Linear Independence of Vectors
Linearly Dependent and Linearly Independent Vectors
A linear combination of n vectors a 1 , a 2 , ..., a n
\[\alpha_1 \mathbf{a}_1 + \alpha_2 \mathbf{a}_2 + \ldots + \alpha_n \mathbf{a}_n,\]
where α 1 , α 2 , ..., α n
Vectors a 1 , a 2 , ..., a n linearly dependent if there are such real numbers α 1 , α 2 , ..., α n a 1 , a 2 , ..., a n
\[\alpha_1 \mathbf{a}_1 + \alpha_2 \mathbf{a}_2 + \ldots + \alpha_n \mathbf{a}_n = \mathbf{0}.\]
Vectors a 1 , a 2 , ..., a n linearly independent . 🤓
More formally, vectors a 1 , a 2 , ..., a n linearly independent if their linear combination is equal to zero only in the case when ALL coefficients α 1 , α 2 , ..., α n
Notice that if at least one of the vectors a 1 , a 2 , ..., a n zero vector , then these vectors are linearly dependent .
If a set of n vectors are linearly dependent, then adding one or more vectors does not change the linear dependence of vectors in the set.
Linear Combinations of Two Vectors
Two vectors are linearly dependent if and only if they are collinear .
Two linearly dependent vectors a and b must satisfy the relation
\[\mathbf{b} = \lambda \mathbf{a},\]
where \(\lambda\) is a real number.
Respectively, if the vectors a and b are not collinear , then they are linearly independent .
Linear Combinations of Three Vectors. Coplanar Vectors
Vectors are called coplanar if they lie in the same plane or in parallel planes.
A system of three vectors is linearly dependent if and only if the vectors are coplanar .
Let vectors a , b and c be coplanar and vectors a and b not collinear. We bring the vectors a , b , c to a common origin O and draw a parallelogram ONCM as shown in the figure.
Figure 1. Express the vector c in terms of a and b . It is clear that
\[\mathbf{c} = \mathbf{OC} = \mathbf{OM} + \mathbf{ON}.\]
The vectors a and OM are collinear. Therefore \(\mathbf{OM} = \lambda\mathbf{a}.\) Similarly, we have \(\mathbf{ON} = \mu\mathbf{b}\) as b and ON are also collinear vectors. This yields:
\[\mathbf{c} = \lambda \mathbf{a} + \mu\mathbf{b}.\]
So if a and b are not collinear vectors then any vector c coplanar to them can be represented as a linear combination of these vectors.
Linear Dependence of Four Vectors
Any four vectors are linearly dependent .
If a , b , c are non-coplanar vectors , then any vector d can be decomposed into vectors a , b and c , that is, it can be represented as
\[\mathbf{d} = \lambda \mathbf{a} + \mu\mathbf{b} + \nu\mathbf{c}.\]
It can be shown that such a decomposition is unique.
Figure 2. Solved Problems
Example 1.
Check if the vectors
\[\mathbf{a} = \left({1,2,3}\right), \mathbf{b} = \left({4,5,6}\right), \mathbf{c} = \left({2,1,0}\right)\]
are linearly independent?
Solution.
Vectors \(\mathbf{a},\) \(\mathbf{b},\) \(\mathbf{c}\) are linearly independent if the vector equation
\[\alpha_1 \mathbf{a} + \alpha_2 \mathbf{b} + \alpha_3 \mathbf{c} = \mathbf{0}\]
has only the trivial solution.
