Finding Basis Help

[fair light]

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[desert copper]

Both are planes, so are of dimension 2, for the first one you can take $(x,y,z)^{T}$ in the plane then you know that $3x+2y=0$ thus $x=-\frac{2}{3}y$ so $(x,y,z)^{T}=((-\frac{2}{3},1,0)y +(0,0,1)z))^{T}$ from this you can deduce a basis (a linearly independent family of vectors that generates the plane)

[cyan pelican]

😑 rotoR

[desert copper]

So the same for the second one

[north knot]

Doesn't the basis only consist of 1 vector tho?

[desert copper]

No here both are planes so the basis consists of two vectors

[north knot]

That doesn't seem right

I put the given equations into a matrix and solved that matrix to get x = -4/3z, y = 2z, and z being the free variable

So that means (x, y, z) = (-4/3z, 2z, z) = z(-4/3, 2, 1)

So the basis should be {(-4/3, 2, 1)}

[desert copper]

That is false though here it’s a plane

Take (0,0,1)

It’s a vector of the plane

But it’s non colinear to your basis vector

[desert copper]

Show what you did

[north knot]

I just set up a matrix with the 2 equations, to get an augmented matrix with the rightmost column all zero and after a few basic row reductions I got that solution

[desert copper]

Wait a second I think I misunderstood the question

I thought you had to find a basis for both planes

not the set of vectors that are in both planes simultaneously

Im sorry it’s my bad

In that case your basis vector does work indeed

because the vector space is basically the orthogonal of $F=Span((3,2,0),(3,1,2))$

[cyan pelican]

😑 rotoR

[desert copper]

Which you can get by just solving a linear system

[viral plank]

To go over it again.

{(3,2,0),(3,1,2)} is linearly independent, hence there is one free variable, and therefore the kernel is one-dimensional, fix any nonzero z to get a basis, e.g z=-3 yields (-4,6,-3) as a generator.

@north knot

one also sometimes calls a basis of Ker A (for a homogeneous system Ax=0) a fundamental system of solutions for Ax = b

[viral plank]

Alternatively, geometrically, the solution is a line (the intersection line of the two nonparallel planes) and we know that the cross product of the normals of the planes is collinear to a direction of this line

[north knot]

+close