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Math Help - Rank of a Matrix

  1. #1
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    Rank of a Matrix

    Every column of AB is a linear combination of the columns of A. Then, Rank (AB) <= Rank (A).

    Prove Rank (AB) <= Rank (B).

    I see how Rank (AB) <= Rank (A) because the rankk of A is the number of its linearly independent columns, and AB will have the same number of less of linealry independent columns. But how can I prove this for B?
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  2. #2
    Member HappyJoe's Avatar
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    Do you know that the rank of a matrix is equal to the rank of the transposed? I.e. for all matrices A, we have \text{Rank}(A) = \text{Rank}(A^T).

    Given that you know the one inequality to be true, you can use this to prove the other.
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  3. #3
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    Here's another, perhaps more basic, way of looking at it. For linear transformation A, from vector space V to vector space U, the "rank of A", rank(A), is the dimension of A(V), a subspace of U. If A is invertible, then A(V) has the same dimension as V. If not, then A(V) has smaller dimension than V. That is, rank(A)\le dim(V). (In fact, Rank(A)+ nullity(A)= dim(V) where " nullity(A)" is the dimension of the nullspace of A but you don't need to use that here.)

    If B is a linear transformation from U to V and A is a linear transformation from V to W, then AB is a linear transformation from U to W. For any vector v, in U, ABv is, by definition, A(Bv) so any vector in AB(U) is A applied to a vector in B(U). The dimension of B(U) is rank(B) and, as we just saw, AB(U) must have dimension no larger than that- rank(AB)\le rank(B).
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