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4.3 Linear Independence Sets and Bases

A set \( \{ \vec{v}_1, \dots, \vec{v}_p \} \) in \( V \) is linearly independent if

\[ c_1 \vec{v}_1 + c_2 \vec{v}_2 + \dots + c_p \vec{v}_p = \vec{0} \]

has only the trivial solution: \( c_1 = c_2 = \dots = c_p = 0 \).

Very straight forward if vectors are in \( \mathbb{R}^n \) (\( A\vec{x} = \vec{0} \)).

An alternate definition (for vectors typically not in \( \mathbb{R}^n \))

An indexed set \( \{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_p \} \) of two or more vectors with \( \vec{v}_1 \neq \vec{0} \) is linearly dependent if and only if some \( \vec{v}_j \) (\( j > 1 \)) is a linear combination of the preceding vectors \( \vec{v}_1, \vec{v}_2, \dots, \vec{v}_{j-1} \).

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Example

\[ \{ 1, t, t^2 \} \]

Where \( \vec{v}_1 = 1 \), \( \vec{v}_2 = t \), and \( \vec{v}_3 = t^2 \).

is linearly independent because \( \vec{v}_j \) is never a linear combo of vectors preceding it.

What does it mean for a set to be a basis of some subspace \( H \) of a vector space \( V \)?

Set is linearly independent: \( \{ \vec{b}_1, \vec{b}_2, \dots, \vec{b}_p \} \)
The subspace spanned by the set must coincide with \( H \):
\[ H = \text{span} \{ \vec{b}_1, \vec{b}_2, \dots, \vec{b}_p \} \]

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Linear Independence and Bases in \(\mathbb{R}^3\)

For example, let \(H = \mathbb{R}^3\).

\[B = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right\} \text{ is clearly linearly independent}\]

but not a basis of \(\mathbb{R}^3\) because the vectors in \(B\) do not span \(\mathbb{R}^3\).

Evaluating a Set of Three Vectors

How about \(B = \left\{ \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} -6 \\ -2 \\ -6 \end{bmatrix} \right\} \)? Basis for \(\mathbb{R}^3\)?

Independent? We check the matrix:

\[\begin{bmatrix} 1 & 2 & -6 \\ 0 & 1 & -2 \\ 2 & 1 & -6 \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & 2 & -6 \\ 0 & 1 & -2 \\ 0 & -3 & 6 \end{bmatrix}\]

2 pivots, no unique solution to \(A\vec{x} = \vec{0}\)

NOT linearly independent

So NOT basis for \(\mathbb{R}^3\)

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\[B = \left\{ \begin{bmatrix} 1 \\ -5 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix}, \begin{bmatrix} 3 \\ -7 \\ 6 \end{bmatrix}, \begin{bmatrix} 0 \\ 4 \\ -4 \end{bmatrix} \right\}\]

indp? We row reduce the augmented matrix:

\[\begin{bmatrix} 1 & 0 & 3 & 0 \\ -5 & 3 & -7 & 4 \\ 3 & -1 & 6 & -4 \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & 0 & 3 & 0 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 8 \end{bmatrix}\]

\(4^{\text{th}}\) vector is a linear combo of the first 3
but \(B\) does span \(\mathbb{R}^3\)

Constructing a Basis

Can we make \(B\) a basis for \(\mathbb{R}^3\)?

Yes, simply throw the \(4^{\text{th}}\) (the one dependent on others) out.

\[C = \left\{ \begin{bmatrix} 1 \\ -5 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix}, \begin{bmatrix} 3 \\ -7 \\ 6 \end{bmatrix} \right\} \stackrel{\text{is}}{=} \text{basis for } \mathbb{R}^3\]

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The Spanning Set Theorem

\[ S = \{ \vec{v}_1, \dots, \vec{v}_p \} \text{ in } V, \quad H = \text{span} \{ \vec{v}_1, \dots, \vec{v}_p \} \]

a).

If one of the vectors in \( S \), for example, \( \vec{v}_k \), is a linear combination of the remaining vectors in \( S \), then the set formed by removing \( \vec{v}_k \) from \( S \) still spans \( H \).

b).

If \( H \neq \{ \vec{0} \} \), some subset of \( S \) is a basis for \( H \).

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Finding basis for \( \text{Nul } A \) is easy.

The standard way of solving \( A\vec{x} = \vec{0} \) always gives us the basis for \( \text{Nul } A \).

Example

\[ A = \begin{bmatrix} -2 & 6 & -2 & -6 \\ 2 & -9 & -2 & 1 \\ -3 & 12 & 1 & -4 \end{bmatrix} \] \[ \begin{bmatrix} A & \vec{0} \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & 0 & 5 & 8 & 0 \\ 0 & 3 & 4 & 5 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]

\( x_3, x_4 \) free

\[ x_1 = -5x_3 - 8x_4 \] \[ x_2 = -4/3 x_3 - 5/3 x_4 \]

\[ \text{Nul } A = x_3 \begin{bmatrix} -5 \\ -4/3 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -8 \\ -5/3 \\ 0 \\ 1 \end{bmatrix} \]

\[ \text{basis for Nul } A = \left\{ \begin{bmatrix} -5 \\ -4/3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -8 \\ -5/3 \\ 0 \\ 1 \end{bmatrix} \right\} \]

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Basis for Column Space

Basis for \(\text{Col } A\) are pivot columns in the original \(A\) (NOT the reduced form).

here, basis for \(\text{Col } A = \left\{ \begin{bmatrix} -2 \\ 3 \\ -3 \end{bmatrix}, \begin{bmatrix} 6 \\ -9 \\ 12 \end{bmatrix} \right\} \)

Two Ways to Look at a Basis

There are two ways to look at a basis:

a minimum or the most efficient spanning set
a largest possible linearly independent set for that subspace

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Example

Find a basis for the set of vectors in \(\mathbb{R}^2\) on the line \(y = 5x\).

Rewrite this as a homogeneous "system" and find the null space.

\(-5x + y = 0\)

\(\underbrace{\begin{bmatrix} -5 & 1 \end{bmatrix}}_{A} \begin{bmatrix} x \\ y \end{bmatrix} = 0\)

Finding \(\text{Nul } A\)

\(\begin{bmatrix} -5 & 1 & 0 \end{bmatrix}\)

(pivot column)(free variable)

\(y\) is free

\(x = \frac{1}{5}y\)

\(\rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = y \begin{bmatrix} 1/5 \\ 1 \end{bmatrix}\)

So the basis is \(\begin{bmatrix} 1/5 \\ 1 \end{bmatrix}\)