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Linear Algebra Question

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Write the general element in terms of aij and bij for (AB)^T [AB transposed].


    2. Relevant equations

    (AB)^T = B^T*A^T; A=[aij]mxn; B=[bij]nxp

    3. The attempt at a solution
    n
    AB= [sigma aik*bkj]mxp. Let this be equal to [xij]mxp
    k=1
    n
    (AB)^T=[[sigma aik*bkj]mxp]^T
    k=1

    =[xji]pxm

    n
    =[sigma aki*bjk]mxp
    k=1
    n
    so the general element xji=[sigma aki*bjk]
    k=1


    My teacher says this is wrong. Where did I go wrong?


    ------------------------
    Alternate way I used to "check" my wrong answer:
    n
    (AB)^T=B^T*A^T=[bji]pxn[aji]nxm=[sigma bjk*aki]pxm=[sigma aki*bjk]pxm
    k=1


    We are using an differential equations/linear algebra textbook for engineers. It never discusses element-by-element proofs, and it leaves out many important differential equations topics, such as exact equations. I have a real diff eq book that my neighbor lent me, but I have to teach myself these types of problems through Wikipedia.
     
  2. jcsd
  3. Oct 9, 2008 #2
    I'm really not certain what all that stuff that's not in LaTeX is, but consider this:

    If [itex] A = \{a_{ij} \} [/itex] then you know that 1) [itex] A^T = \{ a_{ji} \} [/itex]
    and you also know that 2) [itex] (AB)^T = B^T A^T [/itex].

    Now if [itex] B = \{ b_{jk} \} [/itex] (where I've used the index "j" again since I know that j will iterate B in precisely the same manner as A for AB to make sense) you can write the matrix AB as [itex] AB = \displaystyle \left\{ a_{ij} b_{jk} \right\} [/itex]. Now let's say, that without being too rigorous, you were to apply the operations from 1) and 2) to this sum, what would you get?
     
  4. Oct 10, 2008 #3
    [bkj][aij] ?
     
  5. Oct 15, 2008 #4
    So I let A={a[tex]_{}ij[/tex]} and B={b[tex]_{}ij[/tex]}.

    Then I know that A[tex]^{}T[/tex]={a[tex]_{}ji[/tex]} and B[tex]^{}T[/tex]={a[tex]_{}ji[/tex]}.

    A typical element of the product B[tex]^{}T[/tex]A[tex]^{}T[/tex]={b[tex]_{}ji[/tex]a[tex]_{}ji[/tex]}.

    However, B[tex]^{}T[/tex]A[tex]^{}T[/tex]={[tex]\Sigma[/tex][tex]^{}n[/tex][tex]_{}k=1[/tex]b[tex]_{}jk[/tex]a[tex]_{}kj[/tex]}.


    Is this correct?
     
  6. Oct 15, 2008 #5
    So I let A={a[tex]_{}ij[/tex]} and B={b[tex]_{}ij[/tex]}.

    Then I know that A[tex]^{}T[/tex]={a[tex]_{}ji[/tex]} and B[tex]^{}T[/tex]={a[tex]_{}ji[/tex]}.

    A typical element of the product B[tex]^{}T[/tex]A[tex]^{}T[/tex]={b[tex]_{}ji[/tex]a[tex]_{}ji[/tex]}.

    The sum would therefore be: B[tex]^{}T[/tex]A[tex]^{}T[/tex]={[tex]\Sigma[/tex]b[tex]_{}jk[/tex]a[tex]_{}ki[/tex]} from k=1 to n.

    Is this correct?
     
  7. Oct 15, 2008 #6
    Max, I think we are both right.

    What did the professor get?
     
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