Are u, v, w Linearly Independent and a Spanning Set for R2?

[negation]

Homework Statement

In R2 let

u = (4, -2), v = (8, 5), w = (4, 1).

a)Is the set {u, v, w} a spanning set for R2?
b) Are the vectors u, v linearly independent?
c) Are the vectors u, v, w linearly independent?

The Attempt at a Solution

a) u, v and w is a spanning set for the vector space R2 iff every vector in R2 can be expressed as a linear combination of the vectors in {u,v,w}. In other words, and if my defintion is correct, if a vector space V spans set A of vectors, then the set A of vector is a spanning for the vector space V.

(x,y) = λ1(4,-2) + λ2(8,5) + λ3(4,1)
4λ1 + 8λ2 + 4λ3 = x
-2λ1 +5λ2 + λ3 = y

In augmented matrix:

4 8 4 | x
-2 5 1 | y

R2<- R2 +0.5R1

4 8 4 | x
0 9 3 | y + x/2

There is one free variable λ3-so, 1 parameter.
There are 2 pivots and, therefore, dimension, 2.
u, v, and, w are spanning sets for R2.b) u, v are linearly independent iff span {u,v} = 0 where λ1=λ2= 0.4 8 | 0
0 9 | 0

9λ2 = 0
λ2 = 0
4λ1 + 8(0) = 0
λ1 = 0

since λ1 = λ2 = 0, u and v are linearly independent.

c) (0,0) = λ1 ( 4,-2) + λ2 ( 8,5) + λ3 (4,1)

4 8 4 | 0
-2 5 1 | 0

...RREF

4 8 4 | 0
0 9 3 | 0

λ3 = free variables
The corollary from this is that there are infinite solutions. From this, it can be deduced that the solution is non-trivial, and therefore, u,v and w is not linearly independent. I'm making this conclusion based on the definition that for a system of linear equation to be linearly independent, it must be the case that the only solution is such that λ1=λ2=...=λn=0 but infinite solution implies that the solutions are either non-trivial or that a trivial solution is only one of the many infinite possible solutions. (PLEASE correct me if I am wrong)

The computer system states my answer to be correct. However, I would appreciate if my conclusion in part (c) is correct or a fluke.

 

[PeroK]

I didn't look closely at what you did, but there is a much simpler way to do this. Note that:

Two vectors are linearly dependent iff one is a multiple of the other

Any two linearly independent vectors will span R2

No three vectors can be linearly independent in R2.

 

[negation]

I didn't look closely at what you did, but there is a much simpler way to do this. Note that:

Two vectors are linearly dependent iff one is a multiple of the other

Any two linearly independent vectors will span R2

No three vectors can be linearly independent in R2.

Hi, I have reedited my entire OP.

 

[HallsofIvy]

Homework Statement

In R2 let

u = (4, -2), v = (8, 5), w = (4, 1).

a)Is the set {u, v, w} a spanning set for R2?
b) Are the vectors u, v linearly independent?
c) Are the vectors u, v, w linearly independent?

The Attempt at a Solution

a) u, v and w is a spanning set for the vector space R2 iff every vector in R2 can be expressed as a linear combination of the vectors in {u,v,w}. In other words, and if my defintion is correct,

Yes, your definition is correct.

if a vector space V spans set A of vectors, then the set A of vector is a spanning for the vector space V.

(x,y) = λ1(4,-2) + λ2(8,5) + λ3(4,1)
4λ1 + 8λ2 + 4λ3 = x
-2λ1 +5λ2 + λ3 = y

In augmented matrix:

4 8 4 | x
-2 5 1 | y

R2<- R2 +0.5R1

4 8 4 | x
0 9 3 | y + x/2

There is one free variable λ3-so, 1 parameter.
There are 2 pivots and, therefore, dimension, 2.

More fundamentally, if we multiply -2λ1+ 4λ2+ λ3= y by 2 and add it to 4λ1+ 8λ2+ 4λ3= x, we get 16λ2+ 6λ3= x+ 2y. Take λ2= 0 and we have 6λ3= x+ 2y so λ3= (x+ 2y)/6 and then -2λ1+ (x+ 2y)/6= y or λ1= (4y- x)/6. Since we can find λ1, λ2, λ3 for any x, y, yes, it spans R2.

u, v, and, w are spanning sets for R2.

u, v, and w are NOT sets. {u, v, w} is a spanning set for R2.

b) u, v are linearly independent iff span {u,v} = 0 where λ1=λ2= 0.

That is not well stated. The "span {u,v}" consists of all λ1u+ λ2v. That span is NOT "0".
You mean "if λ1u+ λ2v= 0 then λ1= λ2= 0"

4 8 | 0
0 9 | 0

9λ2 = 0
λ2 = 0
4λ1 + 8(0) = 0
λ1 = 0

since λ1 = λ2 = 0, u and v are linearly independent.

Yes, that is correct.

c) (0,0) = λ1 ( 4,-2) + λ2 ( 8,5) + λ3 (4,1)

4 8 4 | 0
-2 5 1 | 0

...RREF

4 8 4 | 0
0 9 3 | 0

λ3 = free variables
The corollary from this is that there are infinite solutions. From this, it can be deduced that the solution is non-trivial, and therefore, u,v and w is not linearly independent. I'm making this conclusion based on the definition that for a system of linear equation to be linearly independent, it must be the case that the only solution is such that λ1=λ2=...=λn=0 but infinite solution implies that the solutions are either non-trivial or that a trivial solution is only one of the many infinite possible solutions. (PLEASE correct me if I am wrong)

The computer system states my answer to be correct. However, I would appreciate if my conclusion in part (c) is correct or a fluke.

Your answer to (c) is completely correct. If λ1u+ λ2v+ λ3w= 0 then
λ1(4,-2)+ λ2(8, 5)+ λ3(4, 1)= (0, 0)
(4λ1+ 8λ2+ 3λ3, -2λ1+ 5λ2+ λ3)= (0, 0)
4λ1+ 8λ2+ 3λ3= 0, -2λ1+ 5λ2+ λ3= 0

Multiply the second equation by 2 and add to the first equation: 18λ2+ 5λ3= 0 so if you take λ1= 0, λ2= 5 and λ3= -18 you have non-zero coefficients that make the linear combination 0. The set of vectors is NOT independent.