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Multiplication of two matrices

  1. Aug 21, 2010 #1
    Hi, thank you for viewing this thread. My question is as follow:

    Suppose A is a n x m matrix and B is a m x n matrix, and we also know that the matrix B has infinite solutions, then what will the solution/s of the matix product AB be? I am thinking that it might be a matrix of infinite solutions, but is there a proof to show this case?

    Now suppose we let A be a n x m matrix with no solution, and the conditions for B in the previous paragraph still hold. Then what will the solution of the matrix product AB be in this case? Just wondering if there is a proof to illustrate this case again?
  2. jcsd
  3. Aug 21, 2010 #2


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    What do you mean by a matrix with no solutions or infinite solutions? Are you referring to a matrix equation Ax=b, and solving for x?
  4. Aug 21, 2010 #3
    I think the result that you need is

    [tex]\operatorname{rank}(AB) \leq \min(\operatorname{rank}\,A, \operatorname{rank}\,B)[/tex]

    If [tex]\operatorname{rank}\,C < \operatorname{dim}\,C[/tex] then the system of equations [tex] C\cdot x = b [/tex] does not have a unique solution, but rather "infinite solutions" that span a space of dimension [tex]\operatorname{dim}\,C-\operatorname{rank}\,C[/tex].

    Now let [tex] C = A\cdot B [/tex] where [tex]B[/tex] is not of maximal rank (as in your question) and you can work out the answer...
  5. Aug 21, 2010 #4
    Hi, sorry for the confusion caused, actually what i meant is having matix B in the form Bx=0 and having infinite solutions, so will the multiplication of matrix A onto Bx=0, ABx=0,result in a product where there are infinite solutions as well, regardless of what A is? Is there a proof in showing it?

    Now if we change the conditions to let the matrix A of the equation Ax=0 be a matrix with no solution, while the conditions for B still hold. So will the product ABx=0 changes anything as compared to the first question?

    Erm,I have just started Linear Algebra and have not gone into rank/ vector spaces yet. Is the mentioned topic a prequsite to understanding/ proving this concept?
  6. Aug 22, 2010 #5


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    If Bv= 0 has an infinite number of solutions, then A(Bv)= A(0)= 0 for every such v and so ABv= 0 also has an infinite number of solutions. In fact, if A itself is singular, then there may exist MORE solution that just those such that B= 0.

    Now, "Now suppose we let A be a n x m matrix with no solution", if you still mean Ax= 0, is impossible. That always has the solution x= 0.
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