A space has many bases, and they may look nothing alike. What they always share is their size, and that shared number is what the word dimension means.

A room can be described by many different sets of three measurements, but never by two and never needing four. The three is a property of the room.

Every basis of a given subspace has the same number of vectors, and that number is the dimension of the subspace. It is what makes the word well defined: without the theorem below, the size of a basis would be a property of the basis rather than of the space.

• $\dim\mathbb{R}^n=n$, because the standard basis has n vectors.
• $\dim\{\mathbf{0}\}=0$: its basis is the empty list.
• More than $\dim W$ vectors in are always dependent, and fewer never span it.

The engine is a lemma: in a space spanned by $k$ vectors, any $m>k$ vectors are dependent. Write each of the $m$ vectors as a combination of the $k$ spanning ones; the weights form a $k\times m$ matrix . A homogeneous system with more unknowns than equations has a non-zero solution , and the combination of the vectors with the weights comes out as . So an independent list is never longer than a spanning one, and two bases, each of them both, have the same size.

Let $\mathbf{v}_1,\ldots,\mathbf{v}_k$ span $W$ and take $$ in with . Each is a combination of the , and its weights form column of a matrix .

$A$ has more columns than rows, so this homogeneous system has a free variable and a non-zero solution, as the chapter on elimination showed.

Substitute and collect the terms on each $\mathbf{v}_i$: its weight is entry $i$ of $A\mathbf{c}$, which is $0$.

The weights $c_j$ are not all zero, so the $\mathbf{w}_j$ are dependent: an independent list is never longer than a spanning one.

$B$ is independent and $C$ spans $W$, and the other way round, so the two sizes are equal.

$\mathbf{e}_1,\ldots,\mathbf{e}_n$ is a basis of $\mathbb{R}^n$

It has n vectors, so the dimension is n.

$\operatorname{span}\{(1,2,3)\}$ has dimension 1

One vector, independent, spanning the line.

$3$ vectors in $\mathbb{R}^2$

Always dependent, whatever the vectors are.

$\mathbf{e}_1,\mathbf{e}_2$ and $(1,0),(1,1)$

Two different bases of $\mathbb{R}^2$, both with two vectors.

Count a basis. $\mathbb{P}_n$ has the basis $1,x,\ldots,x^n$, which has $n+$ members, and has the matrices .

The free-variable method gives a basis of $\operatorname{Nul}A$ with one vector per free variable. So for an $m\times n$ matrix, $\dim\operatorname{Nul}A=n-\operatorname{rank}A$.

• Every basis of a space has the same size, its dimension: an independent list is never longer than a spanning one.
• $\dim\mathbb{R}^n=n$, $\dim\mathbb{P}_n=n+1$, .

Great work! You now know how to:

• find a dimension by counting a basis;
• explain why every basis of a space has the same size;
• compute $\dim\operatorname{Nul}A$ from the rank;
• use dimension to decide spanning and independence at once.