Question

Let S and T be non-empty subsets of \(\R^2\), and W be a non-zero proper subspace of \(\R^2\). Consider the following statements :
I. If span(S) = \(\R^2\) , then span(S ∩ W) = W.
II. span(S ∪ T) = span(S) ∪ span(T).
Then

• A. both I and II are TRUE
• B. I is TRUE but II is FALSE
• C. II is TRUE but I is FALSE
• D. both I and II are FALSE

Answer 1

The Correct Option is D

Let's analyze the given statements and determine their validity:

Statement I: If \(\text{span}(S) = \R^2\), then \(\text{span}(S \cap W) = W\).

This statement suggests that if the span of set \(S\) is the entire space \(\R^2\), then the span of the intersection of \(S\) with a proper subspace \(W\) should be equal to \(W\). Consider the following reasoning:

• If \(\text{span}(S) = \R^2\), it means \(S\) contains vectors that can generate any vector in \(\R^2\).
• However, the intersection of \(S\) with \(W\) (a proper subspace) may not contain sufficient vectors to span the entire subspace \(W\). It is possible for \(\text{span}(S \cap W)\) to be a subset of \(W\), but not exactly equal to \(W\).

Thus, Statement I is false.

Statement II: \(\text{span}(S \cup T) = \text{span}(S) \cup \text{span}(T)\).

Let's explore this statement with the properties of spans:

• The span of a union of two sets \(S\) and \(T\), denoted as \(\text{span}(S \cup T)\), is the smallest subspace containing all linear combinations of vectors from \(S\) and \(T\).
• The union of the spans, \(\text{span}(S) \cup \text{span}(T)\), is not necessarily a valid span, as a union of two subspaces is not always a subspace. In fact, \(\text{span}(S) \cup \text{span}(T)\) is not guaranteed to be closed under addition or scalar multiplication.

Therefore, Statement II is false.

Conclusion: Both Statement I and Statement II are false, confirming the correct choice is both I and II are FALSE.