I know my question seems extremely vague or meta, but I've been studying linear algebra since around the beginning of September and we've recently touched on the concept of Vector Spaces, which changed my idea of what vectors actually are.

I'm of a physics background, so to me, a vector is a dimensioned quantity with magnitude and direction, and the fact that dimensions like forces, velocities and torques have the properties of vector mathematics is basically axiomatic, at least that's how I've understood it as. So to me, they were a projection of a quantity with unique, orthogonal components. In my linear algebra course, I'm interpreting them more generally as any element of a vector space, which means elements of a set which satisfy the vector space axioms, which are extremely and surprisingly general. So, for example, any 3-tuple that satisfies something like, $2x + 4y+ 3z = 0$ (this is completely arbitrary - I don't know if this exact example even satisfies the axioms but bare with me for the sake of the point). So, if those 3-tuples are vectors, and so is $\vec F = 3 \hat i + 4xy \hat j$, I feel like I'm missing some key understanding as to what makes a vector a vector if so many things can be it.

For example, how exactly is a vector different from a point in coordinate space, other than one obviously being a vector and one being a coordinate, one having magnitude and direction, and one not, etc? Plus, consider, say, my point in coordinate space $(x,y,z)$ such that it satisfies the set condition I said previously that $2x+4y+3z=0$. Can I now draw a line to that point and call it a vector?

If anything about my question is confusing, please let me know and I'll try to elaborate.

  • $\begingroup$ Elements of a vector space, where the answer to your question will be found in noticing that nothing in the nature of the vectors is important except the properties that the vector space must satisfy. $\endgroup$
    – Hellen
    Commented Oct 25, 2017 at 20:38
  • $\begingroup$ Vectors are defined by the linear algebra you reference, but are often extended and sometimes constrained to suit random other purposes where their properties are useful, like forces/torques in physics, or like points on a coordinate plane. $\endgroup$ Commented Oct 25, 2017 at 20:42
  • $\begingroup$ "How is a vector different from a point in coordinate space" For one thing, we can talk about vectors without any knowledge of bases or coordinate axes. Further, we can talk about vectors without any knowledge of length, magnitude, or direction. For example, were you aware that the continuous real functions from $[0,1]$ to $\Bbb R$ form a vector space? There isn't really any good way to give a basis to a set like that. $\endgroup$
    – JMoravitz
    Commented Oct 25, 2017 at 20:42
  • 1
    $\begingroup$ The issue here is that the set of "dimensioned quantities with magnitude and direction", furnished with operations given, say, by the parallelogram rule, form a vector space. For this reason, they deserve the name of "vectors". $\endgroup$
    – Ivo Terek
    Commented Oct 25, 2017 at 20:46

2 Answers 2


The ''arrows'' that have magnitude, direction and orientation, used to represent physical quantities as forces, displacements, etc., was historically the starting point for the definition of a vectors space. But, as you have noted, the modern definition is more abstract and general and it is not a definition of the object ''vector'', but a definition of the structure ''vector space'' that is given statring from a set and a field, and defining some operations with suitable properties. A vector is simply an element of the set that subtends the vector space.

Sometimes the elements of this set are called ''points'', but this is an abuse. Strictly speaking ''points'' are elements well defined on other structures as, e.g. affine spaces, where we have '' points'' and ''vectors''.

The ''magnitude'' of a vector, or the angle between vectors are concepts that can be defined for vector spaces that have some additional properties, as normed space or space with an inner product. So we can have vector spaces whose elements does not have any well defined magnitude or direction as the vector spaces of functions.


In mathemarical terms, a vector is an element of a set called a vector space. A vector space over a field $\mathbb{F}$ is an algebraic structure defined with two binary operators (a set $V$ with two binary operations), say $+$ (vectos sum) and $\cdot$ (scalar multiplication) such that if $x,y,z$ are ANY elements of $V$ and $\alpha, \beta \in \mathbb{F} $ then

  1. $x+y \in V$
  2. $x+y=y+x$
  3. $x+(y+z)=(x+y)+z$
  4. Exists an element $0 \in V$ such that $0+x=x=x+0$
  5. For every $x \in V$, there exists $-x \in V$ such that $x+(-x)=0$
  6. $\alpha \cdot x \in V$ for every $x \in V$
  7. $\alpha\cdot (\beta \cdot x)=(\alpha \cdot \beta)\cdot x$
  8. There exists an element $1 \in V$ such that $1\cdot x=x$ for every $x \in V$
  9. $\alpha\cdot (x+y)=\alpha \cdot x+ \alpha \cdot y$ and $(\alpha+\beta)\cdot x=\alpha \cdot x+\beta \cdot x$

For example:

(a) The set $\mathbb{R}^n$ equipped with the usual coordinate sum and escalar product is a vector space over the fiel of real numbers $\mathbb{R}$.

(b) The set of square complex matrices $M_n(\mathbb{C})$ equipped with the usual matrix sum and scalar multiplication is a vector space over the field of complex numbers $\mathbb{C}$.

(c) The set of real functions $\mathscr{F}(\mathbb{R})$ is a vector space over the field of real numbers $\mathbb{R}$ (when equipped with the usual function sum and scalar product).

As for your question. If $S=\{ (x,y,z):2x+4y+3z=0 \}$ equipped with the usual $\mathbb{R}^n$ operations, then $S$ IS a vector space over the real numbers. Note that if $(x_1,y_1,z_1),(x_2,y_2,z_z) \in S$ and $\lambda \in \mathbb{R}$ then $$2(x_1+x_2)+4(y_1+y_2)+3(z_1+z_2)=(2x_1+4y_1+3z_1)+(2x_2+4y_2+3z_1)=0+0=0,$$ hence $(x_1,y_1,z_1)+(x_2,y_2,z_2) \in S$ and $$2(\lambda x)+4(\lambda y)+3(\lambda z)=\lambda (2x+4y+3z)=\lambda \cdot 0=0,$$ and hence $\lambda (x,y,z)\in S$. This is enough to conclude that $S$ is a vector subspace of $\mathbb{R}^3$ and therefore it's a vector space (over $\mathbb{R}$).


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