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 Linear Dependence


Definition

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

(1)


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


                                     a1= a2=...=ak=0



If S = {v1,v2,...,vk} ,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,...,vk, Equation (1) always holds if we choose all the scalars a1,a2,...,ak, 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=...=ak=0 is a solution. However, the main idea from the Definition is 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 + a2a3 = 0

2a1 + 2a2 + 2a3 = 0

a1a3 = 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) $

and

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


linearly dependent or linearly independent?


Solution We 

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood