Subspace & Basis: Proving A is a Subspace of R^3

In summary, the conversation is about finding a basis for the set of all vectors orthogonal to a given vector u = [4 3 1], and showing that this set is a subspace of R^3. The first part involves proving that the set is closed under addition and scalar multiplication, and the second part involves finding a basis for the set. The hint for finding a basis is to use the concept of dot product.
  • #1
tracedinair
50
0

Homework Statement



Let u be a vector where u = [4 3 1]. Let A be the set of all vectors orthogonal to u. Show that A is subspace of R^3. Then find the basis for A.

Homework Equations





The Attempt at a Solution



For showing that A is a subspace...

Zero vector is in A because A(0) = 0

For any u & v, u+v is in A because Au=0, Av=0, and A(u+v) = Au+Av = 0

And for any scalar c, A(cu) = c(Au) = c(o) = 0

As for the basis, I really have no idea where to even start with that.

Thanks for any help.
 
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  • #2
tracedinair said:

Homework Statement



Let u be a vector where u = [4 3 1]. Let A be the set of all vectors orthogonal to u. Show that A is subspace of R^3. Then find the basis for A.

Homework Equations





The Attempt at a Solution



For showing that A is a subspace...

Zero vector is in A because A(0) = 0

For any u & v, u+v is in A because Au=0, Av=0, and A(u+v) = Au+Av = 0

And for any scalar c, A(cu) = c(Au) = c(o) = 0

As for the basis, I really have no idea where to even start with that.

Thanks for any help.

You should do the first part of this problem; namely, finding the set of vectors that are orthogonal to u = (4, 3, 1). How can you tell that an arbitrary vector (x, y, z) is orthogonal to a given vector?
Zero vector is in A because A(0) = 0
For any u & v, u+v is in A because Au=0, Av=0, and A(u+v) = Au+Av = 0

And for any scalar c, A(cu) = c(Au) = c(o) = 0
None of this makes any sense. A is a set, not a matrix, so it doesn't make any sense to multiply a vector by A.

As for finding a basis for A, if you do the first part you will be on your way toward a basis.
 
  • #3
I'm not entirely sure how to show an arbitrary vector is orthogonal to a given vector. I've looked through my text for help, but it's not really helping.
 
  • #4
Is "dot product" a hint?
 

FAQ: Subspace & Basis: Proving A is a Subspace of R^3

What is a subspace in mathematics?

A subspace is a subset of a vector space that satisfies the same vector space properties as the larger space. This means that it is closed under addition and scalar multiplication, and contains the zero vector.

What is a basis in linear algebra?

A basis is a set of linearly independent vectors that span a vector space. This means that any vector in the space can be written as a linear combination of the basis vectors.

How do you prove that A is a subspace of R^3?

To prove that A is a subspace of R^3, you need to show that it satisfies the three conditions of a subspace: closure under addition, closure under scalar multiplication, and containing the zero vector. This can be done by showing that any two vectors in A added together or multiplied by a scalar are also in A, and that the zero vector is in A.

What is the process for proving that a set of vectors is a basis?

To prove that a set of vectors is a basis, you need to show that the vectors are linearly independent and span the vector space. This can be done by setting up a system of equations and showing that the only solution is the trivial solution (all coefficients equal to zero), and also showing that any vector in the space can be written as a linear combination of the basis vectors.

Can a subspace have more than one basis?

Yes, a subspace can have more than one basis. As long as the basis vectors are linearly independent and span the subspace, any set of vectors can be considered a basis for that subspace.

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