Hey guys,

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

Any help?

Peter

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- April 1st 2011, 01:59 PMflybynightCompaing the Rank of Matrices
Hey guys,

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

Any help?

Peter - April 1st 2011, 02:38 PMDeveno
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). - April 2nd 2011, 07:08 AMflybynight
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?

- April 2nd 2011, 07:19 AMHallsofIvy
No, it doesn't. Nothing is said or assumed about .