Prove W = {(a, b) | a = -b} is a Subspace of R^2

TL;DR

Explanation of conditions for W to be a subspace of R^2 and a step-by-step proof using vector addition and scalar multiplication.

Transcript

prove that this set W is a subspace of R squared so proof before we start the proof let me really really quickly remind you what it means for W to be a subspace of R squared so in general a subset W of a vector space V so V here's our vector space is a subspace of V if the following three conditions holds so the first condition is that W must be a ... Read More

Key Insights

• 👾 Demonstrated conditions for a subset to be considered a subspace in a vector space.
• 🚱 A step-by-step proof illustrating non-empty, vector-addition closure, and scalar multiplication closure for a subspace.
• ✖️ Use of specific vector component conditions to show closure under vector addition and scalar multiplication.

Questions & Answers

Q: What are the three conditions for a subset W to be considered a subspace of a vector space V?

The three conditions are that W must be non-empty, closed under vector addition, and closed under scalar multiplication. These conditions are essential for W to qualify as a subspace of V.

Q: How is the non-empty condition of a subset W being a subspace demonstrated in the proof?

The proof shows that W contains the zero vector, satisfying the non-empty condition, as the zero vector can also be represented as the negative of itself, fulfilling the criteria for W to be a subspace.

Q: How is the closure under vector addition condition of a subset W being a subspace proven in the given example?

By showing that the sum of two vectors X and Y from W is also in W, the proof demonstrates closure under vector addition. This is achieved by applying specific conditions related to the vector components.

Q: What does it mean for a subset W to be closed under scalar multiplication, and how is this condition validated in the proof?

Closure under scalar multiplication means that multiplying any scalar alpha with a vector X from W results in a vector still in W. The proof showcases this by manipulating the vector components to show that the product is indeed in W.

Summary & Key Takeaways

• A subspace W of a vector space V must be non-empty, closed under vector addition, and closed under scalar multiplication to be considered a subspace.

• The proof involves showing that W contains the zero vector, is closed under vector addition, and closed under scalar multiplication using specific conditions.

• By following the conditions for a subspace, the proof demonstrates that W is indeed a subspace of V.