- #1

- 40

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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

- Thread starter gunnar
- Start date

- #1

- 40

- 0

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

- #2

matt grime

Science Advisor

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The subspace is a plane. Find two linearly independent vectors lying in the plane.

- #3

HallsofIvy

Science Advisor

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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).