# Are vectors assumed to be with respect to a standard basis?

1. Aug 26, 2013

### chipotleaway

For example, if were given only a vector <5, 3, 1>, is this assumed to be respect with the standard basis of R^3?

And would this mean that any nonstandard basis is with respect to a standard basis?

2. Aug 26, 2013

### Simon Bridge

You are free to expand a vector against any basis set you like.
For your example, any set of three vectors that form a basis for R^3 is fine.

There is a standard representation though - from the rules for vector spaces.

3. Aug 26, 2013

### Office_Shredder

Staff Emeritus
The vector (5,3,1) is defined without reference to any basis at all. R3 is lists of three real numbers, and (5,3,1) is a list of three real numbers. It turns out that this is a vector space and that it has a basis for which (5,3,1) = 5e1 + 3e2 + 1e3 but that's totally incidental to the definition of (5,3,1)

4. Aug 26, 2013

### chipotleaway

So (5,3,1) with no reference to a basis is NOT a vector? And the lists of R^3 happen to be vectors because R^3 satisfy the properties of a vector space?

5. Aug 26, 2013

### economicsnerd

There's no definition of what is/isn't a vector. There's a definition of a vector space, and given a vector space we call its elements "vectors".

It's exactly as Office_Shredder put it:
- We define the set $\mathbb R^3$ as the set of ordered triples of real numbers.
- For any real number $c \in \mathbb R$ and any ordered triple $x\in \mathbb R^3$, we define what we mean by $cx \in \mathbb R^3$.
- For any two ordered triples $x,y\in \mathbb R^3$, we define what we mean by $x+y \in \mathbb R^3$.
Then it turns out that these things together satisfies what we call a vector space.

All of this makes sense without needing to define the word "vector".

6. Aug 26, 2013

### WannabeNewton

A vector exists independent of any basis you choose to represent it in. If $V$ is a vector space then $v\in V$ is a vector, that's all. You can choose a basis $(e_i)$ for $V$ and represent $v$ as $v = \sum v_i e_i$ and further express $v$ relative to this basis in coordinate form $[v]_{(e_i)} = (x_1,...,x_n)$ by passing over to $K^n$ where $K$ is the field that $V$ is over.

7. Aug 26, 2013

### Simon Bridge

I'm wondering if this question is more about notation than basis sets.

if $\small \{\vec{u},\vec{v},\vec{w}\}$ is an arbitrary basis for $\small \mathbb{R}^3$, then is $\small <x,y,z>=x\vec{u}+y\vec{v}+z\vec{w}$ ?

I don't know if I'm putting this well.

If you see <5,3,1> written down, with no other information but that it is a vector, it is safe to expand it as 5<1,0,0>+3<0,1,0>+<0,0,1>.

Another basis, written in terms of the standard one, may be {<1,1,0>,<1,-1,0>,<0,0,1>} ... we would not expect <5,3,1> to mean 5<1,1,0>+3<1,-1,0>+<0,0,1> unless we are told that it does.

However, three numbers written like that could be a list of data, or a spherical-polar position, or just a row of text characters - depending on context. If <5,3,1> is the only thing on the page, you have no information.

8. Aug 27, 2013

### chipotleaway

For a bit of context, this arised from an example in Anton's linear algebra text; in the first part he showed how to find the coordinate vector of a given vector (I'm gonna call it <5, 3, 1>) in terms of a given nonstandard basis S.
Next part, find a vector in R^3 whose coordinate vector with respect to S is <3, 2, 4>.

So a list of numbers in R^3, (5, 3, 1) exists independent of a basis but happens to have that same representation with respect to the standard basis of R^3 (and also happens to be a vector?)

9. Aug 27, 2013

### Office_Shredder

Staff Emeritus
This is correct.

10. Aug 28, 2013

### Stephen Tashi

11. Aug 29, 2013

### Simon Bridge

Yep - I wondered if this could have been the motivation of the question.