# Linear Independence and Dependence Examples 4

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

**Prove that if $\{ v_1, v_2, …, v_n \}$ is a linearly independent set of vectors then for $\lambda \in \mathbb{F}$, $\lambda \neq 0$ we have that $\{ \lambda v_1, \lambda v_2, …, \lambda v_n \}$ is also a linearly independent list. What happens to the linear independence of this list when $\lambda = 0$?**

Let $\{ v_1, v_2, …, v_n \}$ be a linearly independent set of vectors. We want to show that then $\{ \lambda v_1, \lambda v_2, …, \lambda v_n \}$ is a linearly independent set of vectors. Consider the following vector equation for $a_1, a_2, …, a_n \in \mathbb{F}$:

(1)

Since $\lambda \neq 0$, we can divide both sides by it to get that:

(2)

Since $\{ v_1, v_2, …, v_n \}$ is a linearly independent set of vectors, the equation above implies that $a_1 = a_2 = … = a_n = 0$ so $\{ \lambda v_1, \lambda v_2, …, \lambda v_n \}$ is a linearly independent set of vectors.

Note that if $\lambda = 0$ then our set of vectors reduces to $\{ 0, 0, …, 0 \}$, and any set of vectors containing the zero vector is linearly dependent, so we require that $\lambda \neq 0$.

## Example 2

**Consider the vector space $\mathbb{C}$ over the field $\mathbb{R}$. Show that the set of vectors $\{ 1 – i, 1 +i \}$ is linearly independent in $\mathbb{C}$.**

Let $a_1, a_2 \in \mathbb{R}$ and consider the following vector equation:

(3)

We note that the above equation implies that $a_1 + a_2 = 0$ and $a_1 = a_2$. Substituting the second equation into the first and we have that $2a_1 = 0$ which implies that $a_1 = 0$, and substituting this into the second equation gives us that $a_2 = 0$.

So indeed, $\{ 1 – i , 1 + i \}$ is a linearly independent set in $\mathbb{C}$.

## Example 3

**Consider the vector space $\mathbb{C}$ over the field $\mathbb{C}$. Show that the set of vectors $\{ 1 – i, 1 + i \}$ is linearly dependent in $\mathbb{C}$.**

*Note the subtle difference between example 2 and example 3.*

Let $a_1, a_2 \in \mathbb{C}$ and consider the following vector equation:

(4)

In this case we have that $a_1 + a_2 = 0$ and $a_2 – a_1 = 0$, however, this time, the scalars are not real numbers and are instead complex numbers. Let $a_1 = b_1 + c_1i$ and let $a_2 = b_2 + c_2i$ where $b_1, b_2, c_1, c_2 \in \mathbb{R}$. Then we have that:

(5)

The equation above implies that $b_1 + b_2 = c_2 – c_1$ and $c_1 + c_2 = b_1 – b_2$. If we $b_1$ from the second equation then we get $b_1 = c_1 + c_2 + b_2$, and plugging this into the first equation gives us:

(6)

Thus $c_1 = -b_2$. Let $c_1 = 1$, $b_2 = -1$. Then $b_1 = c_2$, so let $b_1 = 2$ and $c_2 = 2$. Thus we have that $a_1 = 2 + i$ and $a_2 = -1 + 2i$. Thus:

(7)

Thus the choice of scalars $a_1, a_2 \in \mathbb{C}$ such that $a_1(1 – i) + a_2(1 + i) = 0$ are not unique so $\{ 1 – i, 1 + i \}$ is linearly dependent in $\mathbb{C}$.