Linear Algebra: Orthogonal basis ERG HELP

[kickthemoon]

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 represents a unique solution, but how does this apply to R4.

I'm interested in the process to finding the basis, more so than the answer. What steps would you use in general, to find the basis of a vector, that satisfy the conditions that the basis must be perpendicular to the vector.

Thanks,

 

[vela]

If x is some vector in R⁴, what equation must it satisfy if it is perpendicular to V?

 

[wisvuze]

Do you know this theorem?

If A := {v1,...,vn} is an orthonormal set in an n-dim inner product space V, then A can be extended to an orthonormal basis for V { v1,..., vn, v n+1 ,..... vz }

and if W = Span(A) , then B : = { v n+1 ,..... vz } is an orthonormal basis for A perp

 

[HallsofIvy]

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 represents a unique solution, but how does this apply to R4.

That is true in any dimension. In this case, since [1 2 3 4] itself, being a single vector spans a 1 dimensional subspace of R⁴. R⁴ can be written as as a direct sum of that and a 3 dimensional subspace. I recommend vela's suggestion: What must be true of [w x y z] in order that it be perpendicular to [1 2 3 4]? Specifically, what does "perpendicular" mean in a general Rⁿ?

Use that equation to write one of w, x, y, or z in terms of the other three.

I'm interested in the process to finding the basis, more so than the answer. What steps would you use in general, to find the basis of a vector, that satisfy the conditions that the basis must be perpendicular to the vector.

Thanks,

 

[kickthemoon]

If x is some vector in R⁴, what equation must it satisfy if it is perpendicular to V?

X dot V = 0 right?

 

[vela]

Yup!

 

[kickthemoon]

Okay, so you mean that:

[w x y z]' dot [1 2 3 4]' = [0 0 0 0]

 

[kickthemoon]

whoooaaa wait a minute, hold up. what wizvuze said...

can you explain that to me? I'm not sure where that comes from...?

 

[Mark44]

Okay, so you mean that:

[w x y z]' dot [1 2 3 4]' = [0 0 0 0]

The dot product does not produce a vector; it produces a number (scalar). So [w x y z]' dot [1 2 3 4]' = 0. Now work with that.

 

[Mark44]

whoooaaa wait a minute, hold up. what wizvuze said...

can you explain that to me? I'm not sure where that comes from...?

You don't need an orthonormal basis. Any old basis for your three-dimensional subspace of R⁴ will do just fine. Heed the advice from HallsOfIvy.

 

[kickthemoon]

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

 

[kickthemoon]

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?

 

[vela]

A basis is a set of linearly independent vectors. In this case, you need to specify three linearly independent such that any linear combination of those vectors is perpendicular to V. So far you have

$$\vec{x}=\begin{pmatrix}-2y-3z-4w\\y\\z\\w\end{pmatrix}$$

You want to rewrite that as a linear combination of three vectors. Those three vectors will form the basis you're trying to find.

 

[vela]

No, those aren't correct. If you multiply any of those by V, you won't get 0.

Start by writing x as a sum of three vectors, one involving just y, one involving just z, and one involving just w. Then pull the variables out of each one so you end up with

x = y(vector 1)+z(vector 2)+w(vector 3)

Those three vectors are a basis for the subspace.

 

[kickthemoon]

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?

 

[kickthemoon]

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

 

[vela]

A basis is a set of linearly independent vectors that span a vector space. In this problem, you have a three-dimensional space, so a basis consists of a set of three vectors. Each of the vectors isn't a basis; the three vectors together are a basis.

 

[Mark44]

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

The three vectors might be a basis for a three-dimensional subspace of R⁴. (I say "might be" because I didn't check your work.)
Since the vectors are in R⁴, they can't possibly have anything to do with R³.

 

[Mark44]

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?

No, you have an error. None of these vectors is perpendicular to <1, 2, 3, 4>.

Edit: I didn't realize that vela had already spotted the error in post 14. Look carefully at what vela is showing in post 13.

 

[kickthemoon]

No, you have an error. None of these vectors is perpendicular to <1, 2, 3, 4>.

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

 

[Mark44]

Look at vela's post #13.

 

[vela]

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

Sorry, I led you astray in post #13. I went back and fixed it. Just do what you did before starting from the corrected expression.

EDIT: Well, now that I read the thread again, you led me astray in post #12! Do you see why your expression in post #12 isn't correct, and the one in #13 (now) is?

 

[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 used GSO to adjust them, would that be "kosher"? GSO is supposed to make all bases orthonormal-- therefore, anything made by the later 3 vectors would result in them being perpendicular to the 1st?

 

[vela]

Yup, that would work too as long as your four vectors are linearly independent to start with and you begin the process with (1,2,3,4). It's a lot more work though.