How Do You Find a Basis for a Subspace Spanned by Vectors in R^3?

[snoggerT]

use rowspace/colspace to determine a basis for the subspace of R^n spanned by the given set of vectors:

{(1,-1,2),(5,-4,1),(7,-5,-4)}

*note: the actual instructions are to use the ideas in the section to determine the basis, but the only two things learned in the section are rowspace and colspace.

The Attempt at a Solution

- I thought you could just find the rowspace, and that would be a subspace of R^n, but the answer in the back of the book isn't at all the same. How would you go about solving this problem?

 

[NateTG]

use rowspace/colspace to determine a basis for the subspace of R^n spanned by the given set of vectors:

{(1,-1,2),(5,-4,1),(7,-5,-4)}
- I thought you could just find the rowspace, and that would be a subspace of R^n, but the answer in the back of the book isn't at all the same. How would you go about solving this problem?

What's the definition of basis in your book?

 

[HallsofIvy]

Since there is no matrix here, I cannot see how it can have anything to do with "rowspace" of "columnspace"! The "rowspace" is, after all, the space spanned by the rows of a matrix (thought of as vectors) and, of course, the "columnspace" is the spanned by the columms- but you have no matrix here.

Can we assume that, since this problem asks about a basis, you know what a "basis" is? Was that in a previous section so you think you shouldn't use the definition? Also, since each of the vectors given is in R³, we are assuming that n= 3.

A set of vectors always spans some vector space. It that set is also [bindependent[/b] then it is a basis for the space. If a set is not independent then one or more of the vectors can be written as a linear combination of the others. Get rid of those and you have a basis.

"Independent" means that a linear combination equal to the 0 vector must have all coefficients 0. Here, that means we must look at a(1,-1,2)+ b(5,-4,1)+ c(7,-5,-4)= (0, 0, 0) so we must have a+ 5b+ 7c= 0, -a- 4b- 5c= 0, and 2a+ b- 4c= 0. If we multiply the second equation by 2 and add to the third, we get -7b- 14c= 0 or b= -2c. If add the first and second equations, we get b+ 2c= 0: again b= -2c. Putting b= -2c into the second equation, -a- 4b+ 10b= -a+ 6b= 0 so a= 6b. Taking b= 1, a= 6, c= -2 and we have 6(1, -1, 2)+ (5, -4, 1)- 2(7, -5, -4)= (0,0,0). We can take anyone of those to the right side and divide by its coefficient to see that it can be replace by a linear combination of the other 2. Any two of the given vectors forms a basis for their span (and show that the span is a two dimension subspace of R³).

NateTG got in just ahead of me! I need to learn to type faster!

 

[snoggerT]

Since there is no matrix here, I cannot see how it can have anything to do with "rowspace" of "columnspace"! The "rowspace" is, after all, the space spanned by the rows of a matrix (thought of as vectors) and, of course, the "columnspace" is the spanned by the columms- but you have no matrix here.

Can we assume that, since this problem asks about a basis, you know what a "basis" is? Was that in a previous section so you think you shouldn't use the definition? Also, since each of the vectors given is in R³, we are assuming that n= 3.

A set of vectors always spans some vector space. It that set is also [bindependent[/b] then it is a basis for the space. If a set is not independent then one or more of the vectors can be written as a linear combination of the others. Get rid of those and you have a basis.

"Independent" means that a linear combination equal to the 0 vector must have all coefficients 0. Here, that means we must look at a(1,-1,2)+ b(5,-4,1)+ c(7,-5,-4)= (0, 0, 0) so we must have a+ 5b+ 7c= 0, -a- 4b- 5c= 0, and 2a+ b- 4c= 0. If we multiply the second equation by 2 and add to the third, we get -7b- 14c= 0 or b= -2c. If add the first and second equations, we get b+ 2c= 0: again b= -2c. Putting b= -2c into the second equation, -a- 4b+ 10b= -a+ 6b= 0 so a= 6b. Taking b= 1, a= 6, c= -2 and we have 6(1, -1, 2)+ (5, -4, 1)- 2(7, -5, -4)= (0,0,0). We can take anyone of those to the right side and divide by its coefficient to see that it can be replace by a linear combination of the other 2. Any two of the given vectors forms a basis for their span (and show that the span is a two dimension subspace of R³).

NateTG got in just ahead of me! I need to learn to type faster!

- Can you explain to me how plugging b=-2c back into the 2nd equation gives you -a-4b+10b?