Results 1 to 2 of 2

Math Help - Prove a(AB)=(aA)B=A(aB) ..

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    133

    Prove a(AB)=(aA)B=A(aB) ..

    Prove a(AB)=(aA)B=A(aB) for any scalar a. At first I started out like:

    a(AB)=(aA)(aB) and quickly noticed that that was very wrong. And then I kept trying to play around with it in different ways and I seem to get tangled and I know this is such an easy concept.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member HappyJoe's Avatar
    Joined
    Sep 2010
    From
    Denmark
    Posts
    234
    I suppose A and B are matrices.

    You need to apply the definition of matrix product. Note that two matrices are equal, precisely when each of the corresponding entries are equal. Let me change the name of the scalar to c.

    Let a_{ij} be the (i,j)-entry of A and let b_{ij} be the (i,j)-entry of B. Suppose also that A is an m \times n matrix, and that B is an n \times p matrix.

    Then the (i,j)-entry of AB is "the i'th row of A times the j'th column of B". So this (i,j)-entry of AB is:

    \sum_{k=1}^n a_{ik}b_{kj},

    and so the (i,j)-entry of c(AB) is

    c\sum_{k=1}^n a_{ik}b_{kj} = \sum_{k=1}^n ca_{ik}b_{kj},

    where the first expression is the entry that follows by applying the definition of the product directly, and where the equality follows from bringing the scalar c into each term of the sum.

    Consider then the product (cA)B. We need to check that the (i,j)-entry of this product is same expression as for c(AB). But cA is the matrix, where the (i,j)-entry is ca_{ij}. The product (cA)B has as its (i,j)-entry the i'th row of this matrix multiplied by the j'th column of B, so

    \sum_{k=1}^n ca_{ik}b_{kj}.

    But this is exactly the same expression as above.

    Similarly you may show that A(cB) has the correct (i,j)-entries.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove that
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 21st 2010, 05:48 AM
  2. Prove n^2<= ......
    Posted in the Advanced Algebra Forum
    Replies: 12
    Last Post: November 17th 2009, 05:52 AM
  3. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  4. prove that
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 7th 2008, 05:14 PM
  5. prove
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 7th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum