# Dimension

The dimension n is, loosely speaking, the number of different things you could observe after making a measurement on the particle.

Let V be an 𝔽-vector space (where 𝔽 is either ℚ or ℝ or ℂ) and let n be a positive interger. If there is a list { v1, v2, … , vn } of vectors that is a basis for V, then V is n-dimensional (or, Vhas dimension n). The zero vector space has dimension zero. If V has dimension n for some nonnegative integer n, then V is finite dimensional; otherwise, V is inifinite dimensional. If V is finite dimensional, its dimension is denoted by dim V.

Theorem 1: Let V be an n-dimensional vector space, and let { v1, v2, … , vn } be any bssis.

1. If a set in V has more than n vectors, then it is linearly dependent.
2. If a set in V has fewer than n vectors, then it does not span V.
Let n = dim V and k = dim ker(T), so 0 ≤ k ≤ n. If n = 0 or if k = n, there is nothing to prove, so we may assume that 0 ≤ k < n.

If k = 0, let α = {w1, w2, ... , wn} be a basis for V. Then

$\mbox{range}\left( T \right) = \mbox{span} \left\{ T{\bf w}_1 , T{\bf w}_2 , \ldots T{\bf w}_n \right\} ,$
so it suffices to show that the set β = {Tw1, Tw2, ... , Twn} is linearly independent. If
${\bf 0} = c_1 T{\bf w}_1 + c_2 T{\bf w}_2 + \cdots + c_n T{\bf w}_n = T \left( c_1 {\bf w}_1 + c_2 {\bf w}_2 + \cdots + c_n {\bf w}_n \right) ,$
then c1w1 + c2w2 + ... + cnwn ∈ ker(T) = {0}, hence, the linear independence of α implies that c1 = c2 = ... = cn = 0 and so β is linearly independent.

If k ≥ 1, let {v1, v2, ... , vk} be a basis for ker(T), which we extend to a basis

$\alpha = \left\{ {\bf v}_1 , {\bf v}_2 , \ldots , {\bf v}_k , {\bf w}_1 , {\bf w}_2 , \ldots , {\bf w}_{n-k} \right\}$
for V. Since Tv1 = Tv2 = ... = Tvk = 0, we have
$\mbox{range}(T) = \mbox{span} \left\{ T{\bf v}_1 , \ldots , T{\bf v}_k , T{\bf w}_1 , \ldots , T{\bf w}_{n-k} \right\} = \mbox{span} \left\{ T{\bf w}_1 , \ldots , T{\bf w}_{n-k} \right\} .$
It suffices to show that β = {Tw1, Tw2, ... ,Twn-k} is linearly independent. If
$c_1 T{\bf w}_1 + c_2 T{\bf w}_2 + \cdots + c_{n-k} T{\bf w}_{n-k} = {\bf 0} ,$
then
$T \left( c_1 {\bf w}_1 + c_2 {\bf w}_2 + \cdots + c_{n-k} {\bf w}_{n-k} \right) = {\bf 0} .$
Therefore, c1w1 + c2w2 + ... + cn-kwn-k ∈ ker(T) and there are scalars a1, a2, ... , ak such that
$c_1 {\bf w}_1 + c_2 {\bf w}_2 + \cdots + c_{n-k} {\bf w}_{n-k} = a_1 {\bf v}_1 + a_2 {\bf v}_2 + \cdots + a_k {\bf v}_k .$
Then
$c_1 {\bf w}_1 + c_2 {\bf w}_2 + \cdots + c_{n-k} {\bf w}_{n-k} - a_1 {\bf v}_1 - a_2 {\bf v}_2 - \cdots - a_k {\bf v}_k = {\bf 0} ,$
so the linear independence of α implies that c1 = c2 = cn-k = a1 = a2 = ... = ak = 0. We conclude that β is linearly independent.

Corollary: Let V and U be finite dimensional vector spaces over the same field of scalars (either real numbers or complex numbers). Suppose that dimV = dim U and let T be a linear transformation from V into U. Then ker(T) = {0} if and only if the range of T is U.

If ker(T) = {0}, then dim ker(T) = 0 and dim range(T) = dimV = dim U. But the image of T is a subspace of U; hence range(T) = U.

Conversely, if range(T) = U, then dim range(T) = dimV = dim U and we get dim ker (T) = 0, so ker(T) = {0}.

Corollary (The dimension Theorem for matrices): Let A be an m×n matrix. Then

$\mbox{dim}\,\mbox{nullspace}({\bf A}) + \mbox{dim}\,\mbox{column}({\bf A}) = n \quad (\mbox{the number of columns in }\,{\bf A} .$
If m = n, then the nullspace of A is {0} if and only if it is a full range matrix.
Apply the preceding theorem to the linear transformation generated by matrix A.