# Linear Independence and Dependence

Definition: A set of vectors $V = \{ \mathbf{x_1}, \mathbf{x_2}, ..., \mathbf{x_n} \}$ and the scalars $k_1, k_2, ..., k_n$ has a solution for the vector equation $k_1\mathbf{x_1} + k_2\mathbf{x_2} + ... + k_n\mathbf{x_n} = 0$, namely when $k_1 = k_2 = ... = k_n = 0$. If this is the only solution to the vector equation, then the set $V$ is said to be Linearly Independent. If there exists other solutions where not all $k_i = 0$, then $V$ is said to be Linearly Dependent. |

We will now look at some examples of vector sets which are either linearly independent or linearly dependent.

Note: Another way to write that a set of vectors $\{v_1, v_2, ..., v_n \}$ is linearly independent is by saying that $\mathrm{span} (v_1, v_2, ..., v_n) = \bigoplus_{i=1}^{n} \mathbb{F}v_i$, which says that any vector that is a linear combination of the set of vectors $v_1, v_2, ..., v_n$ is a unique linear combination. |

## Example 1

**Determine whether or not the vector set $V = \{ (1, 2), (-5, -3) \}$ is linearly independent or linearly dependent.**

We first must see if there exists only one set of scalars $c_1 = c_2 = 0$ (the trivial solution) as a solution to the vector equation $c_1(1, 2) + c_2(-5, -3) = 0$ or if there exists more solutions.

From this vector equation we get the following system of linear equations:

(1)When we reduce this system to RREF, we get that:

(2)Therefore our only set of scalars $c_1, c_2$ are both equal to zero. Therefore, the vector set $V$ is linearly independent.

## Example 2

**Determine whether the vector set $V = \{ (3, 6), (4, 8) \}$ is a linearly independent or a linearly dependent set.**

Once again, for $V$ to be a linearly independent set then the vector equation $c_1(3, 6) + c_2(4, 8) = 0$ must containing only one set of scalars $c_1 = c_2 = 0$. From our vector equation be obtain the following system of linear equations:

(3)When we solve this system of equations we obtain that:

(4)Suppose that $c_2 = s$ so that $c_1 = - \frac{4}{3} s$. We see that for any $s \in \mathbb{R}$, the scalars $c_1$ and $c_2$ satisfy the vector equation and thus there are infinitely many scalars sets. Therefore $V$ is a linearly *dependent* set.

## Example 3

**Show that if $A$ is an $n \times n$ invertible matrix, then the column vectors of $A$ denoted $C_1$, $C_2$, …, $C_n$ are linearly independent.**

Suppose that $A$ is an invertible matrix and suppose that we have the linear system $Ax = 0$. We note that the only solution to this linear system is the trivial solution, namely $x = 0$. This system corresponds to the following vector equations though:

(5)And since $x = (x_1, x_2, ..., x_n) = (0, 0, ..., 0)$, we see that this vector equation only have the scalars $x_1 = x_2 = ... = x_n = 0$, so the column vectors of $A$ are linearly independent.

## Example 4

**Consider the set of vectors $\{1, 1-x, 1-x^2 \}$ from the vector space $\wp_2 (\mathbb{F})$. Determine if this set of vectors is linearly independent or linearly dependent.**

When we expand the vector equation $a(1) + b(1 -x) + c(1 - x^2) = 0$ as follows, notice:

(6)We can clearly see that this equation holds only if $a = b = c = 0$. We can also verify this with the following matrix:

(7)Since this is an upper triangular matrix, its determinant is the product of the entries down the main diagonal and so $\mathrm{det} (A) = 1 \cdot (-1) \cdot (-1) = 1$ which implies $A$ is invertible, and so the homogenous system $\begin{bmatrix}1 & 1 & 1\\ 0 & -1 & 0\\ 0 & 0 & -1\end{bmatrix} \begin{bmatrix}a\\ b\\ c \end{bmatrix} = \begin{bmatrix}0\\ 0\\ 0 \end{bmatrix}$ has only the trivial solution $a = b = c = 0$.

## Example 5

**Show that the set of vectors $\{ m, m - x, m - x^2 \}$ from $\wp_2 (\mathbb{F})$ is a linearly independent set for all nonzero $m$.**

We should note that this example is a more general version of example 4. First, let's expand the vector equation $a_1m + a_2(m - x) + a_3(m - x^2) = 0$ as follows:

(8)Once again, this vector equation is only true if $a_1 = a_2 = a_3 = 0$. The same matrix from example 4 can be used to provide a more extensive argument.