Vector spaces · Definition

Vector space

linear algebravector spaces

Before you start you need to know

A vector space is any set where addition and scaling behave as in \(\mathbb{R}^n\); what matters are the rules, not the objects.

Definition

A real vector space is a set \(V\) with an addition and a scalar multiplication by reals satisfying eight axioms; its elements are called vectors, whether arrows, matrices or functions.

\[u+v\in V,\quad cv\in V,\quad (c+d)v=cv+dv\]

After working with vectors in \(\mathbb{R}^n\) one notices that many other objects are added and scaled in the same way: matrices, polynomials, continuous functions. Axler and Strang proceed like this: list the properties that are really used and call vector space any set satisfying them. Whatever is proved from those axioms holds at once for every example.

A guiding example: polynomials of degree \(\le 2\) form a space, with the polynomial \(0\) as zero vector and \(-p\) as the inverse of \(p\). Polynomials of degree exactly 2 do not, because \((t^2+t)+(-t^2+1)=t+1\) leaves the set. Another odd example: on \(\mathbb{R}_{>0}\) with \(u\oplus v=uv\) and \(c\odot u=u^c\), the zero vector is the number \(1\).

Typical mistakes: thinking the zero vector is always \((0,\dots,0)\) (it depends on the space), checking only a couple of axioms, or confusing the scalar \(0\) with the zero vector. Immediate consequences such as \(0v=0\) and \((-1)v=-v\) follow from the axioms.

It connects with what came before because \(\mathbb{R}^n\) is the first example, and with what follows: a vector subspace asks when a subset of a vector space is, with the same operations, a space by itself. The eight axioms are: commutativity and associativity of addition, existence of the zero vector and of the additive inverse \(-v\), \(1v=v\), \(c(dv)=(cd)v\) and both distributive laws.

Think about it

Is \(\{(x,y)\mid x+y=1\}\) a vector space with the usual operations?

No. It does not contain the zero vector \((0,0)\), and \((1,0)+(0,1)=(1,1)\) has coordinate sum 2, so it is not closed under addition.

Why does \(0\cdot v=0\) follow from the axioms?

Because \(0v=(0+0)v=0v+0v\); adding the additive inverse of \(0v\) to both sides leaves \(0=0v\).

Exercises 10

References

  • Gilbert Strang, Introduction to Linear Algebra, 6th edition, Wellesley-Cambridge Press, 2023. §3.1
  • Sheldon Axler, Linear Algebra Done Right, 4th edition, Springer, 2024. §1.B