Linear Algebra: Linear dependence of vectors?

So they are certainly not linearly independent. In fact, since we can write v1 as a linear combination of v2 and v3 (v1= -v2+ v3), v1, v2, v3 are not linearly independent.
  • #1
SticksandStones
88
0

Homework Statement


Given:
[tex]v_1 = \left(\begin{array}{cc}1\\-5\\-3\end{array}\right)

v_2 = \left(\begin{array}{cc}-2\\10\\6\end{array}\right)

v_3 = \left(\begin{array}{cc}2\\-9\\h\end{array}\right)

[/tex]

For what value of h is [tex]v_3[/tex] in Span{[tex]v_1, v_2[/tex]} and for what value of h is {[tex]v_1,v_2,v_3[/tex]} linearly dependent?

Homework Equations





The Attempt at a Solution


If my understanding is correct, a vector is in a span of another set of vectors if all vectors are in the same plane? If so then my guess is h would have to be a multiple of -3.

The linear dependence part of the question is getting me though. I was under the impression that all three vectors would have to be multiples of each other for it to be linearly dependent?
 
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  • #2
For the first question.
How can you express the vector space [itex]V=Span\{u_1,u_2\}[/itex]?
 
  • #3
Rainbow Child said:
For the first question.
How can you express the vector space [itex]V=Span\{u_1,u_2\}[/itex]?

A plane?
 
  • #4
Yes, and the equation that desribes that plane, is which?
 
  • #5
SticksandStones said:

Homework Statement


Given:
[tex]v_1 = \left(\begin{array}{cc}1\\-5\\-3\end{array}\right)

v_2 = \left(\begin{array}{cc}-2\\10\\6\end{array}\right)

v_3 = \left(\begin{array}{cc}2\\-9\\h\end{array}\right)

[/tex]

For what value of h is [tex]v_3[/tex] in Span{[tex]v_1, v_2[/tex]} and for what value of h is {[tex]v_1,v_2,v_3[/tex]} linearly dependent?

Homework Equations





The Attempt at a Solution


If my understanding is correct, a vector is in a span of another set of vectors if all vectors are in the same plane?
You understanding is not quite correct. A vector is in a span of another set of vectors if it can be written as a linear combination of the vectors in the set. The span of the set of vectors is a plane only if there are exactly two independent vectors in the set. That is not true in this problem!

If so then my guess is h would have to be a multiple of -3.

The linear dependence part of the question is getting me though. I was under the impression that all three vectors would have to be multiples of each other for it to be linearly dependent?

Are these really the problem you were given? Have you copied the vectors correctly? Notice that -2v1= v2. The span of these two vectors has dimension 1, not 2. v3 will be in their span only if it also is a multiple of either v1 or v2. The first component of v3, 2, is 2 times the first component of v1 (and -1 times the first component of v2) but the second component of v3 is -9 which is not 2 times the second component of v1 (and not -1 times the second component of v2).

v3 while not be in the span of v1 and v2 for any value of h.


(If the second component of v3 were -10, then v3 would be in the span of v1 and v2 for h= -6.)
 
  • #6
The span of the set of vectors is a plane only if there are exactly two independent vectors in the set. That is not true in this problem!
Ahhh, Ok. That makes a lot more sense.

Are these really the problem you were given? Have you copied the vectors correctly?
Yes, these really are the vectors.

The span of these two vectors has dimension 1, not 2.
Just because making assumptions has burned me in the past, you mean the span is a line, correct?

v3 while not be in the span of v1 and v2 for any value of h.
After going back and re-reading the chapter this is what I thought. Thanks for confirming it though.

Since v3 isn't a linear combination of v1 and v2, this means that the set of {v1, v2, v3} has to be linearly independent, right?
 
  • #7
SticksandStones said:
Just because making assumptions has burned me in the past, you mean the span is a line, correct?
Yes, the linear span of v1 and v2 is a line in R3.

Since v3 isn't a linear combination of v1 and v2, this means that the set of {v1, v2, v3} has to be linearly independent, right?
What does it mean for the set to be linearly independent? If it is linearly independent, you should not be able to represent any vector as a linear combination of the other 2. But haven't you just shown (or Halls did) that v1 and v2 are multiples of each other?
 

FAQ: Linear Algebra: Linear dependence of vectors?

1. What is linear dependence of vectors?

Linear dependence of vectors refers to the relationship between two or more vectors in a vector space. It describes how one vector can be written as a linear combination of other vectors, meaning it can be expressed as a sum of scalar multiples of those vectors.

2. How is linear dependence determined?

Linear dependence is determined by checking if there exist scalars (coefficients) such that when multiplied by the corresponding vectors and then summed together, the resulting vector is equal to the zero vector. If such scalars exist, then the vectors are linearly dependent.

3. What does it mean for vectors to be linearly independent?

Vectors are linearly independent if there are no scalars (except for the trivial solution of all scalars being zero) that can be multiplied by the vectors and then summed together to equal the zero vector. In other words, each vector in a set of linearly independent vectors contributes unique information to the vector space.

4. How is linear independence related to linear dependence?

Linear independence and linear dependence are two sides of the same coin. If vectors are linearly independent, then they are not linearly dependent, and vice versa. In other words, if one vector can be expressed as a linear combination of other vectors, then those vectors are linearly dependent.

5. Why is linear dependence important?

Linear dependence is important because it helps us understand the relationships between vectors and how they can be used to represent information in a vector space. It also allows us to determine if a set of vectors is a basis (a set of linearly independent vectors that span the entire vector space) or not.

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