Linear independance

  • Thread starter Gregg
  • Start date
  • #1
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linear independence

Homework Statement



given that a,b and c are linearly independant vectors determine if the following vectors are linearly independant.

a) a,0

b) a+b, b+c, c+a

c) a+2b+c, a-b-c, 5a+b-c

The Attempt at a Solution



I'm not sure how to tackle the question in this form.

Edit:

a) a=0a Dependant

[itex]
\text{Det}\left[\left(
\begin{array}{ccc}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{array}
\right)\right]=2[/itex]

Independant

(c)
[itex]
\text{Det}\left[\left(
\begin{array}{ccc}
1 & 1 & 5 \\
2 & -1 & 1 \\
1 & -1 & -1
\end{array}
\right)\right]=0[/itex] Dependant

Is it ok to use those vector co-efficients in a matrix like that?
 
Last edited:

Answers and Replies

  • #2
rock.freak667
Homework Helper
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31
Where'd you get you those co-efficient matrices from?
What is the definition of vectors being linearly dependent?
 
  • #3
34,936
6,699
That actually works. For example, if you row-reduce the matrix in c), you get
[tex]\left(\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0\end{array}\right)[/tex]

This says that c1 = -2c3, c2 = -3c3, and c3 = c3. If you take c3 = 1, then c1 = -2 and c2 = -3. That linear combination of the vectors given in part c results in a sum of 0, thus demonstrating that the set is linearly dependent. Note spelling of "dependent" Gregg. Similar for independent.
 
  • #4
459
0
That actually works. For example, if you Note spelling of "dependent" Gregg. Similar for independent.

whoops
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
964
I think it is always better to use the basic definitions than try to memorize a specific method without understanding it.

The definition of "dependent" for a set of vectors [itex]\left{v_1, v_2, \cdot\cdot\cdot v_n}[/quote] is that there are numbers, [itex]a_1, a_2, \cdot\cdot\cdot a_n[/itex], not all 0, such that \(\displaystyle a_1v_1+ a_2v_2+ \cdot\cdot\cdot a_nv_n= 0[/itex].

For the first problem, { a, 0}, take [itex]a_0= 0[/itex], [itex]a_1= 1[/itex]: [itex]a_0a+ a_10= 0(a)+ 1(0)= 0[/itex].

For (b), with a+b, b+c, c+a, if [itex]a_1(a+b)+ a_2(b+ c)+ a_3(c+a)= 0[/itex] then [itex](a_1+ a_3)a+ (a_1+ a_2)b+ (a_2+ a_3)c= 0[/itex]. Since a, b, and c are independent, we must have [itex]a_1+ a_3= 0[/itex], [itex]a_1+ a_2= 0[/itex], and [itex]a_2+ a_3= 0[/itex]. Obviously, [itex]a_0= a_1= a_2= 0[/itex] satisfies that but is it the only solution?\)
 
  • #6
34,936
6,699
I couldn't agree with HallsOfIvy more, in what he said about the importance of understanding definitions as opposed to memorizing a technique without understanding why you are doing it. To often students get tangled up in the details of calculating a determinant or row reducing a matrix without understanding what it means that the matrix determinant is zero or why the matrix should be row reduced.

The definition of linear independence of a set of vectors is stated very simply, but there is a subtlety to it that escapes many students. The only thing that distinguishes a set of linearly independent vectors from a set that is linearly dependent is whether the equation [itex]c_1 v_1 + c_2 v_2 + c_3 v_3 + ... + c_n v_n + = 0[/itex] has only one solution (independent vectors) or an infinite number of solutions (dependent vectors).
 

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