Consider the augmented matrix and bring it to a triangular form:
\[\left[ \begin{array}{*{20}{r}}
1 & 4 & 2 & 0\\
2 & 5 & 1 & 0\\
3 & 6 & 0 & 0
\end{array} \right], \Rightarrow \left[ \begin{array}{*{20}{r}}
1 & 4 & 2 & 0\\
0 & -3 & -3 & 0\\
0 & -6 & -6 & 0
\end{array} \right], \Rightarrow \left[ \begin{array}{*{20}{r}}
1 & 4 & 2 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & 0
\end{array} \right],\]
or
\[\left\{ \begin{array}{*{20}{r}}
\alpha_1 + 4\alpha_2 + 2\alpha_3 = 0\\
\alpha_2 + \alpha_3 = 0
\end{array} \right..\]
We have two independent equations with three variables. One of the variables, for example, \(\alpha_3\) can be considered free. So let \(\alpha_3 = 1.\) Then
\[\alpha_2 = -\alpha_3 = -1,\;\alpha_1 = -4\alpha_2 - 2\alpha_3 = -4\cdot\left({-1}\right) - 2\cdot1 = 2.\]
So there are non-zero values of \(\alpha_1, \alpha_2, \alpha_3\) such that
\[\alpha_1 \mathbf{a} + \alpha_2 \mathbf{b} + \alpha_3 \mathbf{c} = 2\mathbf{a} - \mathbf{b} + \mathbf{c} = \mathbf{0}.\]
Hence the vectors \(\mathbf{a},\) \(\mathbf{b},\) \(\mathbf{c}\) are linearly dependent .
Example 2.
Given three vectors
\[\mathbf{a}_1 = \left({2,3,4,5}\right), \mathbf{a}_2 = \left({-3,6,-6,-4}\right), \mathbf{a}_3 = \left({-3,3,-6,-5}\right).\]
Determine whether these vectors are linearly dependent.
Solution.
Linearly dependent vectors satisfy the relation
\[\alpha_1 \mathbf{a}_1 + \alpha_2 \mathbf{a}_2 + \alpha_3 \mathbf{a}_3 = \mathbf{0}.\]
We represent this vector equation in matrix form:
\[\alpha_1 \left[ \begin{array}{*{20}{r}}
2\\
3\\
4\\
5
\end{array} \right] + \alpha_2 \left[ \begin{array}{*{20}{r}}
-3\\
6\\
-6\\
-4
\end{array} \right] + \alpha_3 \left[ \begin{array}{*{20}{r}}
-3\\
3\\
-6\\
-5
\end{array} \right] = \left[ \begin{array}{*{20}{r}}
0\\
0\\
0\\
0
\end{array} \right].\]
We obtain the following system of equations:
\[\left\{ \begin{array}{l}
2\alpha_1 - 3\alpha_2 - 3\alpha_3 = 0\\
3\alpha_1 + 6\alpha_2 + 3\alpha_3 = 0\\
4\alpha_1 - 6\alpha_2 - 6\alpha_3 = 0\\
5\alpha_1 - 4\alpha_2 - 5\alpha_3 = 0
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
2\alpha_1 - 3\alpha_2 - 3\alpha_3 = 0\\
\alpha_1 + 2\alpha_2 + \alpha_3 = 0\\
2\alpha_1 - 3\alpha_2 - 3\alpha_3 = 0\\
5\alpha_1 - 4\alpha_2 - 5\alpha_3 = 0
\end{array} \right..\]
Note that the first and third equations are the same, so only 3 equations are independent. Solve the system by elimination.
\[\left\{ \begin{array}{l}
2\alpha_1 - 3\alpha_2 - 3\alpha_3 = 0\\
\alpha_1 + 2\alpha_2 + \alpha_3 = 0\\
5\alpha_1 - 4\alpha_2 - 5\alpha_3 = 0
\end{array} \right..\]
From the second equation we find
\[\alpha_1 = -2\alpha_2 - \alpha_3.\]
Substituting into the other two equations we get
\[\left\{ \begin{array}{l}
2\left({-2\alpha_2 - \alpha_3}\right) - 3\alpha_2 - 3\alpha_3 = 0\\
5\left({-2\alpha_2 - \alpha_3}\right) - 4\alpha_2 - 5\alpha_3 = 0
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
-4\alpha_2 - 2\alpha_3 - 3\alpha_2 - 3\alpha_3 = 0\\
-10\alpha_2 - 5\alpha_3 - 4\alpha_2 - 5\alpha_3 = 0
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
-7\alpha_2 - 5\alpha_3 = 0\\
-14\alpha_2 - 10\alpha_3 = 0
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
7\alpha_2 + 5\alpha_3 = 0\\
7\alpha_2 + 5\alpha_3 = 0
\end{array} \right., \Rightarrow 7\alpha_2 + 5\alpha_3 = 0.\]
We got one independent equation with two variables. Let \(\alpha_3 = 7\) (free variable). Then \(\alpha_2 = -5.\) The third coefficient is
\[\alpha_1 = -2\alpha_2 - \alpha_3 = -2\cdot\left({-5}\right) - 7 = 3.\]
So the vectors \(\mathbf{a}_1,\) \(\mathbf{a}_2,\) \(\mathbf{a}_3\) are linearly dependent since
\[3\mathbf{a}_1 - 5\mathbf{a}_2 + 7\mathbf{a}_3 = \mathbf{0}.\]
Consider the augmented matrix and bring it to a triangular form:
\[3\mathbf{a}_1 - 5\mathbf{a}_2 + 7\mathbf{a}_3 = \mathbf{0}.\]
Example 3.
