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14 June, 14:30

3. Let U and V be subspaces of a vector space W. Prove that their intersection UnV is also a subspace of W

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  1. 14 June, 14:42
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    Answer: The proof is done below.

    Step-by-step explanation: Given that U and V are subspaces of a vector space W.

    We are to prove that the intersection U ∩ V is also a subspace of W.

    (a) Since U and V are subspaces of the vector space W, so we must have

    0 ∈ U and 0 ∈ V.

    Then, 0 ∈ U ∩ V.

    That is, zero vector is in the intersection of U and V.

    (b) Now, let x, y ∈ U ∩ V.

    This implies that x ∈ U, x ∈ V, y ∈ U and y ∈ V.

    Since U and V are subspaces of U and V, so we get

    x + y ∈ U and x + y ∈ V.

    This implies that x + y ∈ U ∩ V.

    (c) Also, for a ∈ R (a real number), we have

    ax ∈ U and ax ∈ V (since U and V are subspaces of W).

    So, ax ∈ U∩ V.

    Therefore, 0 ∈ U ∩ V and for x, y ∈ U ∩ V, a ∈ R, we have

    x + y and ax ∈ U ∩ V.

    Thus, U ∩ V is also a subspace of W.

    Hence proved.
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