That won't work. If the vectors in S are parallel to the hyperplane, then all of these vectors are perpendicular to a normal to the hyperplane. If you're given the equation of a hyperplane, do you know how to find a normal to this hyperplane?quanta13 said:Homework Statement
Can you help please? I have this problem:
Let S be the set of all vectors parallel to the hyper-plane 4x +2y+z + 3r =0 in R^4 .
(a) Show that S is a subspace, (b) Determine a basis for S , (c) Find its dimension
Homework Equations
The Attempt at a Solution
S= { u=(x, y,z,r) | 4x+2y+z+3r=0} u is in R^4.
A vector subspace is a subset of a vector space that is closed under vector addition and scalar multiplication. This means that if you take any two vectors in the subspace, their sum and any scalar multiple of the vectors will also be in the subspace.
To show that a set is a subspace, you must prove that it satisfies the two conditions of closure under vector addition and scalar multiplication. This can be done by showing that the set contains the zero vector, and that it is closed under vector addition and scalar multiplication.
The basis of a vector subspace is a set of linearly independent vectors that span the entire subspace. To determine the basis, you can use the process of Gaussian elimination to reduce the vectors to their simplest form, and then choose the vectors that make up the basis.
The dimension of a vector subspace is equal to the number of vectors in its basis. So, to find the dimension, you can simply count the number of vectors in the basis set.
Yes, a vector subspace can have more than one basis. This is because there can be multiple sets of linearly independent vectors that span the same subspace. However, all of these bases will have the same number of vectors, which is equal to the dimension of the subspace.