Linear Independence and Dependence Examples 4

# 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)
\begin{align} \quad a_1\lambda v_1 + a_2 \lambda v_2 + ... + a_n \lambda v_n = 0 \\ \quad \lambda (a_1v_1 + a_2v_2 + ... + a_nv_n) = 0 \end{align}

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

(2)
\begin{align} \quad a_1v_1 + a_2v_2 + ... + a_nv_n = 0 \end{align}

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)
\begin{align} \quad a_1(1 - i) + a_2(1 + i) = 0 \\ \quad a_1 - a_1i + a_2 + a_2i = 0 \\ \quad (a_1 + a_2) + (a_2 - a_i)i = 0 + 0i \end{align}

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)
\begin{align} \quad a_1(1 - i ) + a_2(1 + i) = 0 \\ \quad (a_1 + a_2) + (a_2 - a_1)i = 0 \end{align}

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)
\begin{align} \quad (b_1 + b_2) + (c_1+c_2)i + ((b_2 - b_1) + (c_2 - c_1)i)i = 0 \\ \quad (b_1 + b_2) + (c_1 + c_2)i + (b_2 - b_1)i + (c_2 - c_1)i^2 = 0 \\ \quad (b_1 + b_2) + (c_1 + c_2)i + (b_2 - b_1)i - (c_2 - c_1) = 0 \\ \quad [(b_1 + b_2) - (c_2 - c_1)] + [(c_1 + c_2) + (b_2 - b_1)]i = 0 \end{align}

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)
\begin{align} c_1 + c_2 + b_2 + b_2 = c_2 - c_1 \\ c_1 + 2b_2 = - c_1\\ 2c_1 + 2b_2 = 0 \end{align}

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)
\begin{align} \quad (a_1 + a_2) + (a_2 - a_1)i \\ \quad = ([2 + i] + [-1 + 2i]) + ([-1 + 2i] - [2 + i])i \\ \quad = 1 + 3i + (-1 + 2i)i - (2 + i)i \\ \quad = 1 + 3i -i + 2i^2 - 2i -i^2 \\ \quad = 1 + 3i - i - 2 - 2i + 1 \\ \quad = 0 \end{align}

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 $\mathbb{C}$.