Multiplication of two matrices

[Zeato]

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?

 

[Office_Shredder]

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?

 

[Simon_Tyler]

I think the result that you need is

\operatorname{rank}(AB) \leq \min(\operatorname{rank}\,A, \operatorname{rank}\,B)

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

Now let C = A\cdot B where B is not of maximal rank (as in your question) and you can work out the answer...

 

[Zeato]

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?

 

[HallsofIvy]

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.