Recent content by kickthemoon

ah, so it's got an identity matrix on the bottom... curious: would Gram-Schmidt Orthonormalization work to get these vectors too? Say, if I just arbitrarily picked: [1 0 0 0], [0 1 0 0] and [ 0 0 1 0] and put them together to the vector [ 1 2 3 4] to get an arbitrary basis for R4 and then...

so, um, i don't understand then...yes i see that they're not perpendicular...i followed all of the outlined steps...what happened?

* mean the three vectors are a basis for R3, the subspace of R4 that are all perpendicular to V

So then X= y[-2 2 0 0] + z[-3 0 3 0] + w[ -4 0 0 4] ? and each of those vectors represents the basis?

okay, so if i wrote X in terms of y,z,w (e.g. x= -2y-3z-4w) and substituted that into X, which then becomes [-2y-3z-4w 2y 3z 4w], then that is basis consisting of the span of all perpendicular vectors to [1 2 3 4]? In other words, if I took the dot product of those, then it would be zero?

whoops. I was trying to be smart and let that slip.

whoooaaa wait a minute, hold up. what wizvuze said... can you explain that to me? I'm not sure where that comes from...?

Okay, so you mean that: [w x y z]' dot [1 2 3 4]' = [0 0 0 0]

X dot V = 0 right?

Homework Statement Consider the vector V= [1 2 3 4]' in R4, find a basis of the subspace of R4 consisting of all vectors perpendicular to V. Homework Equations I mean, I'm just completely stumped by this one. I know that in R2, any V can be broken down to VParallel + VPerp, which...