# Linear Independence and Dependence Examples 3

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

**Suppose that $\{ v_1, v_2, …, v_n \}$ is a linearly independent set of vectors. Determine whether or not $\{ v_1 – v_2, v_2 – v_3, …, v_{n-1} – v_n, v_n \}$ is a linearly independent set of vectors.**

For $a_1, a_2, …, a_n \in \mathbb{F}$, consider the following vector equation:

(1)

Since $\{ v_1, v_2, …, v_n \}$ is a linearly independent set of vectors, we have that the equation above implies that:

(2)

This system of equations is easily solved for using forward substitution to get that $a_1 = a_2 = … = a_n = 0$. So indeed $\{ v_1 – v_2, v_2 – v_3, …, v_{n-1} – v_n, v_n \}$ is a linearly independent set of vectors.

## Example 2

**Show that any set of three distinct vectors in $\mathbb{R}^2$ must be a linearly dependent set of vectors.**

Let $(a, b), (c, d), (e, f) \in \mathbb{R}^2$. We want to show that $\{ (a, b), (c, d), (e, f) \}$ must form a linearly dependent set of vectors. Consider the following vector equation where $q_1, q_2, q_3 \in \mathbb{F}$:

(3)

If one of the vectors $(a, b), (c, d), (e, f)$ is identically the zero vector, then the set $\{ (a, b), (c, d), (e, f) \}$ is automatically linearly dependent. Assume that none of these vectors are identically the zero vector. Then we have that:

(4)

We thus obtain the following system of equations where the coefficients $q_1, q_2, q_3$ are unknown:

(5)

This a system of two equations in three unknowns, which is guaranteed a nontrivial solution for the constants $q_1, q_2, q_3$, and so $\{ (a, b), (c, d), (e, f) \}$ form a linearly dependent set of vectors.

## Example 3

**Let $\{ v_1, v_2, …, v_n \}$ be a linearly independent set of vectors in $V$. Prove that if $\{ v_1 + w, v_2 + w, …, v_n + w \}$ is a linearly dependent set of vectors in $V$ then $w \in \mathrm{span} (v_1, v_2, …, v_n)$.**

Let $\{ v_1, v_2, …, v_n \}$ be a linearly independent set of vectors in $V$, and suppose that $\{ v_1 + w, v_2 + w, …, v_n + w \}$ is a linearly dependent set of vectors in $V$. Then for constants $a_1, a_2, …, a_n \in \mathbb{F}$ that are not all zero we have that, and for letting $S = a_1 + a_2 + … + a_n$ we have that:

(6)

We note that $S = a_1 + a_2 + … + a_n \neq 0$. If so, then the equation above reduces to $a_1v_1 + a_2v_2 + … + a_nv_n = 0$ which implies by the linear independence of $\{ v_1, v_2, …, v_n \}$ that $a_1 = a_2 = … = a_n = 0 4$ which is a contradiction as $\{ v_1 + w, v_2 + w, …, v_n + w \}$ is linearly dependent.

Therefore the linear combination for $w$ in terms of the vectors $\{ v_1, v_2, …, v_n \}$ is well defined and $w \in \mathrm{span} (v_1, v_2, …, v_n )$.