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Math Help - Linear Algebra II

  1. #1
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    Linear Algebra II

    Need some help on this problem:

    Prove that if V is finite dimensional and U1....Um are subspaces of V, then

    dim(U1+...+Um) is less than or equal to dim U1+...+Um.

    Thanks
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  2. #2
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    Quote Originally Posted by taypez View Post
    Need some help on this problem:

    Prove that if V is finite dimensional and U1....Um are subspaces of V, then

    dim(U1+...+Um) is less than or equal to dim U1+...+Um.

    Thanks
    Again, I will assume that U_1+...+U_m means the direct sum of vector spaces, as in the Cartesian product.
    For simplicity sake, I will work with only two vector spaces. U_1,U_2\leq V. And the more general case will follow easily and similarly.
    I did not confirm this because of laziness, but I believe that if B_1=\{u_1,u_2,...,u_n\} is a basis for U_1, and B_2=\{v_1,v_2,...,v_m\} is a basis for V_2. Then, B_1+B_2=\{(b_1,b_2)|b_1\in B_1\mbox{ and }b_2\in B_2\} is not necessarily a basis for V_1+V_2 but it is a spanning set of vectors. And hence, since the size of the basis is less than the size of the spanning set we have,
    \dim (V_1+V_2)\leq \dim V_1+\dim V_2=n+m
    Because n+m is the size of spanning set and \dim (V_1+V_2) is the size (cardinality) of the dimension (basis) of those spanning vectors.
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