To prove that the union of two subsets X and Y of a vector space is linearly dependent if either X or Y is linearly dependent, we can follow these steps:

1. Assume without loss of generality that X is the set that is linearly dependent. This means that there exists a non-trivial linear combination of vectors in X that equals the zero vector.

2. This non-trivial linear combination can be written as a sum of scalar multiples of vectors in X, which equals the zero vector.

3. Since every vector in X is also in X ∪ Y, this non-trivial linear combination is also a non-trivial linear combination of vectors in X ∪ Y.

4. Therefore, there exists a non-trivial linear combination of vectors in X ∪ Y that equals the zero vector, which means that X ∪ Y is linearly dependent.

This completes the proof.

Question

To prove that the union of two subsets X and Y of a vector space is linearly dependent if either X or Y is linearly dependent, we can follow these steps:

1. Assume without loss of generality that X is the set that is linearly dependent. This means that there exists a non-trivial linear combination of vectors in X that equals the zero vector.

2. This non-trivial linear combination can be written as a sum of scalar multiples of vectors in X, which equals the zero vector.

3. Since every vector in X is also in X ∪ Y, this non-trivial linear combination is also a non-trivial linear combination of vectors in X ∪ Y.

4. Therefore, there exists a non-trivial linear combination of vectors in X ∪ Y that equals the zero vector, which means that X ∪ Y is linearly dependent.

This completes the proof.

Knowee AI · Accepted Answer

To prove that the union of two subsets X and Y of a vector space is linearly dependent if either X or Y is linearly dependent, we can follow these steps:

1. Assume without loss of generality that X is the set that is linearly dependent. This means that there exists a non-trivial linear combination of vectors in X that equals the zero vector.

2. This non-trivial linear combination can be written as a sum of scalar multiples of vectors in X, which equals the zero vector.

3. Since every vector in X is also in X ∪ Y, this non-trivial linear combination is also a non-trivial linear combination of vectors in X ∪ Y.

4. Therefore, there exists a non-trivial linear combination of vectors in X ∪ Y that equals the zero vector, which means that X ∪ Y is linearly dependent.

This completes the proof.

If X and Y are subsets of a vector space, and either X or Y is linearly dependent, show thatX ∪ Y is linearly dependent.

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