Vectors

Axioms of a vector space; bases and coordinates.

about 7 min

Start from a problem

Problem

Arrows in the plane, polynomials, matrices, continuous functions: their addition and scaling obey exactly the same rules. Proving the "same" theorem separately for each kind of object is wasted work.

We need a definition that depends only on the rules of the operations, not on what the objects "are". Then one proof holds everywhere.

Observe

Arrows in the plane can be added (tip to tail) and stretched (scalar multiplication). So can polynomials. So can $2\times2$ matrices, continuous functions, sequences…

These objects look completely different, yet the rules obeyed by their addition and scaling are identical: addition commutes and associates, scaling distributes, there is a zero, there are negatives.

Conjecture

If a theorem uses only these rules, then whenever it holds for arrows it holds for polynomials, matrices and functions too. So it pays to isolate those rules, name every object satisfying them, and study only the name.

Definition

Definition 1.1Vector space

Let $V$ be a set with an addition $u+v$ and a real scalar multiplication $cv$ such that for all $u,v,w\in V$ and $a,b\in\mathbb R$ :

$u+v=v+u$ ;　2. $(u+v)+w=u+(v+w)$ ;
there is $0\in V$ with $v+0=v$ ;　4. each $v$ has $-v$ with $v+(-v)=0$ ;
$a(u+v)=au+av$ ;　6. $(a+b)v=av+bv$ ;
$(ab)v=a(bv)$ ;　8. $1v=v$ .

Then $V$ is a (real) vector space and its elements are vectors.

$\mathbb R^2$ , $\mathbb R^n$ , polynomials of degree at most $n$ , all $m\times n$ matrices, continuous functions on $[0,1]$ : all vector spaces. The exercises contain a non-example: polynomials of degree exactly 2 are not, since $x^2+(-x^2+x)$ has degree 1 and leaves the set.

Definition 1.2Linear combination, span, independence

A linear combination of $v_1,\dots,v_k$ is $c_1v_1+\cdots+c_kv_k$ . The set of all of them is $\operatorname{span}(v_1,\dots,v_k)$ . If $c_1v_1+\cdots+c_kv_k=0$ only for $c_1=\cdots=c_k=0$ , the vectors are linearly independent.

Independence means no vector is redundant: none can be built from the others.

Definition 1.3Basis and dimension

If $v_1,\dots,v_n$ are independent and $\operatorname{span}(v_1,\dots,v_n)=V$ , they form a basis of $V$ . The number of vectors in a basis is the dimension $\dim V$ .

For dimension to make sense we must prove that any two bases have the same size.

Theorems and proofs

Theorem 1.1Uniqueness of coordinates

If $v_1,\dots,v_n$ is a basis of $V$ , every $v\in V$ can be written uniquely as $v=c_1v_1+\cdots+c_nv_n$ . The tuple $(c_1,\dots,c_n)$ is the coordinate vector of $v$ in this basis.

Proof

Existence is spanning. For uniqueness, if $v=\sum c_iv_i=\sum d_iv_i$ then $\sum(c_i-d_i)v_i=0$ , so $c_i=d_i$ by independence.

This theorem is the entire justification for the word "coordinates". Once a basis is fixed, every vector in any vector space corresponds to one list of numbers, a point of $\mathbb R^n$ .

Theorem 1.2Dimension is well defined

Any two finite bases of a vector space have the same number of elements.

Proof

It suffices to show: if $u_1,\dots,u_m$ are independent and $v_1,\dots,v_n$ span $V$ , then $m\le n$ (swap the roles to get equality).

Exchange argument. $u_1\in\operatorname{span}(v_1,\dots,v_n)$ ; write $u_1=\sum a_iv_i$ . Since $u_1\ne0$ some $a_i\ne0$ , say $a_1$ . Then $v_1$ is a combination of $u_1,v_2,\dots,v_n$ , so these still span $V$ . Now expand $u_2$ in this new spanning set; some $v_j$ ( $j\ge2$ ) must have nonzero coefficient, otherwise $u_2\in\operatorname{span}(u_1)$ , contradicting independence. Replace that $v_j$ by $u_2$ . Continue. Each step trades one $v$ for one $u$ , and the $v$ 's cannot run out before the $u$ 's do (or some $u_k$ would lie in the span of earlier $u$ 's). Hence $m\le n$ .

Common mistake

Linear independence is not "no two are collinear". $(1,0,1),(0,1,1),(1,1,2)$ are pairwise non-collinear yet dependent, since the third is the sum of the first two. The definition is about the whole family having only the trivial zero combination; it must be checked all at once.

Application

ApplicationWhy pictures in ℝ² speak for every 2-dimensional space

Any two-dimensional vector space, once a basis is chosen, corresponds one-to-one with $\mathbb R^2$ , with addition and scaling preserved. So every picture drawn in the plane in Linear maps is true for every two-dimensional space, whether its vectors are arrows, linear polynomials, or solutions of a second-order linear differential equation.

Remark

A vector is not "a quantity with magnitude and direction"; that is one concrete picture valid in $\mathbb R^2,\mathbb R^3$ . A vector is an element of a vector space, and a vector space is defined by eight axioms. Defining things by the rules their operations obey, rather than by what they "are", is the basic move of all of algebra.

Exercises

What is the dimension of the subspace of

\mathbb R^3

spanned by

(1,0,1),(0,1,1),(1,1,2)

Under the usual operations, which of these is not a vector space?