Recall from Lecture 10that the homogeneous vector equation \( x_1 \bbm v_1 + x_2 \bbm v_2 + \cdots x_p \bbm v_p = \bbm 0 \) always has the "trivial" solution \( \bbm x = \bbm 0 \), and may have other solutions in general.
Definition. A set of vectors \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \) in \( \mathbb R^n \) is said to be linearly independent if the vector equation \( x_1 \bbm v_1 + x_2 \bbm v_2 + \cdots x_p \bbm v_p = \bbm 0 \) has only the trivial solution \( \bbm x = \bbm 0 \).
Example 1. Show that the set \( \left \{ \vecfour 2 0 {-1} 1, \vecfour 1 1 0 {-3}, \vecfour 0 0 4 5 \right \} \) is linearly independent.
We consider the homogeneous vector equation \( x_1 \vecfour 2 0 {-1} 1 + x_2 \vecfour 1 1 0 {-3} + x_3 \vecfour 0 0 4 5 = \vecfour 0000 \). Set up the augmented matrix and row-reduce: \[ \begin{bmatrix} 2 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 4 & 0 \\ 1 & -3 & 5 & 0 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
Since our matrix has a pivot in every column (except the last column), the vector equation has no free variables. The only solution is \( x_1 = x_2 = x_3 = 0 \), and so this set of vectors is linearly independent. \( \Box \)
Example 2. Is the set \( \left \{ \vectwo 1 {-2}, \vectwo {-3} 7, \vectwo 2 2, \vectwo {-6} 0 \right \} \) linearly independent?
The question asks us to think about the vector equation \( x_1 \vectwo 1 {-2} + x_2 \vectwo {-3} 7 + x_3 \vectwo 2 2 + x_4 \vectwo {-6} 0 = \vectwo 00 \).
We set up and row-reduce the corresponding augmented matrix: \[ \begin{bmatrix} 1 & -3 & 2 & -6 & 0 \\ -2 & 7 & 2 & 0 & 0 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 20 & -42 & 0 \\ 0 & 1 & 6 & -12 & 0 \end{bmatrix} \]
Since this matrix has columns (other than the last one) that do not contain pivots, the vector equation has free variables. This means that the equation has infinitely many solutions, and so the given set of vectors is not linearly independent. \( \Box \)
Definition. A set of vectors \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \) in \( \mathbb R^n \) is said to be linearly dependent if it is not linearly independent.
As an alternative to the above defintion, we can say that a set of vectors \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \) is linearly dependent if the vector equation \( x_1 \bbm v_1 + x_2 \bbm v_2 + \cdots x_p \bbm v_p = \bbm 0 \) has solutions other than \( \bbm x = \bbm 0 \).
When a set of vectors \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \) is linearly dependent, there will exist scalars \( c_1, c_2, \ldots, c_p \) that are not all zero and for which \( c_1 \bbm v_1 + c_2 \bbm v_2 + \cdots c_p \bbm v_p = \bbm 0 \). In this case, we call such an equation a dependence relation for the set.
Example 3. Find a dependence relation for the set \( \{ \bbm v_1, \bbm v_2, \bbm v_3 \} \), where \( \bbm v_1 = \vecthree 123 \), \( \bbm v_2 = \vecthree 456 \), and \( \bbm v_3 = \vecthree 210 \).
We are looking for a nontrivial solution of the equation \( x_1 \bbm v_1 + x_2 \bbm v_2 + x_3 \bbm v_3 = \bbm 0 \). Set up and row-reduce the augmented matrix: \[ \begin{bmatrix} 1 & 4 & 2 & 0 \\ 2 & 5 & 1 & 0 \\ 3 & 6 & 0 & 0 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
We write the general solution, which is \( x_1 = 2 x_3 \), \( x_2 = -x_3 \), and \( x_3 \) is free. We can choose any value for \( x_3 \) other than zero, since we are looking for a nontrivial solution. For example, if \( x_3 = 5 \), then this gives the dependence relation \( 10 \bbm v_1 + (-5) \bbm v_2 + 5 \bbm v_3 = \bbm 0. \Box \)
Example 4. Let \( \bbm u_1 = \vecthree 1 0 {-2} \), \( \bbm u_2 = \vecthree {-1} 1 0 \), \( \bbm u_3 = \vecthree {-1} 2 {-1} \), and \( \bbm u_4 = \vecthree 0 1 0 \). Is the set \( \{ \bbm u_1, \bbm u_2, \bbm u_3, \bbm u_4 \} \) linearly independent? If not, find a dependence relation.
As usual, we consider the homogeneous vector equation \( x_1 \bbm u_1 + x_2 \bbm u_2 + x_3 \bbm u_3 + x_4 \bbm u_4 = \bbm 0 \). The augmented matrix is: \[ \begin{bmatrix} 1 & -1 & -1 & 0 & 0 \\ 0 & 1 & 2 & 1 & 0 \\ -2 & 0 & -1 & 0 & 0 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & -3 & 0 \\ 0 & 0 & 1 & 2 & 0 \end{bmatrix} \]
Since there is no pivot in the fourth column, the vector equation has infinitely many solutions, and so the given set of vectors is linearly dependent. To find a dependence relation, we find any nontrivial solution.
We write the general solution, which is \( x_1 = x_4 \), \( x_2 = 3x_4 \), \( x_3 = -2x_4 \), and \( x_4 \) is free. We can now choose any nonzero value for the free variable, \( x_4 \). For example, if \( x_4 = 1 \), we get the dependence relation \( 1\bbm u_1 + 3\bbm u_2 + (-2)\bbm u_3 + 1\bbm u_4 = \bbm 0. \Box \)
Given a set of vectors \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \) in \( \mathbb R^n \), we consider the homogeneous vector equation \( x_1 \bbm v_1 + x_2 \bbm v_2 + \cdots + x_p \bbm v_p = \bbm 0 \).
If this equation has a non-trivial solution, then the set \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \) is linearly dependent. Any particular non-trivial solution gives a dependence relation for these vectors.
If this equation has only the trivial solution, then the set \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \) is linearly independent.
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