greendays said:Find a basis of the given span{[2,1,0,-1],[-1,1,1,1],[2,7,4,5]}
So I got the RREF, and found the basis to be two rows of the RREF, which are [1,0,-1/3,0] and [0,1,2/3,1], but the answer is [2,1,0,-1],[-1,1,1,1]
Where did I do wrong?
jbunniii said:As Robert1986 indicated, the solution is not unique, so you can have a correct answer that doesn't match the book.
However, your answer is incorrect. You didn't show your work, so I can't say what you did wrong. But your vectors do not span the same subspace as the given vectors. There is no linear combination of your two vectors that will give you [2, 1, 0, -1].
If
a[1, 0, -1/3, 0] + b[0, 1, 2/3, 1] = [2, 1, 0, -1]
then equality of the first coordinate forces a = 2, and equality of the second coordinate forces b = 1. But then you don't get equality in the fourth coordinate.
A basis of a subspace is a set of linearly independent vectors that span the subspace. This means that any vector in the subspace can be written as a linear combination of the basis vectors.
To find a basis of a subspace, you can use the process of elimination. Start with a set of vectors that span the subspace and then eliminate any linearly dependent vectors until you are left with a set of linearly independent vectors. These vectors will form the basis of the subspace.
Finding a basis of a subspace is important because it allows us to represent any vector in the subspace as a linear combination of the basis vectors. This makes it easier to perform calculations and understand the properties of the subspace.
Yes, a subspace can have more than one basis. This is because there can be multiple sets of linearly independent vectors that span the subspace. However, all bases of a subspace will have the same number of vectors, known as the dimension of the subspace.
The number of vectors in a basis is equal to the dimension of the subspace. This means that finding a basis is one way to determine the dimension of a subspace. Additionally, the basis vectors can be used to create a basis matrix, which can be used to find the dimension of the subspace as well as perform other calculations.