# Linear Algebra - Dimension of a vector space

## About

The dimension of a vector space V is the size of a basis for that vector space written:

• dim V.

dim Span = rank

## Dimension Lemma

If U is a subspace of W then

• D1: $dim U <= dimW$ (or $dim U <= |W|$ ) and
• D2: if $dim U = dimW$ then $U = W$

Example:

• Suppose V = Span {[1, 2], [2, 1]}. Clearly V is a subspace of R2. However, the set {[1, 2], [2, 1]} is linearly independent, so dim V = 2. Since $dim \mathbb{R}^2 = 2$ , D2 shows that V = R2.
• S = {[−0.6,−2.1,−3.5,−2.2], [−1.3, 1.5,−0.9,−0.5], [4.9,−3.7, 0.5,−0.3], [2.6,−3.5,−1.2,−2.0], [−1.5,−2.5,−3.5, 0.94]} Since every vector in S is a 4-vector, Span S is a subspace of R4. Since $dim \mathbb{R}^4 = 4$ , D1 shows $dimSpan S <= 4$ .

## Theorem

### Kernel-Image

#### Definition

For any linear function $f : V \mapsto W$ : $$\text{dim } \href{function#kernel}{Ker} f + \text{dim } \href{function#image}{lm} f = \text{dim V}$$

#### Rank-Nullity

Apply Kernel-Image Theorem to a matrix function f (x) = Ax:

• Ker f = Null A
• dim lm f = dim Col A = rank A

For any n-column matrix A, $$\href{matrix#nullity}{nullity } A + \href{rank}{rank } A = n$$

### Dual Dimension

Example

• over R: Let V = Span {[1, 0, 1], [0, 1, 0]}. Then V* = Span {[1, 0,-1]}. dim V + dimV* = 3
• over R: Let V = Span {[1, 0, 1, 0], [0, 1, 0, 1]}. Then V* = Span {[1, 0,-1, 0], [0, 1, 0,-1]} dim V + dimV* = 4
• over GF(2): Let V = Span {[1, 0, 1], [0, 1, 0]}. Then V* = Span {[1, 0, 1]}. dim V + dimV* = 3

Dual Dimension Theorem: $$dim V + dim V^* = n$$

Proof:

Let $a_1, \dots , a_m$ be generators for V.

Let $A = \begin{bmatrix} \begin{array}{r} a_1 \\ \hline \\ \vdots \\ \hline \\ a_m \end{array} \end{bmatrix}$ then $V^* = \href{matrix#null space}{Null} A$

Rank-Nullity Theorem states that

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