Hi. I need to find a base for the subspace V in R3 which has the equation
x+2y+3z=0
Can someone please tell my if the space I'm looking for is
[1 0 0;0 2 0;0 0 3] ????
If not, please explain what I'm doing wrong
matt grime
Oct24-04, 09:47 AM
The subspace is a plane. Find two linearly independent vectors lying in the plane.
HallsofIvy
Oct24-04, 01:07 PM
Your answer can't be correct because it has 3 basis vectors and so would span all of R3.
You have one equation is 3 unknowns. Solve for ONE of the unknowns and use the other two as parameters.
x+ 2y+ 3z= 0 so x= -2y- 3z. If you take y= 1, z= 0, then x= -2. One basis vector is (-2, 1, 0). If you take y= 0, z= 1, then x= -3. Another basis vector is (-3, 0, 1).
The basis is [(-2, 1, 0), (-3, 0, 1)].
Of course, a basis is not unique. There are many possible solutions (but they will all contain 2 basis vectors).
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