Linear Independence to Determining Vectors' Dependence"

[Gregg]

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

<br /> \text{Det}\left[\left(<br /> \begin{array}{ccc}<br /> 1 &amp; 0 &amp; 1 \\<br /> 1 &amp; 1 &amp; 0 \\<br /> 0 &amp; 1 &amp; 1<br /> \end{array}<br /> \right)\right]=2

Independant

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

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

 

[rock.freak667]

Where'd you get you those co-efficient matrices from?
What is the definition of vectors being linearly dependent?

 

[Mark44]

That actually works. For example, if you row-reduce the matrix in c), you get
\left(\begin{array}{ccc} 1 &amp; 0 &amp; 2 \\ 0 &amp; 1 &amp; 3 \\ 0 &amp; 0 &amp; 0\end{array}\right)

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.

 

[Gregg]

That actually works. For example, if you Note spelling of "dependent" Gregg. Similar for independent.

whoops

 

[HallsofIvy]

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 \left{v_1, v_2, \cdot\cdot\cdot v_n}[/quote] is that there are numbers, a_1, a_2, \cdot\cdot\cdot a_n, <b>not</b> all 0, such that [math]a_1v_1+ a_2v_2+ \cdot\cdot\cdot a_nv_n= 0.

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

For (b), with a+b, b+c, c+a, if a_1(a+b)+ a_2(b+ c)+ a_3(c+a)= 0 then (a_1+ a_3)a+ (a_1+ a_2)b+ (a_2+ a_3)c= 0. Since a, b, and c are independent, we must have a_1+ a_3= 0, a_1+ a_2= 0, and a_2+ a_3= 0. Obviously, a_0= a_1= a_2= 0 satisfies that but is it the only solution?

 

[Mark44]

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 c_1 v_1 + c_2 v_2 + c_3 v_3 + ... + c_n v_n + = 0 has only one solution (independent vectors) or an infinite number of solutions (dependent vectors).