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:
Articles Related
Dimension Lemma
If U is a subspace of W then
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 <math>dim \mathbb{R}^2 = 2</math>
, 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 <math>dim \mathbb{R}^4 = 4</math>
, D1 shows <math>dimSpan S <= 4</math>
.
Theorem
Kernel-Image
Definition
For any linear function <math>f : V \mapsto W</math>
:
<MATH>
\text{dim } \href{function#kernel}{Ker} f + \text{dim } \href{function#image}{lm} f = \text{dim V}
</MATH>
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,
<MATH>
\href{matrix#nullity}{nullity } A + \href{rank}{rank } A = n
</MATH>
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:
<MATH>dim V + dim V^* = n</MATH>
Proof:
Let <math>a_1, \dots , a_m</math>
be generators for V.
Let
<math>A =
\begin{bmatrix}
\begin{array}{r}
a_1 \\
\hline \\
\vdots \\
\hline \\
a_m
\end{array}
\end{bmatrix}
</math>
then <math>V^* = \href{matrix#null space}{Null} A</math>
Rank-Nullity Theorem states that