Results 1 to 7 of 7

Math Help - Linear algebra proof - Elementary matricies

  1. #1
    Newbie
    Joined
    Mar 2014
    From
    Bangladesh
    Posts
    3

    Linear algebra proof - Elementary matricies

    I'm having trouble with this proof we need to learn -

    Let E = (I). Prove that (A) = EA for every A ∈ Mnn

    A hint some guy gave is to begin with the ij entry of EA and prove that it equals the ij entry of (A), by using the
    definition of matrix multiplication and the definitions of E and (A).

    Any help? Thanks a bunch
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,380
    Thanks
    745

    Re: Linear algebra proof - Elementary matricies

    What is ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2014
    From
    Bangladesh
    Posts
    3

    Re: Linear algebra proof - Elementary matricies

    Let Mnn be the set of all real n n matrices, let 1 ≤ k, l ≤ n be fixed integers with k != l, and let c (!= 0) be
    a real number. Consider the function : Mnn → Mnn, whose image (A) for each A ∈ Mnn is defined
    by

    Linear algebra proof - Elementary matricies-math.png

    for 1 ≤ i, j ≤ n

    Oops sorry, this was some information supplied with the theories.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,350
    Thanks
    903

    Re: Linear algebra proof - Elementary matricies

    This is just showing that elementary row operations can be cast as matrix multiplication. A pretty standard part of any linear algebra course.

    http://en.wikipedia.org/wiki/Elementary_matrix
    Last edited by romsek; March 31st 2014 at 04:35 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2014
    From
    Bangladesh
    Posts
    3

    Re: Linear algebra proof - Elementary matricies

    Thanks for the reply!

    Yeah I kinda thought that was the case, although the problem I'm having is the actual index notation in trying to prove the fact :P

    Not too sure on how to start it even
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,350
    Thanks
    903

    Re: Linear algebra proof - Elementary matricies

    Quote Originally Posted by Goatman View Post
    Thanks for the reply!

    Yeah I kinda thought that was the case, although the problem I'm having is the actual index notation in trying to prove the fact :P

    Not too sure on how to start it even
    The row addition transformation section of the wiki page I linked has the exact matrix you need. They call it $T_{i,j}(m)$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,380
    Thanks
    745

    Re: Linear algebra proof - Elementary matricies

    To be more precise, $\mu$ should be "tagged" with the subscripts k,l because you get a DIFFERENT function for each pair.

    I will suppose that k and l are fixed, and unequal. So, $\mu$ is the function that adds c times the k-th row of a matrix A, to the l-th row, and leaves all other rows unchanged.

    We wish to show that $\mu(A) = \mu(I)A$. For all rows but the l-th row, A is left unchanged by $\mu$, so we need to show that likewise for any row but the l-th row, the i-th row of $\mu(I)A$ is the same as the i-th row of A.

    By the definition of matrix multiplication, the i-th row of $\mu(I)A$ is the dot product of the i-th row of $\mu(I)$ with each column of A in turn, that is:

    $$(\mu(I)A)_{ij} = \sum_{r = 1}^n (\mu(I))_{ir}(A)_{rj}$$

    If i is not equal to l, then the i-th row of $\mu(I)$ is just the i-th standard (row) vector: (0,...,0,1,0,...,0) with 1 in the i-th place. That is:

    $(\mu(I))_{ir} = 0, i \neq r$, and $(\mu(I))_{ii} = 1$. In this case, the only term that survives in the sum is $(\mu(I))_{ii}(A)_{ij} = (A)_{ij}$, so we see that for any row but the l-th, A is indeed left unchanged.

    Now we consider what happens for our "exceptional" l-th row. In this case, the l-th row of $\mu(I)$ is: $e_l + ce_k$. For example, if l < k, this looks like:

    (0,...,0,1,0,...,0,c,0,...,0).

    So now we get TWO terms that survive in the dot product, one is $(A)_{lj}$, and the other is $c(A)_{kj}$, the sum of which is precisely the l,j-th entry of $\mu(A)$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: July 18th 2010, 09:48 AM
  2. Replies: 7
    Last Post: August 30th 2009, 10:03 AM
  3. Help with elementary Linear Algebra proof?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 7th 2009, 01:53 PM
  4. Elementary linear algebra college work
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 26th 2008, 09:20 AM
  5. Linear Algebra-Matricies/Word Problems?
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: February 2nd 2008, 11:22 AM

Search Tags


/mathhelpforum @mathhelpforum