# I have a question about the definition of a vector

1. Jan 3, 2016

### Chump

Why is it that a vector can be described in terms of a simple linear combination, like v = xi + yj + zk, where v is a vector, and i, j, and k are all unit vectors. It just seems a bit convenient that it's as simple as just adding the components this way. It seems like there should be a bit more to it. I guess I'm just looking to get as fundamental with the concept as possible. Is it just a matter of this being the definition of a vector by convention?

2. Jan 3, 2016

### jfizzix

The most general (mathematical) definition of a vector is any element of a vector space.
A vector space is a set of objects satisfying some simple properties, such as closure under addition and scalar multiplication (there are more, which can be found in any linear algebra textbook, or Wikipedia)
This definition of a vector is so broad as to include objects like matrices, tensors, and even polynomial curves.

Adding the components is as simple as you say because the unit vectors are chosen to be orthogonal to one another, so that no one vector can be expressed as a linear combination of the others. Indeed, this is not the only way to describe a vector $\vec{v}$. We can use any number of vectors we like, so long as adding them tip to tail puts us at the same position, but it is a lot easier to use an orthogonal set of unit vectors, because $\vec{v}$ has a unique expression for each set of such vectors.

3. Jan 3, 2016

### andrewkirk

It's not clear what you're asking. Why should it be surprising that a vector can be written as a sum of other vectors? After all, any number can be written as a sum of other numbers, and vectors are in a sense just a generalisation of the notion of number. What is it about the decomposition v = xi + yj + zk that you find surprising?

4. Jan 4, 2016

### Samy_A

You are right, there is more to it.

As @jfizzix said, in general a vector is by definition an element of a Vector space.
In the study of vector spaces, one discovers that some vector spaces are finite dimensional. The example you gave concerns a vector space with dimension 3. That means that there exists a (non-unique) set of just 3 vectors, that has the property that every vector v can be written as v = xi + yj + zk, where i, j, and k are vectors, x, y and z are scalars. In the case of $\mathbb R³$ (as in other finite dimensional vector spaces), where we have the notions of orthogonality and length of vectors, i, j and k can even be chosen to be mutually orthogonal units vectors.
That is indeed convenient, but it is not true for all vector spaces.

For example, the vector space of all real polynomials is not finite dimensional.

Last edited: Jan 4, 2016
5. Jan 4, 2016

### andrewkirk

Although finite dimensionality of a space may be a sufficient condition for the property that any vector in the space can be written as a finite linear combination of unit vectors, I don't think it's a necessary condition. Any vector $\vec v$ in any normed vector space (ie Banach Space) can trivially be written that way as
$$\vec v=\|\vec v\|\left(\frac{\vec v}{\|\vec v\|}\right)$$
because the item inside the parentheses is a unit vector.

For a vector space of real polynomials, if the domain of $x$ is compact, such as a finite interval $[a,b]$, the real polynomials on that domain are a subspace of the Banach Space $L^2([a,b])$, and hence any such polynomial can be thus expressed. Further, although the space is infinite-dimensional, it has a countable basis $\{x^n\ |\ n\in\mathbb{N}\}$, which can be converted to a countable normalized basis by dividing each element by its norm. Any polynomial can be written as a finite sum of elements of that normalized basis.

I suspect (but am not sure) that what the OP may have been getting at was the property that, for any normalised basis of the vector space, any vector in the space can be written as a finite linear combination of elements of that basis. For that property, infinite-dimensionality would be a problem and I'm pretty sure the property would not hold for all normalized bases of all vector spaces.

6. Jan 4, 2016

### Samy_A

Ok.

I meant that only in a finite dimensional vector space you can write every vector as a linear combination of the same n (=dimension of the vector space) basis vectors.

7. Jan 4, 2016

### lavinia

$i$, $j$, and $k$ form what is called a basis for the vectors in the space. This means that any vector can be written uniquely as a linear combination of them. and further, one can not remove one of them and still be able to express every vector. For instance if you remove $i$ then you are left with only $j$ and $k$ and these can never combine to give you a vector with non-zero $i$ coordinate.

But there are many other bases for the vectors in 3 space. For instance the 3 vectors $v = i + j$, $w = i - j$ and $q = i + j + k$ also form a basis. Any vector in 3 space can be written as a unique linear combination of them. To prove this note that $i = 1/2(v+w)$, $j = 1/2(v - w)$ and $k = q - v$. And again, if you remove any one of them, then you can not write any vector as a linear combination of the remaining two. For instance, if you remove, $v$ then you cannot express $v$ as a linear combination of $w$ and $q$.

If you write a vector as a linear combination $av + bw + cq$ then $a$, $b$, and $c$ are the coordinates of the vector with respect to the basis $v$,$w$, $q$. If you choose a different basis then the coordinates will be different. For instance, in the $i$, $j$, $k$ basis of orthogonal unit vectors, the coordinates of $i$ are $a = 1$, $b=0$, $c = 0$. In the $v$,$w$,$q$ basis the coordinates of $i$ are $a = 1/2$,$b= 1/2$, $c = 0$.

