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Math Help - rank theory

  1. #1
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    rank theory

    literally have no clue on how to do this one... :'(


    Prove that if A and B are n x n matrices and if rank(A) = n and rank(B) = n then rank(AB) = n.

    any help would be much appreciated here as i have a test tomorrow!
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  2. #2
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    Quote Originally Posted by situation View Post
    literally have no clue on how to do this one... :'(


    Prove that if A and B are n x n matrices and if rank(A) = n and rank(B) = n then rank(AB) = n.

    any help would be much appreciated here as i have a test tomorrow!
    I don't know what tools you have exactly but here is a way.

    A square matrix has full rank if and only if it has non zero determinant. (only the zero vector can be in the nullspace) We also know that

    \det(AB)=\det(A)\det(B)
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  3. #3
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    Another way- the "rank" of a linear transformation, A, from vector space V to itself is the dimension of A(V). Here, A is given by an n by n matrix so it is from R^n to R^n. Since the rank of A is n, A(R^n)= R^n. Similarly, the rank of B is n so that B(R^n)= R^n. Putting those together, AB maps R^n into all of R^n: AB(R^n)= R^n and so AB has rank n.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Another way- the "rank" of a linear transformation, A, from vector space V to itself is the dimension of A(V). Here, A is given by an n by n matrix so it is from R^n to R^n. Since the rank of A is n, A(R^n)= R^n. Similarly, the rank of B is n so that B(R^n)= R^n. Putting those together, AB maps R^n into all of R^n: AB(R^n)= R^n and so AB has rank n.
    please could you just briefly explain what is meant by the dimension, a linear transformation, and a vector space? i have so pretty awful lecture notes which just arent making sense to me in working out what your explanation means.

    thank you both very much!
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