Bases

Basis

A basis is a linearly independent AND spanning set. They can be thought of as coordinates and a way to frame linear algebra into a coordinate system.

The best way to find a basis is to get a set of vectors and cut it down until the vectors are linearly independent.

Theorem

Any subspace of Rⁿ has a basis and any bases have the same number of elements. The dimension of a subspace (notated as dim(H)) is equal to the number of elements of a basis.

Examples

$\text{Q: What is } dim(\mathbb{R}^n) \text{?} \\ \text{A: } dim(\mathbb{R}^n) = n \text{ because } \{\hat{e}_1, ..., \hat{e}_n,\} \text{ is a basis.}$

$$\text{Q: Is it true that } dim(Span\{\overrightarrow{v}_1, ..., \overrightarrow{v}_n\}) = k \text{ is always true?} \\ \text{A: No. It is only true if } Span\{\overrightarrow{v}_1, ..., \overrightarrow{v}_n\} \text{ is linearly independent.}$$ $$dim(Span\{ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} \}) = ? \\ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 1 & 2 & 4 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 2 & 3 \\ 0 & -2 & -2 \\ 0 & 0 & 1 \end{pmatrix}\\ \text{Because the matrix has a pivot in every row, the dimension of the span is 3.}$$

Rank Nullity Theorem

$$\text{Suppose we have an m x n matrix A where rank(A) = dim(Col(A)).} \\ \text{THEOREM: } dim(Nul(A)) + rank(A) = n \\ \text{THEOREM: The pivot columns form a basis for Col(A). Therefore, rank(A) is equal to the number of pivot columns.} \\ \text{Also, dim(Nul(A)) is equal to the number of non-pivot columns or number of free variables.} \\$$

To find the basis for the nullspace, you have to find the homogenous solution and get the free variables

Example

$$A = \begin{pmatrix} 1 & 3 & 6 & -8 \\ 4 & 3 & 2 & -3 \\ 2 & 4 & -1 & -1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & \frac{19}{73} \\ 0 & 1 & 0 & -\frac{47}{73} \\ 0 & 0 & 1 & -\frac{27}{73} \end{pmatrix}\\ rank(A) = 3, dim(Nul(A)) = 1$$ $$A = \begin{pmatrix} 1 & 0 & 1 & -3 \\ 0 & 1 & 0 & 2 \\ 1 & 2 & 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 1 & -3 \\ 0 & 1 & 0 & 2 \\ 0 & 2 & -1 & 4 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 1 & -3 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & -1 & 0 \end{pmatrix} \\ rank(A) = 3, dim(Null(A)) = 1$$ $$A = \begin{pmatrix} 2 & 5 & -3 & -4 & 8 \\ 4 & 7 & -4 & -3 & 9 \\ 6 & 9 & -5 & 2 & 4 \\ 0 & -9 & 6 & 5 & -6 \end{pmatrix} \rightarrow \begin{pmatrix} 2 & 5 & -3 & -4 & 8 \\ 0 & -3 & 2 & 5 & -7 \\ 0 & 0 & 0 & 4 & -6 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \\ rank(A) = 3 \text{, where columns 1, 2, and 4 form the basis for our column space.} \\ dim(Nul(A)) = 2 \\ \begin{cases} 2x_1 + 5x_2 -3x_3 -4x_4 + 8x_5 = 0 \\ -3x_2 + 2x_3 + 5x_4 - 7x_5 = 0 \\ 4x_4 - 6x_5 = 0 \\ x_3, x_5 \text{ free} \end{cases} \rightarrow \begin{cases} x_1 = -\frac{1}{6}x_3 - \frac{17}{12}x_5 \\ x_2 = \frac{2}{3}x_3 + \frac{1}{6}x_5 \\ x_4 = \frac{3}{2}x_5 \\ \end{cases} \\ \overrightarrow{x_0} = \begin{pmatrix} -\frac{1}{6}s - \frac{17}{12}t \\ \frac{2}{3}s + \frac{1}{6}t \\ s \\ \frac{3}{2}t \\ t \end{pmatrix} = s \begin{pmatrix} -\frac{1}{6} \\ \frac{2}{3} \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} -\frac{17}{12} \\ \frac{1}{6} \\ 0 \\ \frac{3}{2} \\ 1 \end{pmatrix}$$

Additions to the Invertible Matrix Theorem

Suppose A is an n x n matrix.

1. Col(A) = Rⁿ
2. rank(A) = n
3. Nul(A) = {0}
4. dim(Nul(A)) = 0