Any basis for 3 space must contain three vectors and no one of them can lie in the plane spanned by the other two. They must be "linearly independent" and the dimension of 3 space is three because any basis requires 3 linearlly independent vectors. You do not need more than three and you can not do with less.

So $i$ $j$ and $k$ are just a convenient example of a basis for a three dimensional vector space. But in many applications, other bases are used. For instance, if one considers the linear transformation, $i \rightarrow j$,$j \rightarrow i$, $k \rightarrow k$ then its eigenvectors form the basis, $v = i + j$ , $w = i -j$, and $q = k$.

As one can see, 3 space exists as a vector space without reference to any particular basis. A basis is just a linearly independent set that spans the vector space through linear combinations. For instance, one can think of the vectors in 3 space as arrows pointing from the origin and vector addition as parallel translating the second arrow so that its base touches the end point of the first. One multiplies an arrow by a number by scaling its length. This definition is basis free. It does not require $i$,$j$, and $k$ or any other basis. In fact, one needs to prove that this vector space even has a basis and that its dimension(the number of vectors required for a basis) is three.

If a vector space has a basis with a finite number of vectors in it, then its said to be finite dimensional and its dimension is the number of vectors in the basis. (This only makes sense after you prove that any basis must have the same number of vectors in it.) So one defines finite dimension as the span of finitely many vectors and then one must prove as a theorem that there is a basis and any two bases have the same number of vectors in them.

But there are some vector spaces that are not finite dimensional. This means that there is no finite linearly independent set of vectors that spans the whole space. For example the vector space, $R^{∞}$ of vectors, $(x_1, x_2, ...,x_n, ...) with infinitely many coordinates is infinite dimensional. (Just as in 3 space, vector addition is defined coordinate wise and scalar multiplication is, " Multiply each coordinate by the scalar".). For an infinite dimensional vector space one must again to ask whether it has a basis and even to ask what a basis means in infinite dimensions. The definition of basis is a set of vectors that span the whole space and which is linearly independent. Span here means that any vector can be expressed as a linear combination of finitely many vectors in the basis. Linearly independent means that no vector in the basis can be expressed as a finite linear combination of any of the others. For infinite dimensional vectors spaces, there is generally no procedure for finding a basis. Try to find a basis for$R^{∞}$. ( Note that all linear combinations of finitely many of the vectors (1,0,0,...) (0,1,0,...), ((0,0,0,1,0,0..) with a 1 in one coordinate and zeros everywhere else, does not span all of$R^{∞}$. )Generally, mathematicians assume that every vector space has a basis but this assumption is equivalent logically to the Axiom of Choice. An abstract vector space is defined by forgetting the geometry of Euclidean space and keeping only the algebraic rules of vector addition and scalar multiplication. The algebraic definition is that vector addition forms an abelian group (addition is commutative and associative, and every vectors has an additive inverse) and vectors can be multiplied by a field of scalars so that the multiplication distributes over vector addition,$r(v + w) = rv + rw$and$(rs)v = r(sv)$. These algebraic rules allow for many possibilities. For instance, as has been already mentioned, the polynomials of one variable are a vector space. Another example is the real numbers considered as a vector space with only the rational numbers as the field of scalars. Vector addition in this case is just addition of numbers, and multiplication by a scalar is just multiplication of numbers. This vector space is infinite dimensional. Also a vector space can be finite, for instance a finite dimensional vector space whose field of scalars is the field$Z_2$. The idea of a vector space does not include any idea of length of vectors or angle between them. These ideas are additional structure. The vector space part is just the algebraic rules. So it is irrelevant that$i$,$j$,and$k## are orthogonal unit vectors from the point of view of the algebraic structure of 3 space as a vector space. In general, if one has an idea of length and angle the vector space is called a Hilbert space. If there is an idea of length but not necessarily of angle, the vector space is called a Banach space. 3 space with Euclidean geometry is a Hilbert space.

Last edited: Jan 4, 2016
8. Jan 4, 2016

### Krylov

Adding to @lavinia 's nice exposition, I cannot resist mentioning that in infinite dimensional Banach spaces, one often encounters the alternative notion of a Schauder basis. This notion is more useful in analysis, for example in approximation theory. By contrast, the basis mentioned in lavinia's post is called a Hamel basis. Of course, in order to speak about a Schauder basis we require additional structure, namely a (complete) norm.

Not every Banach space has a Schauder basis, with or without AC. What is more, not every separable Banach space has a Schauder basis. This latter statement is a profound result proved by Per Enflo.

9. Jan 4, 2016

### WWGD

To add to what Krylov and Lavinia said, the main difference between Hamel Bases and Schauder Basis is that Schauder bases, since they use infinite sums require a notion of convergence, which itself requires a topology ; the infinite sum _ converges_ to the value of a vector, while Hamel bases use finite sums which equal the value of the vector. Pure Algebra is not equipped to make sense of infinite sums, and then you need to bring in convergence , which makes use of topology.