Verifying Linear Independence

Linear dependence

We defined linear dependence as a collection with linearly dependent vectors if, given a collection of vectors $\bold x_1, \bold x_2, \cdots, \bold x_n$, there exists a combination of scalars $w_1,w_2,\cdots,w_n$ with at least one nonzero value such that

$$w_1\bold{x_1}+w_2\bold{x_2}+...+w_n\bold{x_n}=\bold{0}$$

There could be infinitely many possible linear combinations, and it’s not possible to check them one by one. We’ve already discussed a few special cases and tricks to check for linear dependence in a combination. This lesson will use elimination as a formal method to solve the vector equation and find the possible combinations.

Procedure

Our goal is to find the combination of scalars $w_1,w_2,\cdots,w_n$, such that

$$w_1\bold{x_1}+w_2\bold{x_2}+...+w_n\bold{x_n}=\bold{0}$$

Expanding the vectors, we get

$$w_1 \begin{bmatrix} x_{11} \\ x_{12} \\ \vdots \\ x_{1m} \end{bmatrix} + w_2 \begin{bmatrix} x_{21} \\ x_{22} \\ \vdots \\ x_{2m} \end{bmatrix} + \cdots + w_n \begin{bmatrix} x_{n1} \\ x_{n2} \\ \vdots \\ x_{1m} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$$

The equation can be converted to a linear system of $m$ equations in $n$ variables. The augmented matrix representation of the system is as follows:

$$\left( \begin{array}{cccc|c} x_{11} & x_{21} & \cdots & x_{n1} & 0 \\ x_{12} & x_{22} & \cdots & x_{n2} & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots\\ x_{1m} & x_{2m} & \cdots & x_{nm} & 0 \\ \end{array}\right)$$

Because the vector $\bold b$ is a zero vector, the system is of the form $A\bold{w}=\bold{0}$ and is always consistent. The two possible solution sets, “only trivial solution” and “infinite solutions,” provide evidence for linear independence and linear dependence, respectively.

Only trivial solution

Suppose the $rref$ of $A$ for the system $A\bold w=\bold 0$ results in an identity matrix (that is, it has no free variable). In such a case, the only possible solution is the trivial solution, $\bold w=\bold 0$. This means that the only combination of scalars satisfying the vector equation is all zeros, and the vectors included in the collection are independent of each other.

Example

Consider the following set of vectors:

$$\bold {x_1} = \begin{bmatrix} 5 \\ 2 \\ 4 \end{bmatrix}, \bold {x_2} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \bold {x_3} = \begin{bmatrix} 10 \\ 9 \\ 15 \end{bmatrix}$$

Writing the vector equation, we get

$$w_1 \begin{bmatrix} 5 \\ 2 \\ 4 \end{bmatrix} + w_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + w_3 \begin{bmatrix} 10 \\ 9 \\ 15 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

Our next step is to write an augmented matrix and apply elimination.

$$\left( \begin{array}{ccc|c} 5 & 0 & 10 & 0 \\ 2 & 1 & 9 & 0 \\ 4 & 1 & 15 & 0 \\ \end{array}\right)$$