Check if the following vectors are coplanar:
\[\mathbf{a} = \left({2,3,-1}\right), \mathbf{b} = \left({2,1,3}\right), \mathbf{c} = \left({0,-1,2}\right).\]
Solution.
Coplanar vectors must satisfy the relation
\[\mathbf{c} = \lambda\mathbf{a} + \mu\mathbf{b}.\]
Let's see if there are such numbers \(\lambda\) and \(\mu:\)
\[\left({0,-1,2}\right) = \lambda\left({2,3,-1}\right) + \mu\left({2,1,3}\right), \Rightarrow \left\{ \begin{array}{l}
0 = 2\lambda + 2\mu\\
-1 = 3\lambda + \mu\\
2 = -\lambda + 3\mu
\end{array} \right..\]
Solve the first two equations of the system and find \(\lambda\) and \(\mu:\)
\[\left\{ \begin{array}{l}
0 = 2\lambda + 2\mu\\
-1 = 3\lambda + \mu
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
\lambda = -\mu\\
-1 = -3\mu + \mu
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
\lambda = -\mu\\
-1 = -2\mu
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
\lambda = -\frac{1}{2}\\
\mu = \frac{1}{2}
\end{array} \right..\]
Check if the third equation is true:
\[-\lambda + 3\mu = -\left({-\frac{1}{2}}\right) + 3\frac{1}{2} = \frac{1}{2} + \frac{3}{2} = 2.\]
As you can see, the third equation is satisfied, so all three vectors are coplanar.
Example 4.
Find out if these points are located in the same plane:
\[A\left({6,2,7}\right), B\left({3,0,2}\right), C\left({8,3,14}\right), D\left({5,1,9}\right).\]
Solution.
The points A , B , C , D belong to the same plane if and only if the vectors \(\mathbf{AB},\) \(\mathbf{AC}\) and \(\mathbf{AD}\) are coplanar. So let's find the given vectors and check their coplanarity.
\[\mathbf{AB} = \left({-3,-2,-5}\right),\;\mathbf{AC} = \left({2,1,7}\right),\;\mathbf{AD} = \left({-1,-1,2}\right).\]
Figure 3. If the vectors are coplanar, then we have
\[\mathbf{AB} = \lambda\mathbf{AC} + \mu\mathbf{AD}.\]
This yields:
\[\left({-3,-2,-5}\right) = \lambda\left({2,1,7}\right) + \mu\left({-1,-1,2}\right), \Rightarrow \left\{ \begin{array}{l}
-3 = 2\lambda - \mu\\
-2 = \lambda - \mu\\
-5 = 7\lambda + 2\mu
\end{array} \right..\]
From the first two equations we get
\[\left\{ \begin{array}{l}
-3 = 2\lambda - \mu\\
-2 = \lambda - \mu
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
\mu = 2\lambda + 3\\
-2 = \lambda - \left({2\lambda + 3}\right)
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
\mu = 2\lambda + 3\\
-2 = \lambda - 2\lambda - 3
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
\mu = 2\lambda + 3\\
\lambda = - 1
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
\mu = 1\\
\lambda = - 1
\end{array} \right..\]
Plugging this in the third equation, we obtain:
\[7\lambda + 2\mu = 7\cdot\left({-1}\right) + 2\cdot1 = -5,\]
which is the correct answer. Therefore, these four points lie in the same plane.
