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    Consider the following subset B of R^3 (real Euclidean space):

    B={(x,y,z) in R^3 such that x=y+z}

    With the usual "component wise" addition and scalar multiplication in R^3, is B a vector space? Prove or disprove your answer.

    © BrainMass Inc. brainmass.com December 24, 2021, 10:01 pm ad1c9bdddf
    https://brainmass.com/math/real-analysis/real-euclidean-space-434595

    SOLUTION This solution is FREE courtesy of BrainMass!

    Proposition: The subset B <= R^3 with B = {(x, y, z)|x = y + z|} is a vector subspace of R^3.

    Proof: We will show that B is closed under both scalar multiplication and subtraction.
    Let (x, y, z) <- B, (u, v, w) <- B and r <- R be arbitrary. Then
    x - y - z = 0 and u - v - w = 0. Hence,
    (x - u) - (y - v) - (z - w) = (x - y - z) - (u - v - w) = 0

    Therefore, (x, y, z) - (u, v, w) = (x - u, y - v, z - w) <- B. Moreover,
    rx - ry - rz = r(x - y - z) = r . 0 = 0

    Therefore, (rx, ry, rz) <- B. Hence, B is a subspace of R^3.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:01 pm ad1c9bdddf>
    https://brainmass.com/math/real-analysis/real-euclidean-space-434595

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