Find a basis for vector space

  • Thread starter vsage
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vsage
Ok another question:

13. The set of solutions to the system of linear equations

a - 2b + c = 0
2a - 3b + c = 0

is a subspace of R^3. Find a basis for this subspace

The book claims one of the possible bases is (1, 1, 1) but I don't see how. I mean I realize a = b = c from the above just am not making the connection why there has to be THREE 1's. Originally I had (1, 1) but I really can't remember why. Is it because since a, b and c are equal, all the solutions to the above in (a, b, c) format are given by d(1, 1, 1) where d is a scalar?
 

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  • #2
cepheid
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vsage said:
Ok another question:

13. The set of solutions to the system of linear equations

a - 2b + c = 0
2a - 3b + c = 0

is a subspace of R^3. Find a basis for this subspace

Makes perfect sense. Think about it...the set of solutions is a set of real numbers (a, b, c) that satisfy both linear equations. It sounds like such a set has three components, necessitating that it be a vector in R3. In fact, a = b = c is satistied by vectors of the form (1, 1, 1)...(2, 2, 2), ...(3, 3, 3). But all of these can be written as linear combinations of (1, 1, 1). Which is exactly what you said...all others are scalar multiples of that vector. The subspace is the line through the origin. One vector is sufficient to span a 1-dimensional subspace, but that subspace is located in R3, so the basis vector must have three components.

Another way to look at it...those two equations define two planes in R3, (where in this case our coordinate space is formed from the mutually orthogonal a, b, and c axes) and evidently they intersect in the line a = b = c.
 
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  • #3
HallsofIvy
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If you were just trying to SOLVE the equations, you would subract one from the other to cancel the "c" terms and get -a+ b= 0. Okay, a= b. Now put that back into the first equation: setting a= b gives -a+ c= 0 so a= c. If you put a= b= c into the second equation, you find it is automatically solved: any numbers a= b= c satisfy both equations so (a, b, c)= (a, a, a)= a(1, 1, 1). The have to be THREE numbers in the "basis" because your solution consists of 3 numbers, a, b, c.

(By the way, your book says "one of the possible bases" because, of course, (2, 2, 2) or (3, 3, 3) or any set of 3 identical numbers would work.
 

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