Basis for a subspace in R^4

In summary, the task is to find a basis for the subspace F, which is a three-dimensional subspace of R4 in the form of \mathbf{x}^T\mathbf{n}=0. The basis will consist of three vectors that are linearly independent and span F. To prove that the basis is correct, one can take the dot product of each basis vector with the normal vector to the hyperplane and show that it equals 0.
  • #1
saifatlast
1
0

Homework Statement


Find a basis for [tex]F=\left\{(x,y,z,w): -x+y+2z-w=0\right\}[/tex]

The Attempt at a Solution



So this looks like a plane to me, but I find 4-d space confusing, so that might be wrong. It does have the form [tex]\mathbf{x}^T\mathbf{n}=0[/tex], so that's kind of where I'm getting the idea that it's a plane. Assuming it is a plane (and if not, I'd appreciate some help getting moving in the right direction), my approach is to find 3 vectors in the plane. The plane should be spanned by the linear combinations of these vectors.

This gives

[tex]

F: \left\{
\begin{bmatrix}1\\1\\0\\0\end{bmatrix},
\begin{bmatrix}2\\0\\1\\0\end{bmatrix},
\begin{bmatrix}-1\\0\\0\\1\end{bmatrix}
\right\}
[/tex]

My questions is, assuming this is correct, how do I prove to myself that it is? I guess if I take the dot product of each of the basis vectors with the normal vector, n, below, I do get 0.

[tex]\mathbf{n}=\begin{bmatrix}-1\\1\\2\\-1\end{bmatrix}[/tex]

I guess what it comes down to is that I feel like I'm on shaky ground here, and I'm not quite sure how to firm things up.
 
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  • #2
Your basis for F is the same one I found, so I would imagine we both used the same technique.

F is not a plane; it is a hyper-plane, a three-dimensional subspace of R4. It is a higher-dimension analog of a plane embedded in R3.

You can convince yourself that the three vectors you found are a basis for F by showing that they are linearly independent, and that they span F. These are actually pretty simple to do, if you have chosen your basis vectors wisely (and you did).

Your idea of dotting the three vectors with the normal to this hyperplane is interesting, and probably reasonable to do. It's a little hard to imagine the normal to a three-D "plane" though.
 

What is a subspace in R^4?

A subspace in R^4 is a subset of the vector space R^4 that satisfies three conditions: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication. This means that any linear combination of vectors in the subspace will also be in the subspace.

How can I determine if a set of vectors is a basis for a subspace in R^4?

A set of vectors is a basis for a subspace in R^4 if it is linearly independent and spans the entire subspace. This means that none of the vectors in the set can be written as a linear combination of the others, and any vector in the subspace can be written as a linear combination of the basis vectors.

What is the dimension of a subspace in R^4?

The dimension of a subspace in R^4 is the number of vectors in a basis for that subspace. This is equivalent to the number of linearly independent vectors in the subspace.

Can a subspace in R^4 have more than four dimensions?

No, a subspace in R^4 can only have four dimensions because it is a subset of the four-dimensional vector space R^4. This means that any vector in the subspace can be represented by four coordinates.

How is a subspace in R^4 different from a subspace in other dimensions?

The main difference between a subspace in R^4 and a subspace in other dimensions is the number of coordinates needed to represent a vector in the subspace. In R^4, a vector can be represented by four coordinates, while in a subspace of a different dimension, it may require a different number of coordinates.

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