# Linear Independence and Dependence Examples 1

Recall from the Linear Independence and Dependence page that a set of vectors $\{ v_1, v_2, ..., v_n \}$ is said to be **Linearly Independent** in $V$ if the vector equation $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$ implies that $a_1 = a_2 = ... = a_n = 0$, that is, the zero vector is uniquely expressed as a linear combination of the vectors in $\{ v_1, v_2, ..., v_n \}$ with the coefficients all being zero.

If a set of vectors $\{ v_1, v_2, ..., v_n \}$ is not linearly independent then we say the set if **Linearly Dependent** and that there exists scalars $a_1, a_2, ..., a_n \in \mathbb{F}$, not all zero, such that $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$.

We will now look at some more examples to regarding the linear independence / dependence of a set of vectors.

## Example 1

**Consider the set of vectors $\left \{ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & -1\\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \right \}$ from the vector space $M_{22}$ of $2 \times 2$ matrices. Determine if this set is linearly independent or linearly dependent.**

We first consider the following vector equation:

(1)This equation equals zero if and only if $a_1 + a_3 = 0$, $-a_2 = 0$, and $a_1 = 0$, and we deduce that therefore the only set of scalars for which the above vector equation is true is $a_1 = a_2 = a_3 = 0$. We can verify this by showing the following matrix representing this system is invertible, $A = \begin{bmatrix}1 & 0 & 1\\ 0 & -1 & 0\\ 1 & 0 & 0 \end{bmatrix}$ is invertible by cofactor expansion along the third row to get that $\mathrm{det} A = 1 \cdot \begin{vmatrix}0 & 1\\ -1 & 0 \end{vmatrix} = 1$, and so the homogenous system $\begin{bmatrix} 1 & 0 & 1\\ 0 & -1 & 0\\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} a_1\\ a_2\\ a_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$ has only the trivial solution $a_1 = a_2 = a_3 = 0$.

## Example 2

**Consider the set of vectors $\{ 1 + x - x^2, 2 + x^2, x^3 - 2x^4 \}$ from the vector space $\wp_{4} (\mathbb{F})$ of polynomials of degree $4$ or less. Determine if this set is linearly independent or linearly dependent.**

Let's first consider the following vector equation:

(2)This equation is zero if and only if:

(3)And the only solution to this system is $a_1 = a_2 = a_3 = a_4 = 0$, and so this set of vectors is linearly independent.

## Example 3

**Consider the set of vectors $\{ (1, 2, 0, 0), (0, 4, 4, 0), (2, 0, 3, 0) \}$ from the vector space $\mathbb{R}^4$ of 4-component standard vectors. Determine if this set is linearly independent or linearly dependent.**

Let's first look at the following vector equation:

(4)We must check to see if the following homogenous system has more than just the trivial solution.

(5)We will do this, once again, by determining if the following matrix representing this system is invertible $A = \begin{bmatrix}1 & 0 & 2\\ 2 & 4 & 0\\ 0 & 4 & 3 \end{bmatrix}$. Using cofactor expansion along row 1 we have that $\det A = 1 \cdot \begin{vmatrix}4 & 0\\ 4 & 3\ \end{vmatrix} + 2 \cdot \begin{vmatrix}2 & 4\\ 0 & 4\ \end{vmatrix} = 12 + 16 = 28$, and so $A$ is invertible while implies that the system $\begin{bmatrix}1 & 0 & 2\\ 2 & 4 & 0\\ 0 & 4 & 3 \end{bmatrix} \begin{bmatrix}a_1\\ a_2\\ a_3\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ has only the trivial solution $a_1 = a_2 = a_3 = 0$.