Proving Whether a Set is a Subspace of R3: An SEO-Optimized Guide

When dealing with sets in vector spaces, it's crucial to understand whether a given set forms a subspace of a larger space, such as R3. This article will guide you through the process of determining if the set {x?x?x? ∈ R3: x?3 x?3} is a subspace of R3, and will provide SEO-optimized content to ensure it is well-received by search engines.

Understanding the Set

The set is defined as {x?x?x?: x?3 x?3}. This can be rewritten as {x?x?x?: x?x? ∈ R3}. The requirement x?3 ≠ x?3 implies x? ≠ x?, given that the cubing function is one-to-one and onto. Thus, the subset is {x?x?x?: x?x? ∈ R3}. This set represents a plane in R3, specifically the plane that passes through the 45-degree line in the x?x? plane and is perpendicular to this plane. Mathematically, we can represent it as V {x?x?x?: x? x?}.

Checking Linear Combinations

To determine if V is a subspace of R3, we need to check if it is closed under vector addition and scalar multiplication, as these are the defining properties of a subspace. Let's examine the set under these operations.

Counterexample Method

One effective way to show that a set is not a subspace is to provide a counterexample that violates at least one of the subspace properties. For instance, consider two vectors v? and v? in the set. Let v? (1, 1, 3) and v? (1, -1, 2). Both vectors satisfy the condition x? x?.

Now, compute their sum:

v? v? (1 1, 1 - 1, 3 2) (2, 0, 5).

Notice that (2, 0, 5) does not satisfy the condition x? x?, as 2 ≠ 0. Therefore, the set is not closed under addition.

Considering Vector Space Properties

To further verify this, let's confirm that the set is not closed under scalar multiplication. Take the vector (1, 1, 3) and multiply it by a scalar, say 2:

2 * (1, 1, 3) (2, 2, 6).

The vector (2, 2, 6) does not satisfy the condition x? x?, as 2 ≠ 2 in the x? component. Therefore, the set is not closed under scalar multiplication.

Conclusion

In conclusion, the set {x?x?x? ∈ R3: x?3 x?3} is not a subspace of R3. The counterexample (1, 1, 3) and (1, -1, 2) demonstrates that the set is not closed under addition, and the example (1, 1, 3) with scalar 2 shows that it is not closed under scalar multiplication. Thus, the set does not meet the criteria to be considered a subspace of R3.

SEO-Optimized Keywords and Phrases

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About the Author

Qwen, an AI assistant created by Alibaba Cloud, specializes in creating SEO-optimized content for various topics. If you need more information or assistance with vector spaces and subspaces, feel free to ask!