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Math Help - Compaing the Rank of Matrices

  1. #1
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    Compaing the Rank of Matrices

    Hey guys,
    Having a bit of trouble with: Prove that, for m x n matrices A and B:

    rank(A+B) \leq rank(A)+rank(B)

    Any help?
    Peter
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  2. #2
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    suppose rank(A) = r, and rank(B) = s.

    then for any vector v in R^n, we can write A(v) = a1v1 +a2v2 +....arvr, where the v's are linearly independent column vectors of A.

    similarly we can write B(v) = b1u1 + b2u2 +...+ bsus, where the u's are linearly independent column vectors of B.

    then A+B(v) = A(v) + B(v) = a1v1 + a2v2 +...+ arvr + b1u1 + b2u2 +...+ bsus

    so {v1,v2,...,vr,u1,u2,...,us} is a spanning set for the column space of A+B.

    thus dim((A+B)(R^n)) ≤ r + s.

    (the (disjoint) union of a basis for col(A) and a basis for col(B) might not be a basis for col(A+B), but it certainly contains one, just take out the linearly dependent elements).
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  3. #3
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    Does this proof assume that the columns of the matrices span all of R^n? If so, how do I know that's a reasonable assumption to make?
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  4. #4
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    No, it doesn't. Nothing is said or assumed about R^n.
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