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Math Help - Basis

  1. #1
    Super Member Deadstar's Avatar
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    Basis

    Can someone explain how to get this answer please.

    Let U_1 and U_2 be the subspaces of R^4 defined by

    U_1 = (x = (x_1,x_2,x_3,x_4)|x_1 + 2x_2 - x_3 - x_4 = 0)

    and

    U_2 = (x = (x_1,x_2,x_3,x_4)|x_1 - x_2 + x_3 + x_4 = 0)

    Find basis for U_1 and U_2 and  U_1 \cap U_2

    Now i can find basis for U_1 and U_2 fairly easily tho they were different from the ones given in the solutions. Problem is i have no idea how to calculate U_1 \cap U_2. This is the solution given.

    (-2, 1, 0, 0), (1, 0, 1, 0) and (1, 0, 0, 1) form a basis for U_1

    (1, 1, 0, 0), (-1, 0, 1, 0) and (-1, 0, 0, 1) form a basis for U_2

    (-1, 2, 3, 0), (-1, 2, 0, 3) form a basis for U_1 \cap U_2
    can someone explain this? It just says clearly this is the basis in the solution... Maybe clear for someone whos been studying it for 30+ years... I can sorta see where the -1 and 2 come from but not the 3's. Any help please?

    Thanks
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Deadstar View Post
    Can someone explain how to get this answer please.

    Let U_1 and U_2 be the subspaces of R^4 defined by

    U_1 = (x = (x_1,x_2,x_3,x_4)|x_1 + 2x_2 - x_3 - x_4 = 0)

    and

    U_2 = (x = (x_1,x_2,x_3,x_4)|x_1 - x_2 + x_3 + x_4 = 0)

    Find basis for U_1 and U_2 and  U_1 \cap U_2

    Now i can find basis for U_1 and U_2 fairly easily tho they were different from the ones given in the solutions. Problem is i have no idea how to calculate U_1 \cap U_2. This is the solution given.

    (-2, 1, 0, 0), (1, 0, 1, 0) and (1, 0, 0, 1) form a basis for U_1

    (1, 1, 0, 0), (-1, 0, 1, 0) and (-1, 0, 0, 1) form a basis for U_2

    (-1, 2, 3, 0), (-1, 2, 0, 3) form a basis for U_1 \cap U_2
    can someone explain this?
    If (x_1,x_2,x_3,x_4)\in U_1\cap U_2 then x_1 + 2x_2 - x_3 - x_4 = 0 and x_1 - x_2 + x_3 + x_4 = 0. The way to find solutions to a set of equations like that is to form the matrix of coefficients and row-reduce it.

    In this case, the matrix is \begin{bmatrix}1&2&-1&-1\\1&-1&1&1\end{bmatrix}. To row-reduce it, subtract the top row from the bottom row, getting \begin{bmatrix}1&2&-1&-1\\0&3&-2&-2\end{bmatrix}. That should help you to see where the 2s and 3s come from.
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  3. #3
    Super Member Deadstar's Avatar
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    sorry but i still have no idea. Can someone please just explain this fully. I have an exam on linear algebra in 2 days and i really dont wanna waste hours trying to figure out something thats only gonna get me a mark or two in the exam if its even in it. And i cant just ignore it cos it'll bug me to the point ill get nothing else done!
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