Row = R 11.Determine whether the following are spanning sets for R 2 : vectors:
ID: 2942316 • Letter: R
Question
Row = R
11.Determine whether the following are spanning sets for R2:
vectors: (note: they all are 2x1 vectors, just no way to put them in brackets. did best I could example: (a) 2 and 1 is one vector and 3 and 2 are another vector. but are all part of the ONE question.) b through e is spaced out. cramster does that for some reason not sure how to fix. example(b) vectors are 2 and 3 for first one and 4 and 6 for the second but part of the same question. I hope this makes since.)
(a) {2}, {3}
{1}, {2}
(b) {2}, {4}
{3}, {6}
(c) {-2}, {1}, {2}
{1}, {3}, {4}
(d) {-1}, {1}, {2}
{2}, {-2}, {-4}
(e) {1}, {-1}
{2}, {1}
Explanation / Answer
No worries about the notation. I'll write the vectors horizontally to avoid confusion -- for example, (a) is (2, 1) and (3, 2).
The standard basis for R2 is (1,0) and (0,1). This spans R2 because we can write any vector x = (x1,x2) as a linear combination of (1,0) and (0,1); just write
(x1,x2) = x1(1,0) + x2(0,1)
So any (x1,x2) is in the span of (1,0), (0,1). We can show that if a family of vectors spans this standard basis, i.e. the family of vectors spans (1,0) and (0,1), then it spans R2. Proof: Let y = (y1,y2) and z = (z1,z2) be vectors that span (1,0) and (0,1). Then we have scalars a,b,c,d such that
(1,0) = a(y1,y2) + b(z1,z2)
(0,1) = c(y1,y2) + d(z1,z2)
Then we can write any vector (x1,x2) as
(x1,x2) = x1(1,0) + x2(0,1)
= x1 (a(y1,y2) + b(z1,z2)) + x2 (c(y1,y2) + d(z1,z2))
= (x1 a) (y1,y2) + (x1 b) (z1,z2) + (x2 c) (y1,y2) + (x2 d) (z1,z2)
= [(x1 a) + (x2 c)] (y1,y2) + [(x1 b) + (x2 d)] (z1,z2).
This is a linear combination of (y1,y2) and (z1,z2) with scalar coefficients ax1+cx2 and bx1+dx2. So if two vectors y and z span the standard basis, they span every vector in R2. (And, additionally, we can say that if a family of vectors doesn't span the standard basis, then clearly it does not span R2, as (1,0) and (0,1) are in R2 themselves.)
We use all this to prove which families of vectors span R2. Note that this is far from the only method you can use to show whether vectors span a space. If you are comfortable with linear dependence and dimension, you can use what you know about those to figure out whether vectors span a space. Or you could put them in a matrix and reduce. There are many ways of looking at spanning sets. We use this one as it's pretty easy to apply in problems like this once you understand why it works.
a) (2,1) and (3,2)
We want to show whether these span (1,0) and (0,1). If they do, then we have scalars a,b,c,d such that
a(2,1) + b(3,2) = (1,0)
and
c(2,1) + d(3,2) = (0,1).
We simplify each equation and break it down component-wise.
a(2,1) + b(3,2) = (1,0)
(2a,a) + (3b,2b) = (1,0)
(2a+3b, a+2b) = (1,0)
Equating components,
2a + 3b = 1 and a + 2b = 0.
From the second equation, a = -2b. Plug that into the first equation to get
2(-2b) + 3b = 1
-4b + 3b = 1
-b = 1
b = -1
a = -2(-1) = 2.
We have found a,b such that the two vectors span (1,0). Now check that they span (0,1):
c(2,1) + d(3,2) = (0,1)
(2c,c) + (3d,2d) = (0,1)
(2c+3d, c+2d) = (0,1)
2c + 3d = 0 and c + 2d = 1.
From the second equation, c = 1 - 2d. Plug this into the first:
2(1 - 2d) + 3d = 0
2 - 4d + 3d = 0
2 - d = 0
d = 2
c = 1 - 2(2) = -3.
So we have found c,d such that the two vectors span (0,1). Since the two vectors span the standard basis, they span R2.
(b) You might see right away that (2,3) and (4,6) are multiples of each other, and are therefore linearly dependent. They cannot, therefore, be a spanning set for R2. We'll check the same way we did part (a) anyway:
a(2,3) + b(4,6) = (1,0)
(2a+4b, 3a+6b) = (1,0)
2a+4b = 1 and 3a+6b = 0
The second equation divided by 3 reduces to a+2b = 0, or a = -2b. Plug this into the first to get
2(-2b) + 4b = 1
0 = 1
which is impossible. Therefore there is no solution for a and b, and these two vectors do not span (1,0) so are not a spanning set for R2.
(c) If any two of these vectors span R2, so does the whole family of vectors. We'll look at (-2,1) and (1,3). If they span R2, then we are done. (We could also check all three at once -- we do that in a later question -- but here it's easier to check two and show they span R2 and therefore the whole family does.)
a(-2,1) + b(1,3) = (1,0)
(-2a+b, a+3b) = (1,0)
-2a + b = 1, a + 3b = 0
From the first equation, b = 1+2a. Plug that into the second:
a + 3(1+2a) = 0
a + 3 + 6a = 0
7a = -3
a = -3/7
b = 1+2(-3/7) = 1 - 6/7 = 1/7.
Check the other vector in the standard basis:
c(-2,1) + d(1,3) = (0,1)
(-2c+d, c+3d) = (0,1)
-2c+d = 0, c+3d = 1
From the first equation, d = 2c. Plug into the second:
c + 3(2c) = 1
c + 6c =1
c = 1/7
d = 2(1/7) = 2/7.
So a,b,c,d exist such that (-2,1) and (1,3) span the standard basis, and therefore the family of vectors spans R2.
d) Once again, you might notice that these vectors are all multiples of each other so they will only span one dimension and cannot span R2. But let's test using this method anyway. We can check whether the whole family spans (1,0) by taking a linear combination of all three vectors:
a(-1,2) + b(1,-2) + c(2,-4) = (1,0)
(-a+b+2c, 2a-2b-4c) = (1,0)
-a+b+2c = 1, 2a-2b-4c = 0
Divide the second equation through by -2 to get -a+b+2c = 0. But by the first equation, -a+b+2c = 1, so these equations are inconsistent and these three vectors do not span (1,0) and therefore do not span R2.
e) We check the same way we have been:
a(1,2) + b(-1,1) = (1,0)
(a-b, 2a+b) = (1,0)
a-b = 1, 2a+b = 0
By the first equation, a = 1+b. Plug this into the second equation to get
2(1+b)+b = 0
2 + 2b + b = 0
2 + 3b = 0
3b = -2
b = -2/3
a = 1+b = 1/3.
So a,b exist such that the vectors span (1,0). Check (0,1):
c(1,2) + d(-1,1) = (0,1)
(c-d, 2c+d) = (0,1)
c-d = 0, 2c+d = 1
By the first equation, c = d. Plug this into the second equation to get
2c + c = 1
3c = 1
c = 1/3
d = 1/3.
So a,b,c,d exist such that the two vectors span the standard basis, and they therefore span R2.
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