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4.4 Bases and Dimension of Vector Space

\[ \mathbb{R}^2 \text{ is spanned by } \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\} = \{ \vec{i}, \vec{j} \} \]

because every \( \mathbb{R}^2 \) vector is a linear combo of \( \vec{i} \) and \( \vec{j} \)

\[ \begin{bmatrix} 5 \\ 7 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 7 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = 5\vec{i} + 7\vec{j} \]

spanning set : enough vectors to build a vector space

\[ \mathbb{R}^2 \text{ is also spanned by } \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \end{bmatrix} \right\} \]

(we can simply ignore the extra one)

\[ \mathbb{R}^2 \text{ is not spanned by just } \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\} \text{ or just } \vec{i} \]

(missing \( y \) component)

\[ \mathbb{R}^2 \text{ is not spanned by } \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \end{bmatrix} \right\} \text{ (missing } y \text{)} \]

\[ \text{but we can span } \mathbb{R}^2 \text{ using } \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\} \]

for \( \mathbb{R}^2 \), we need a minimum of two linear indp vectors in the "spanning set"

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\[ \text{Similarly, in } \mathbb{R}^3 : \text{span} \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} = \mathbb{R}^3 \]

\[ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = 1\vec{i} + 2\vec{j} + 3\vec{k} \]

\[ \text{also spanned by } \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} \pi \\ e \\ 10 \end{bmatrix} \right\} \]

\[ \text{but not by } \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right\} \]

\[ \text{but not by } \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \right\} \]

\[ \text{this is ok: } \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ \pi \end{bmatrix} \right\} \]

Generalization for \( \mathbb{R}^n \)

for \( \mathbb{R}^n \), we need a minimum of \( n \) linear indp vectors to span.

the smallest spanning set (\( n \) linear indp vectors) is called a basis (the vectors in the set are called basis vectors or bases)

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for a given vector space, the basis is NOT unique

for example, in ℝ², the standard basis is \( \vec{i}, \vec{j} \)

\[ \begin{bmatrix} 5 \\ 7 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 7 \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

but \( \left\{ \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 5 \end{bmatrix} \right\} \) is also a basis for ℝ²

\[ \begin{bmatrix} 5 \\ 7 \end{bmatrix} = \frac{57}{10} \begin{bmatrix} 2 \\ 0 \end{bmatrix} - \frac{7}{5} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \]

also ok: \( \left\{ 3\vec{i}, -10\vec{j} \right\} \)

ℝ³: standard basis \( \vec{i}, \vec{j}, \vec{k} \)

\[ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = 1\vec{i} + 2\vec{j} + 3\vec{k} \]

also a basis: \( \left\{ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 5 \\ 2 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 4 \end{bmatrix} \right\} \)

\[ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + \frac{3}{5} \begin{bmatrix} 0 \\ 5 \\ 2 \end{bmatrix} - \frac{1}{2} \begin{bmatrix} -3 \\ 0 \\ 4 \end{bmatrix} \]

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the dimension of a vector space is the number of vectors in its basis

ℝ²: needs two basis vectors → Two-dimensional
ℝⁿ: needs n basis vectors → n-dim

M₂: all possible 2×2 matrices

one easy basis: \( \left\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right\} \)

so M₂ is a 4-D vector space

P₂: all possible 2nd-deg polynomials

\[ ax^2 + bx + c \]

basis: \( \left\{ 1, x, x^2 \right\} \) → 3-D vector space

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Example: Subspace of \(\mathbb{R}^3\)

The subspace of \(\mathbb{R}^3\) given by the plane \(3x + 2y + z = 0\).

All vectors in this subspace are \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\) such that they satisfy the plane equation.

\[\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x \\ y \\ -3x - 2y \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix}\]

Linear combinations of \(\begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix}\) and \(\begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix}\) span (and is a basis for) this subspace.

One Possible Basis

One possible basis is \(\left\{ \begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix} \right\}\). This is a 2-D subspace.

Another Basis

\[\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -\frac{2}{3}y - \frac{1}{3}z \\ y \\ z \end{bmatrix} = y \begin{bmatrix} -2/3 \\ 1 \\ 0 \end{bmatrix} + z \begin{bmatrix} -1/3 \\ 0 \\ 1 \end{bmatrix}\]

Another basis: \(\left\{ \begin{bmatrix} -2/3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1/3 \\ 0 \\ 1 \end{bmatrix} \right\}\)

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Note:

\(\left\{ \begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 3 \end{bmatrix} \right\}\) is also a basis.

The first vector is -3 times the first vector of the last set.
The second vector is 3 times the second vector of the last set.

And many other possible sets.

Example: Solution Space

Solution space of:

\[\begin{aligned} x_1 - 2x_2 - 5x_3 &= 0 \\ 2x_1 - 3x_2 - 13x_3 &= 0 \end{aligned}\]

Solution space: \(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\) that satisfies the equations above.

It is also the null space of the coefficient matrix:

\[\begin{bmatrix} 1 & -2 & -5 \\ 2 & -3 & -13 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

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What does this space look like?

Solve the system:

\[ \begin{bmatrix} 1 & -2 & -5 & 0 \\ 2 & -3 & -13 & 0 \end{bmatrix} \]

\[ \rightarrow \dots \rightarrow \begin{bmatrix} 1 & -2 & -5 & 0 \\ 0 & 1 & -3 & 0 \end{bmatrix} \]

3 variables, 2 pivots

\(\rightarrow\) one free variable

choose \(x_3\) to be free (no pivot in its col)

\(x_3 = r\)

row 2: \(x_2 = 3r\)

row 1: \(\dots \quad x_1 = 11r\)

solution:

\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11r \\ 3r \\ r \end{bmatrix} = r \begin{bmatrix} 11 \\ 3 \\ 1 \end{bmatrix} \]

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null space / solution space contains all possible scalar multiples of \[ \begin{bmatrix} 11 \\ 3 \\ 1 \end{bmatrix} \]

space is spanned by \[ \begin{bmatrix} 11 \\ 3 \\ 1 \end{bmatrix} \] which is also a basis

so, the null space in this example is ONE-dimensional

here, the intersection of the two planes is a line