Linear Independence and Dependence Examples 3

# 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)
\begin{align} \quad a_1(v_1 - v_2) + a_2(v_2 - v_3) + ... + a_{n-1}(v_{n-1} - v_n) + a_nv_n = 0 \\ \quad a_1v_1 - a_1v_2 + a_2v_2 - a_2v_3 + ... + a_{n-1}v_{n-1} - a_{n-1}v_n + a_nv_n = 0 \\ \quad a_1v_1 + (a_2 - a_1)v_2 + ... + (a_{n-1} - a_{n-2})v_{n-1} + (a_n - a_{n-1})v_n = 0 \end{align}

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

(2)
\begin{align} \quad a_1 = 0 \\ \quad a_2 - a_1 = 0 \\ \vdots \\ \quad a_{n-1} - a_{n-2} = 0 \\ \quad a_n - a_{n-1} = 0 \end{align}

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)
\begin{align} \quad q_1(a, b) + q_2(c, d) + q_3(e, f) = 0 \end{align}

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)
\begin{equation} (q_1a + q_2c + q_3e, q_1b + q_2d + q_3f) = (0, 0) \end{equation}

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

(5)
\begin{align} \quad q_1a + q_2c + q_3e = 0 \\ \quad q_1b + q_2d + q_3f = 0 \end{align}

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)
\begin{align} \quad a_1(v_1 + w) + a_2(v_2 + w) + ... + a_n(v_n + w) = 0 \\ \quad a_1v_1 + a_2v_2 + ... + a_nv_n + (a_1 + a_2 + ... + a_n)w = 0 \\ \quad a_1v_1 + a_2v_2 + ... + a_nv_n = -(a_1 + a_2 + ... + a_n)w \\ \quad -\frac{a_1}{S}v_1 - \frac{a_2}{S}v_2 - ... - \frac{a_n}{S} = w \end{align}

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 $w \in \mathrm{span} (v_1, v_2, ..., v_n )$.