(New page:  Linear Dependence <br> Definition The vectors v<sub>1</sub>,v<sub>2</sub>,...,v<sub>k</sub> in a vector space V are said to be linearly dependent if there exist constans a...)
 
Line 1: Line 1:
&nbsp;Linear Dependence  
+
&nbsp;<u>'''Linear Dependence'''</u>
  
 
<br>  
 
<br>  
  
Definition  
+
'''Definition'''
  
 
The vectors v<sub>1</sub>,v<sub>2</sub>,...,v<sub>k</sub>&nbsp;in a vector space V are said to be linearly dependent if there exist constans a<sub>1</sub>,a<sub>2</sub>,...,a<sub>k</sub>, not all zero, such that<br>  
 
The vectors v<sub>1</sub>,v<sub>2</sub>,...,v<sub>k</sub>&nbsp;in a vector space V are said to be linearly dependent if there exist constans a<sub>1</sub>,a<sub>2</sub>,...,a<sub>k</sub>, not all zero, such that<br>  
  
(1)
+
(1)  
  
 +
<br>
 +
 +
Otherwise, v1,v2,...,vk are called linearly independent. That is, v1,v2,...,vk are linearly independent if, whenever a1,v1+a2v2+...+akvk=0,
 +
 +
<br>
 +
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; a1=a2=...=ak=0
 +
 +
<br>
 +
 +
<br>
 +
 +
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.
 +
 +
<br>
 +
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;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.
 +
 +
<br>
 +
 +
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.
 +
 +
<br>
 +
 +
&nbsp; &nbsp; &nbsp; &nbsp; 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 (q) 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.
 +
 +
<br>
 +
 +
Example 1
 +
 +
Determine whether the vectors
 +
 +
<math>v1=\left(\begin{array}{cccc}3\\2\\1\end{array}\right)</math>
  
 +
<math>v2=\left(\begin{array}{cccc}1\\2\\0\end{array}\right)</math>
  
Otherwise,&nbsp;v<sub>1</sub>,v<sub>2</sub>,...,v<sub>k are called linearly independent.</sub>
+
<math>v3=\left(\begin{array}{cccc}-1\\-2\\-1\end{array}\right)</math>

Revision as of 16:20, 7 December 2011

 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 a1,v1+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 (q) 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) $

Alumni Liaison

Meet a recent graduate heading to Sweden for a Postdoctorate.

Christine Berkesch