Linear Independence

28. Linear Independence

Definition of Linear Independence

The set of vectors

𝑆 = {𝑣_{1}, 𝑣_{2}, \dots, 𝑣_{𝑛}}

is said to be linearly independent if the only solution to the equation

𝑐_{1} 𝑣_{1} + 𝑐_{2} 𝑣_{2} + \dots + 𝑐_{𝑛} 𝑣_{𝑛} = 𝟎

is $𝑐_{1} = 𝑐_{2} = \dots = 𝑐_{𝑛} = 0$ . In other words, no vector in the set can be written as a linear combination of the others.

If at least one constant $𝑐_{𝑖}$ is non-zero, the set is linearly dependent.

Example

Example 1: Testing for Linear Independence

Problem: Is the following set of vectors linearly dependent?

𝑆 = {{\vec{𝑣}}_{1} {\vec{𝑣}}_{2}}

Where:

{\vec{𝑣}}_{1} = [\begin{matrix} 2 \\ 1 \end{matrix}] and {\vec{𝑣}}_{2} = [\begin{matrix} 3 \\ 2 \end{matrix}]

For a set of vectors to be linearly independent, the only solution to the equation:

𝑐_{1} {\vec{𝑣}}_{1} + 𝑐_{2} {\vec{𝑣}}_{2} = 𝟎

must be $𝑐_{1} = 0$ and $𝑐_{2} = 0$

In this case:

𝑐_{1} [\begin{matrix} 2 \\ 1 \end{matrix}] + 𝑐_{2} [\begin{matrix} 3 \\ 2 \end{matrix}] = 𝟎

If not only the zero solution exists (i.e., if $𝑐_{1}$ or $𝑐_{2}$ can be non-zero), the set is linearly dependent.

Step 1. Set up the system of equations:

$2 𝑐_{1} + 3 𝑐_{2} = 0$
$1 𝑐_{1} + 2 𝑐_{2} = 0$

2. Eliminate one variable

2 \times (1 𝑐_{1} + 2 𝑐_{2} = 0) \Rightarrow 2 𝑐_{1} + 4 𝑐_{2} = 0

Now the system is

\begin{matrix} 2 𝑐_{1} + 3 𝑐_{2} = 0 \\ 2 𝑐_{1} + 4 𝑐_{2} = 0 \end{matrix}

3. Subtract the equations

(2 𝑐_{1} + 3 𝑐_{2}) - (2 𝑐_{1} + 4 𝑐_{2}) = 0

Simplifies to:

(2 𝑐_{1} - 2 𝑐_{1}) + (4 𝑐_{2} - 3 𝑐_{2}) = 0

So:

𝑐_{2} = 0

4. Substitute back to find $𝒄_{𝟏}$

Now that we know $𝑐_{2} = 0$ , substitute this value into one of the original equations. Let’s use the second equation:

1 𝑐_{1} + 2 𝑐_{2} = 0

Substitute $𝑐_{2} = 0$ :

\begin{matrix} 1 𝑐_{1} + 2 (0) = 0 \\ 𝑐_{1} = 0 \end{matrix}

Conclusion:

Since $𝑐_{1} = 0 𝑐_{2} = 0$ , the set of vectors $𝑆$ is linearly independent. These vecots span $ℝ^{2}$ .

Example 2: Testing for Linear Dependence

Problem: Is the following set of vectors linearly dependent?

𝑆 = {{\vec{𝑣}}_{1} {\vec{𝑣}}_{2}}

Where:

{\vec{𝑣}}_{1} = [\begin{matrix} 2 \\ 3 \end{matrix}] and {\vec{𝑣}}_{2} = [\begin{matrix} 4 \\ 6 \end{matrix}]

The span of this set is the collection of all vectors that can be formed by linear combinations of ${\vec{𝑣}}_{1}$ and ${\vec{𝑣}}_{2}$ :

𝑐_{1} 𝑣_{1} + 𝑐_{2} 𝑣_{2}

Since $𝑣_{2} = 2 𝑣_{1}$ , the linear combination becomes:

\begin{matrix} 𝑐_{1} 𝑣_{1} + 𝑐_{2} (2 𝑣_{1}) = (𝑐_{1} + 2 𝑐_{2}) 𝑣_{1} \\ 𝑐_{1} [\begin{matrix} 2 \\ 3 \end{matrix}] + 𝑐_{2} [\begin{matrix} 4 \\ 6 \end{matrix}] \\ 𝑐_{1} [\begin{matrix} 2 \\ 3 \end{matrix}] + 𝑐_{2} 2 [\begin{matrix} 2 \\ 3 \end{matrix}] \\ (𝑐_{1} + 2 𝑐_{2}) [\begin{matrix} 2 \\ 3 \end{matrix}] \\ 𝑐_{3} [\begin{matrix} 2 \\ 3 \end{matrix}] \end{matrix}

Thus, any linear combination of these vectors is just a scalar multiple of $𝑣_{1}$ . The span is a single line in $ℝ^{2}$ , and the vectors are linearly dependent.

For any two colinear vectors in $ℝ^{2}$ , their span reduces to a single line.

One vector in the set can be represented by some combination of other vectors in the set

2. General Rule

In $ℝ^{𝑛}$ , if you have more than $𝑛$ vectors, at least one vector must be linearly dependent on the others, meaning the set cannot be linearly independent.