Verifying if Vector b is in Span of Vectors v1 and v2

In summary, the homework statement is that vector b in the span of vectors v1,v2. However, the attempt at a solution gives a different result numerically. Specifically, R3=R3+R1 which seems wrong because that gives you 1 in the first column of R3, which you do not want.
  • #1

Homework Statement



is vector b in the span of vectors v1,v2? Give reasons.



Homework Equations





The Attempt at a Solution



v1=(1,4,0,-1)

v2=(2,7,-2,-3)

b= (-1,-1,7,4)

set up in matrix

(v1,v2| b)

and after row reduction I have

(1,0,0,0),(2,-1,0,0)|(-1,3,6,6)

matrix has no sltn

since 0x + 0y = 6

therefore b does not span v1 ,v2.

my row reduction steps are..

R4= R4+R1
R2=R2-4R1


R4=R4+R2
R3=R3+R1


...


is this right?
 
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  • #2
The general result (negative) is correct. However, I get a different result numerically. Specifically, R3 = R3 + R1 seems wrong because that gives you 1 n the first column of R3, which you do not want.
 
  • #3
voko said:
The general result (negative) is correct. However, I get a different result numerically. Specifically, R3 = R3 + R1 seems wrong because that gives you 1 n the first column of R3, which you do not want.

ohh thanks I did not see that.
 
  • #4
A more fundamental resolution is that a vector, b, is in the span of vectors v1 and v2 if and only if there exist scalars, a and b, such that b= av1+ bv2 (that's the definition of 'span').

That is, there must exist a and b such that a(1, 4, 0, -1)+ b(2, 7, -2, -3)= (a+ 2b, 4a+ 7b, -2b, -a- 3b)= (-1, -1, 7, 4) so that we must have
a+ 2b= -1
4a+7b= -1
-2b= 7
-a- 3b= 4.

Of course, what you are doing is using the 'augented' matrix to try to solve those equations. But in this simple case, you can recognize that the third equation, -2b= 7 gives b= -7/2 and so the fourth equation becoms -a+ 21/2= 4 so that a= 21/2- 4= (21- 8)/2= 13/2. Now put those values of x and y into the first and second equations to see if they satisfy them: a+ 2b= -1 becomes 13/2+ (-14/2)= -1/2, NOT -1 so we can stop.
 

1. How do you determine if a vector is in the span of two other vectors?

To determine if a vector b is in the span of vectors v1 and v2, you can perform the following steps:

  • Set up a system of equations using the components of the vectors v1, v2, and b.
  • Use Gaussian elimination to put the system of equations into reduced row echelon form.
  • If the system has a solution, then vector b is in the span of v1 and v2.
  • If the system has no solution, then vector b is not in the span of v1 and v2.

2. What does it mean for a vector to be in the span of two other vectors?

A vector b is in the span of vectors v1 and v2 if it can be written as a linear combination of v1 and v2. This means that there exist scalars c1 and c2 such that b = c1v1 + c2v2.

3. Can a vector be in the span of two linearly dependent vectors?

Yes, a vector can be in the span of two linearly dependent vectors. This is because linear dependence means that one vector can be written as a multiple of the other, so any vector in their span can also be written as a linear combination of the two vectors.

4. How does the dimension of the span of two vectors relate to the vectors themselves?

The dimension of the span of two vectors v1 and v2 is either 1 or 2. If the two vectors are linearly independent, then the span has dimension 2 and any vector in the span is uniquely determined by its components. If the two vectors are linearly dependent, then the span has dimension 1 and any vector in the span can be written as a multiple of the single vector.

5. Is verifying if a vector is in the span of two other vectors the same as checking if it is in the subspace spanned by those vectors?

Yes, verifying if a vector b is in the span of vectors v1 and v2 is equivalent to checking if b is in the subspace spanned by v1 and v2. This is because the span of two vectors is the set of all possible linear combinations of those two vectors, which forms a subspace.

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