Revision as of 17:50, 16 December 2011

Linear Dependence

Definition

The vectors v₁,v₂,...,v_k in a vector space V are said to be linearly dependent if there exist constans a₁,a₂,...,a_k not all zero, such that

                                                                           $ \sum_{j=1}^k a_1v_1+a_2v_2+...+a_kv_k=0 $                       (1)                                 

Otherwise, v₁,v₂,...,v_k are called linearly independent. That is v₁,v₂,...,v_k are linearly independent if, whenever a₁v₁+ a₂v₂+...+a_kv_k=0

                                                                       a₁= a₂=...=a_k=0

If S = {v₁,v₂,...,v_k} ,then we also say that the set S is linearly dependent or linearly independent if the vectors have the corresponding property.

         It should be emphasized that for any vectors v₁,v₂,...,v_k, Equation (1) always holds if we choose all the scalars a₁,a₂,...,a_k, equal to zero. The important point in this definition is whether it is possible to satisfy (1) with at least one of the scalars different from zero.

Remark: Definition is started for a finite set of vectors, but it also applies to an infinite set S of a vector space, using corresponding notation for infinite sums.

        To determine whether a set of vectors is linearly independent or linearly dependent, we use Equation (1). Regardless of the form of the vectors, Equation (1) yields a homogeneous linear system of equations. it is always consistent, since a₁=a₂=...=a_k=0 is a solution. However, the main idea from the Definitionis whether there is a nontrivial solution.

Example 1

Determine whether the vectors

$ v1=\left[\begin{array}{cccc}3\\2\\1\end{array}\right] $

$ v2=\left[\begin{array}{cccc}1\\2\\0\end{array}\right] $

$ v3=\left[\begin{array}{cccc}-1\\-2\\-1\end{array}\right] $

are linearly independent.

Solution

Forming Equation (1)

$ a1\left[\begin{array}{cccc}3\\2\\1\end{array}\right] $$ +a2\left[\begin{array}{cccc}1\\2\\0\end{array}\right] $$ +a3\left[\begin{array}{cccc}-1\\2\\-1\end{array}\right] $$ =\left[\begin{array}{cccc}0\\0\\0\end{array}\right] $

we obtain the homogeneous system

3a₁ + a₂ − a₃ = 0

2a₁ + 2a₂ + 2a₃ = 0

a₁               − a₃ = 0

The corresponding augmented matrix is

$ \left[\begin{array}{cccc}3&1&-1&|0\\2&2&2&|0\\1&0&-1&|0\end{array}\right] $

Whose reduced row echelon form is

$ \left[\begin{array}{cccc}1&0&-1&|0\\0&1&2&|0\\0&0&0&|0\end{array}\right] $

This there is a nontrivial solution

$ \left[\begin{array}{cccc}k\\-2k\\k\end{array}\right] $       k is not equal to 0

so the vectors are linearly dependent.

Example 2

Are the vectors

$ v1=\left[\begin{array}{cccc}1&0&1&2\end{array}\right] $

$ v2=\left[\begin{array}{cccc}0&1&1&2\end{array}\right] $

$ v3=\left[\begin{array}{cccc}1&1&1&3\end{array}\right] $     in R₄

linearly dependent or linearly independent?

Solution

We form Equation (1),

a₁v₁+a₂v₂+a₃v₃=0

and solve for a₁,a₂,a₃.

The resulting homogeneous system is

a₁ + a₃ = 0

a₂ + a₃ = 0

a₁ + a₂ + a₃ = 0

2a₁ + 2a₂ + 3a₃ = 0

The corresponding augmented matrix is

$ \left[\begin{array}{cccc}1&0&1&|0\\0&1&1&|0\\1&1&1&|0\\2&2&3&|0\end{array}\right] $

and its reduced row echelon form is

$ \left[\begin{array}{cccc}1&0&0&|0\\0&1&0&|0\\0&0&1&|0\\0&0&0&|0\end{array}\right] $

Thus the only solution is the trivial solution

a₁ = a₂ = a₃ = 0

so the vectors are linearly independent.

Theorem</u

<u>

Example 3

Is     $ S=\left[\begin{array}{cccc}1&2&3\end{array}\right] $ $ ,\left[\begin{array}{cccc}0&1&2\end{array}\right] $ $ ,\left[\begin{array}{cccc}3&0&-1\end{array}\right] $

a linearly independent set of vectors in R³?

Solution

We form the matrix A whose rows are the vectors in S:

$ A=\left[\begin{array}{cccc}1&2&3\\0&1&2\\3&0&-1\end{array}\right] $

Since det(A) = 2, we conclude that S is linearly independent.

Theorem

Let S₁ and S₂ be finite susbsets of a vector space and let S₁ be a subset of S₂. Then the following statements are true:

(a) If S₁ is linearly dependent, so is S₂.

(b) If S₂ is linearly independent, so is S₁.

Theorem</u

The nonzero vectors v₁,v₂,...,v_n in a vector space V are linearly dependent if and only if one of the vectors v_j $ (2 \le j) $ is a linear combination of the preceding vectors v₁,v₂,...,v_{j − 1.}

Example 4

Let V=R₃ and also 

$ v1=\left[\begin{array}{cccc}1&2&-1\end{array}\right] $

$ v2=\left[\begin{array}{cccc}1&-2&1\end{array}\right] $

$ v3=\left[\begin{array}{cccc}-3&2&-1\end{array}\right] $

$ v4=\left[\begin{array}{cccc}2&0&0\end{array}\right] $

We find that

                                            v₁ + v₂ + 0v₃ − v₄ = 0,

so v_!,v₂,v₃, and v₄ 

are linearly dependent. We then have

                                             v₄ = v₁ + v₂ + 0v₃

Remarks

1. We observe that the previous Theorem does not say that every vector v is a linear combination of the preceding vectors. Thus, in the previous example, we also have v₁ + 2v₂ + v₃ + 0v₄ = 0.

We cannot solve, in this equation, for v₄ as a linear combination of

v₁,v₂, and v₃, since its coefficient is zero.

2. We can also prove that if S = {v₁,v₂,...,v_k} is a set of vectors in a vector space V, then S is linearly dependent if and only if one of the vectors in S is a linear combination of all the other vectors in S. For instance in the previous example,

                                                   v₁ = − v₂ − 0v₃ + v₄;      $ v_2= \frac{-1}{2} v1 - \frac{1}{2} v_3 -0v_4 $

3. Observe that if v₁,v₂,...,v_k

are linearly independent vectors in a vector space, then they must be distinct and nonzero.