^n$, whose elements are $n$-tuples of elements of$~K$ (scalars). I've emphasised "are", because in any $n$-dimension $K$-vector space you can represent vectors by $n$-tuples of scalars (after having chosen an ordered basis; the scalars are the coordinates of these vectors in this basis); however in general a vector and its $n$-tuple of coordinates remain two different things.

You know what the standard basis of $K^n$ is (and it is actually an ordered basis: more than just a set of vectors, it is a list where each basis vector has its own place). The one point I would like to add is mention the property that makes this particular basis stand out among other bases, either of $K^n$ or of other spaces.

Property. Any $v\in K^n$ is equal to the $n$-tuple of coordinates of $v$ with respect to the standard basis of$~K^n$.

Clearly for such a property to hold, it is necessary that such vectors $v$ be $n$-tuples of scalars in the first place, in other words that the vector space in question be $K^n$. Moreover, the coordinates $(c_1,\ldots,c_n)$ of $v$ with respect to an ordered basis $(\mathbf e_1,\ldots,\mathbf e_n)$ are by definition the scalars such that $v=c_1\mathbf e_1+\cdots+c_n\mathbf e_n$ holds; if (with $v\in K^n$) one requires that moreover $v=(c_1,\ldots,c_n)$ as stated in the property, this requires that $$(c_1,\ldots,c_n)=c_1\mathbf e_1+\cdots+c_n\mathbf e_n.$$ This should hold for all possible values of $c_1,\ldots,c_n\in K$. In particular one can take some $c_i=1$ and all other $c_j$ zero, and this leads to the conclusion that $\mathbf e_i\in K^n$ has to be $(0,\ldots,0,1,0,\ldots,0)$ with the nonzero component at position$~i$; this should be so for every$~i$. Once these cases are checked, all others follow by linearity. So the stated property holds for the standard basis, and also characterises the standard basis of$~K^n$.

• Why did my lecturer not take the time to explain this? Thanks for this. It is actually quite subtle. Nov 20, 2015 at 3:43
• @user85798 It is indeed very subtle, and while a good starting point, this answer doesn't cover everything Oct 27, 2021 at 13:49
• @étale-cohomology Thanks for the praise. I'm looking forward to your answer covering the points I missed. Oct 31, 2021 at 5:09
• I didn't meat to irritate you, I'm sorry, I learnt a good deal from your answer and I can't improve on it Oct 31, 2021 at 10:08
• @GaurangAgrawal: Yes $K$ is supposed to be a field when one is talking about $K$-vector spaces. But there is no such thing as "the coordinates" (presumably living in $K$) in a general $K$-vector space (instead the elements of $K$ are used when forming linear combinations). And $K^n$ is the space whose elements are $n$-tuples of elements of $K$ (i.e., of scalars), and as my answer says it it this (and only this) $K$-vector space that has a standard basis. What is more general (than you might think) is not the notion of scalar, but the notion of vector space. Jan 8 at 15:32

The term standard basis only applies to vector spaces of the form $\Bbb F^n$, when every vector is of the form $(x_1, x_2, ..., x_n)^T$. You then stipulate $e_i := (0, ..., 0, 1, 0, ...,0)^T$ ($1$ in the i$^{th}$ place), which is independent of a choice of basis. For arbitrary vector spaces it doesn't really make sense to talk about a standard bases.

Edit: If $V$ is an $n$-dimensional vector space, then the coordinate vector $[v]_B$ of $v \in V$ w.r.t. a basis $B = \{v_1, ..., v_n\}$ of $V$ is defined as follows: Let $f: B \to \Bbb F^n$ be the map $v_i \mapsto e_i$. Extending $f$ to a linear map $V \to \Bbb F^n$, we define $[v]_B := f(v)$. Of course, by this definition $[v_i]_B = e_i$, but that does not mean $v_i = e_i$.

• I thought I understood but then I thought about it again and now I don't: Isn't $(0,...0,1,0,...0)$ the coordinates of a basis vector $b_i$ with respect to the basis $\{b_i\}_i$? Maybe the meaning of stipulate is not too clear to me.
– blue
Aug 14, 2013 at 12:42
• @blue: $(0, \ldots, 0, 1, 0, \ldots, 0)$ is the coordinate vector of $b_i$ in the basis $\{b_i\}$, you're right about that. But for the standard basis $\{e_i\}$ of $\mathbb{F}^n$, it is not only the coordinate vector of $e_i$ in the basis $\{e_i\}$ that is equal to $(0, \ldots, 0, 1, 0, \ldots, 0)$, even $e_i$ itself is equal to $(0, \ldots, 0, 1, 0, \ldots, 0)$. Aug 14, 2013 at 15:25

"A finite $n$-dimensional vector space $V$" doesn't have a standard basis. The specific $n$-dimensional vector space $\mathbb{R}^n$ does have a standard basis: the one whose vectors are the $n$-tuples that have zero entries except for a single one.

There are two senses of "coordinates". One is defined with respect to a basis on a general $n$-dimensional vector space. The other refers to ordered tuplets of numbers $(a_1,\ldots,a_n)\in \mathbb R^n$. (Or whatever your base field is.) The standard basis is the unique basis on $\mathbb R^n$ for which these two kinds of coordinates are the same.

Edit: Other concrete vector spaces, such as the space of polynomials with degree $\leq n$, can also have a basis that is so canonical that it's called the standard basis. In this case, the standard basis vectors are $1, x, x^2, \ldots, x^n$.

I'm not sure that it's common to speak of a "standard basis" on a completely abstract $n$-dimensional vector space. If you have a textbook that does so, it would help to quote the context.

• Just a note: sometimes the basis ${1,x,x^2,...,x^n}$ of the space of polynomials with degree $\leq n$ is called the "standard" basis. Aug 14, 2013 at 8:37
• @pppqqq Excellent point, I'll edit my answer. Aug 14, 2013 at 8:37
• I understand now I think! Standard basis only has meaning in a concrete example of vector space! Thank you!!
– blue
Aug 14, 2013 at 8:42
• @blue: To emphasize, "most" concrete vector spaces do not have a "standard basis". Think of a line through the origin in $\mathbf{R}^2$, for example. As others note, the term "standard basis" really refers to very specific spaces, particularly $\mathbf{R}^n$ with its standard vector space operations. Aug 14, 2013 at 10:14
• @AndrewD.Hwang After thinking about it some more I realize I don't understand after all. The expression $(0,0,....,0,1,0,...0)$ seems to be coordinates. In particular $(1,0,0,....)$ is the coordinates of a basis vector $b_1$ with respect to basis $\{b_i\}_i$. But what are standard basis vectors then?
– blue
Aug 14, 2013 at 12:45

The standard basis arises when we identify a finite-dimensioned vector space $V$ with $\Bbb R^n$. We take any basis in $V$, say, $\vec v_1,\dots,\vec v_n$. Then we can say that any vector $\vec w\in V$ is a linear combination of basis vectors: $$\vec w=\sum_{j=1}^n\alpha_{j}\vec v_j.$$ In other words, instead of saying "a vector $\vec w$" we can say "a vector whose coordinates are $(\alpha_1,\dots,\alpha_n)$", which is clearly can be seen as an element in $\Bbb R^n$. Then, in $\mathbb R^n$ we introduce a standard basis.

• I am sorry, it was a typo in the question. I wanted to write with respect to $v_i$.
– blue
Aug 14, 2013 at 8:43