Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook