Linear Independence

Suppose $v_1,\dots ,v_m\in V$ and $v\in \text{span}(v_1,\dots ,v_m)$. By the definition of span, there exist $a_1,\dots ,a_m\in \mathbb{F}$ such that

$$v=a_1{v}_1+\cdots +a_m{v}_m$$

Is the choice of scalars in the equation above unique? Suppose $c_1,\dots ,c_m$ is another set of scalars such that

$$v=c_1{v}_1+\cdots +c_m{v}_m$$

Subtracting the last two equations, we have

$$0=(a_1-c_1)v_1+\cdots +(a_m-c_m)v_m$$

Thus we have written $0$ as a linear combination of $(v_1,\dots ,v_m)$. If the only way to do this is by using $0$ for all the scalars in the linear combination, then each $a_k-c_k=0$, which means that $a_k=c_k$, and thus the choice of scalars was indeed unique. This brings us to the definition of linear independence.

Definition: Linearly Independent

A list $v_1,\dots ,v_m$ of vectors in $V$ is called linearly independent if the only choice of $a_1,\dots ,a_m\in \mathbb{F}$ that makes

$$a_1{v}_1+\cdots +a_m{v}_m=0$$

is $a_1=\cdots =a_m=0$.

The empty list $()$ is also declared to be linearly independent.

The reasoning above shows that $v_1,\dots ,v_m$ is linearly independent if and only if each vector in $\text{span}(v_1,\dots ,v_m)$ has only one representation as a linear combination of $v_1,\dots ,v_m$.

Linear Independence Examples

Example (a): To see that the list $(1,0,0,0),(0,1,0,0),(0,0,1,0)$ is linearly independent in ${\mathbb{F}}^4$, suppose $a_1,a_2,a_3\in \mathbb{F}$ and

$$a_1(1,0,0,0)+a_2(0,1,0,0)+a_3(0,0,1,0)=(0,0,0,0)$$

Thus

$$(a_1,a_2,a_3,0)=(0,0,0,0)$$

Hence $a_1=a_2=a_3=0$. Thus, the list $(1,0,0,0),(0,1,0,0),(0,0,1,0)$ is linearly independent in ${\mathbb{F}}^4$.

Example (b): Suppose $m$ is a nonnegative integer. To see that the list $1,z,\dots ,z^m$ is linearly independent in $\mathcal{P}(\mathbb{F})$, suppose $a_0,a_1,\dots ,a_m\in \mathbb{F}$ and

$$a_0+a_{1}z+\cdots +a_m{z}^m=0$$

where we think of both sides as elements of $\mathcal{P}(\mathbb{F})$. Then

$$a_0+a_{1}z+\cdots +a_m{z}^m=0$$

for all $z\in \mathbb{F}$. This implies that $a_0=a_1=\cdots =a_m=0$. Thus, $1,z,\dots ,z^m$ is a linearly independent list in $\mathcal{P}(\mathbb{F})$.

Example (c): A list of length one in a vector space is linearly independent if and only if the vector in the list is not $0$. If the only vector in the list is $u=0$, then any $a_k\in \mathbb{F}$ would make $a_{k}u=0$.

Example (d): A list of length two in a vector space is linearly independent if and only if neither of the two vectors in the list is a scalar multiple of the other.

If some vectors are removed from a linearly independent list, the remaining list is also linearly dependent.

Definition: Linearly Dependent

A list of vectors in $V$ is called linearly dependent if it is not linearly independent. In other words, a list $v_1,\dots ,v_m$ of vectors in $V$ is linearly dependent if there exist $a_1,\dots ,a_m\in \mathbb{F}$, not all $0$, such that $a_1{v}_1+\cdots +a_m{v}_m=0$.

Linear Independence Examples

Example (a): $(2,3,1),(1,-1,2),(7,3,8)$ is linearly dependent in ${\mathbb{F}}^3$ because

$$2(2,3,1)+3(1,-1,2)+(-1)(7,3,8)=(0,0,0)$$

Example (b): The list $(2,3,1),(1,-1,2),(7,3,c)$ is linearly dependent in ${\mathbb{F}}^3$ if and only if $c=8$. This is the solution we get when solving the system of equations

$$\begin{array}{r}2{\lambda }_1+{\lambda }_2+7{\lambda }_3=0\\ 3{\lambda }_1-{\lambda }_2+3{\lambda }_3=0\\ {\lambda }_1+2{\lambda }_2+c{\lambda }_3=0\end{array}$$

Example (c): If some vector in a list of vectors in $V$ is a linear combination of the other vectors. Then the list is linearly dependent. Proof: After writing one vector in the list as equal to a linear combination of the other vectors, move that vector to the other side of the equation, where it would be multiplied by $-1$, i.e:

$${\lambda }_{1}a+{\lambda }_{2}b={\lambda }_{3}c⟹{\lambda }_{1}a+{\lambda }_{2}b-{\lambda }_{3}c=0$$

Example (d): Every list of vectors in $V$ containing the $0$ vector is linearly dependent. This is the special case of the above.

See also: Linear Dependence Lemma