Prove that W = {(x_1,...,x_n)| sum(x_i) = 0} is a Subspace of the Vector Space R^n

TL;DR

Demonstration of proving a set is a subspace by satisfying three conditions - non-empty, closed under vector addition, and closed under scalar multiplication.

Transcript

so we have a set W this is actually a subset of RN and we're asked to prove it's a subspace of RN so recall that W is a subspace of a vector space V if we have three conditions the first condition is that W is non empty so there has to be something in there there has to be a vector in W - W is closed under what's called vector addition so symbolica... Read More

Key Insights

• 🚾 To prove a set is a subspace, it must satisfy non-empty, closed under vector addition, and closed under scalar multiplication conditions.
• 🚾 Being closed under vector addition ensures that the sum of two vectors in the set remains in the set.
• 🚾 Scalar multiplication being closed ensures that multiplying a vector by a scalar keeps the result in the set.
• 😫 Showing a set is non-empty involves finding a vector in the set with the sum of its components equal to zero.

Questions & Answers

Q: What are the three conditions that a set must satisfy to be considered a subspace?

A set must be non-empty, closed under vector addition, and closed under scalar multiplication to be classified as a subspace.

Q: How do you show that a set is non-empty in the context of proving it is a subspace?

To show a set is non-empty, find a vector in the set where the sum of its components equals zero.

Q: Why is it important to demonstrate that a set is closed under vector addition in proving it is a subspace?

Proving that a set is closed under vector addition ensures that the sum of any two vectors in the set remains in the set, a crucial property of subspaces in linear algebra.

Q: How is the concept of scalar multiplication relevant in proving a set is a subspace?

By showing that a set is closed under scalar multiplication, it is proven that when a vector in the set is multiplied by a scalar, the result remains in the set, confirming it as a subspace.

Summary & Key Takeaways

• To prove a set W is a subspace, it must be non-empty, closed under vector addition, and closed under scalar multiplication.

• First, show that W is non-empty by finding a vector in W with the sum of its components equal to zero.

• Next, demonstrate that W is closed under vector addition by showing that the sum of two vectors in W is also in W.