# Theorems Regarding Linear Independence and Dependence

Recall from the Linear Independence and Dependence page the definition of linear independence/dependence:

Definition: A set of vectors $V = \{ \mathbf{x_1}, \mathbf{x_2}, …, \mathbf{x_n} \}$ and the scalars $k_1, k_2, …, k_n$ has a solution for the vector equation $k_1\mathbf{x_1} + k_2\mathbf{x_2} + … + k_n\mathbf{x_n} = 0$, namely when $k_1 = k_2 = … = k_n = 0$. If this is the only solution to the vector equation, then the set $V$ is said to be Linearly Independent. If there exists other solutions where not all $k_i = 0$, then $V$ is said to be Linearly Dependent. |

We will now look at some important theorems regarding linear independence/dependence.

Theorem 1: The vector set containing only the zero vector from the vector space $V$ is linearly dependent. |

**Proof:**Consider the set of vector set only containing the zero vector, $\{ 0 \}$. We note that the vector equation $a_1\vec{0} = 0$ is satisfied for all $a_1 \in \mathbb{F}$, and so $\{ 0 \}$ is a linear dependent set. $\blacksquare$

Corollary 1: Any set of vectors $\{ v_1, v_2, …, v_n \}$ containing the zero vector is linearly dependent. |

**Proof:**Without loss of generality suppose that $v_1$ is the zero vector. Then the vector equation $a_1v_1 + a_2v_2 + … + a_nv_n = 0$ has infinitely many sets of scalars $a_1, a_2, …, a_n$ that satisfy this equation, for example, $100v_1 + 0v_2 + … + 0v_n = 0$ since $v_1$ is the zero vector, and so $\{ v_1, v_2, …, v_n \}$ is a linearly dependent set. $\blacksquare$

Theorem 2: If $v_1 \neq 0$ is a vector from the vector space $V$, then the set containing just the vector $v_1$ is linearly independent. |

**Proof:**Suppose that $v_1 \neq 0$ (otherwise we know this set would be linearly dependent from theorem 1). Then if we look at the vector equation, $kv_1 = 0$ if and only if $k = 0$ and so $\{ v_1 \}$ is a linearly independent set. $\blacksquare$

Theorem 3: The set of vectors $\{ v_1, v_2 \}$ from the vector space $V$ is linearly independent if $v_1$ is not a scalar multiple of $v_2$. |

**Proof:**We will show this by proof by contradiction. Suppose that $\{ v_1, v_2 \}$ is a linearly independent set of vectors and $v_1$ is a scalar multiple of $v_2$, that is $kv_1 = v_2$ for some $k \in \mathbb{F}$. Then it follows from the vector equation that:

(1)

\begin{equation} a_1v_1 + a_2v_2 = 0 a_1v_1 + ka_2v_1 = 0 \end{equation}

- Therefore if $a_1 = 1$ and $a_2 = -\frac{1}{k}$, then $a_1v_1 + a_2v_2 = 0$, so then $\{ v_1, v_2 \}$ would not be a linearly independent set of vectors which is a contradiction. So then $\{v_1, v_2 \}$ is linearly independent if $v_1$ is not a scalar multiple of $v_2$. $\blacksquare$

Note: It is important to recognize that theorem 3 cannot necessarily be expanded to a set of three or more vectors. That is, a set of three or more vectors is not necessarily guaranteed to be linearly independent if none of the vectors are scalar multiples of one another. For example, consider the set of vectors $(1,0,0)$, $(0,0,1)$ and $(1,0,1)$, and notice that $(0,0,0) = 1(1,0,0) + 1(0,0,1) – 1(1, 0, 1) = 0$, and so $\{ (1, 0, 0), (0, 0, 1), (1, 0, 1) \}$ is a linearly dependent set, however, none of the vectors in this set are scalar multiples of one another. |

Theorem 4: Let $\{ v_1, v_2, …, v_n \}$ be a set of vectors from the vector space $V$. If the vector $v_i$ is a linear combination of the set of vectors $\{ v_1, v_2, …, v_{i-1}, v_{i+1}, …, v_n \}$ then $\{ v_1, v_2, …, v_n \}$ is a linearly dependent set. |

**Proof:**Let $\{ v_1, v_2, …, v_n \}$ be a set of vectors from $V$. Without loss of generality, assume that $v_i = v_1$ is a vector that is a linear combination of $\{ v_2, v_3, …, v_m \}$. Then for some set of scalars $v_1 = a_2v_2 + a_3v_3 + … + a_nv_n$, and rewriting this equation we get:

(2)

\begin{equation} 1v_1 -a_2v_2 – a_3v_3 – … – a_nv_n = 0 \end{equation}

- Since $a_1 = 1$, we have that this set of vectors is linearly dependent. $\blacksquare$

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