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B Proof of elementary row matrix operation.

  1. Jun 6, 2017 #1
    Prove that interchange of two rows of a matrix can be accomplished by a finite sequence of elemenatary row operations of the other two types.

    My proof :-

    If ##A_k## is to be interchanged by ##A_l## then,

    ##\displaystyle \begin{align} A_k &\to A_l + A_k \\ A_l &\to - A_l \\ A_l &\to A_k + A_l \\ A_l &\to A_l + A_k \\ A_l &\to \dfrac12 A_l \\ A_k &\to - A_k \\ A_k &\to A_k + A_l \\ A_k &\to - A_k \end{align}##

    I think this now interchanges original ##A_l## with ##A_k##.
    Is this correct ?
     
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  3. Jun 6, 2017 #2

    andrewkirk

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    Let ##K## and ##L## denote the original rows ##A_k,A_l##. Then we want to stepwise change ##A_k## and ##A_l## so that we end up with ##A_k=L,\ A_l=K##.

    I suggest you re-write your proof, showing in each line the updated values of ##A_k## and ##A_l##, in terms of ##K## and ##L##. For instance line 1 would read:
    $$
    A_k\to A_l+A_k;\quad\quad A_k=L+K;\ A_l=L\quad\quad\quad(1)
    $$

    On my calculation, your sequence does not end up with the desired result.

    However, given your first three steps, the desired result can be reached by three carefully chosen subsequent steps. I think if you write it out the way I suggest, those steps will not be hard to find.
     
    Last edited: Jun 6, 2017
  4. Jun 6, 2017 #3

    mathwonk

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    hey! that's really cool! I never knew that. of course it is quite useles in practice since it took me 6 steps to accomplish one. but nice to know in theory. since the OP's steps did reveal how to do it, I presume his "error" is possibly a matter of miscommunication. Even if not, his steps, repeated, do work as pointed out subsequently by andrew.
     
  5. Jun 7, 2017 #4
    ##\displaystyle \begin{align} A_k &\to A_l + A_k \qquad &K &= L + K
    \\ A_l &\to - A_l \qquad &L &= -L
    \\ A_l &\to A_k + A_l \qquad &L &= L + K - L = K
    \\ A_k &\to -A_l + A_k \qquad &K &= K+ L - K = L
    \end{align}##

    Now ?
     
  6. Jun 7, 2017 #5

    andrewkirk

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    Writing things like ##K=L+K## doesn't make sense. Have another look at what I wrote in post 2. Think of ##A_l## and ##A_k## as the variable values of the rows, which get updated at each step, and write things like ##A_k=L+K;\ A_l=L##, on each line, which tells us the current values of both ##A_l## and ##A_k##, in terms of the constants ##K## and ##L##.
     
  7. Jun 7, 2017 #6
    ##\displaystyle \begin{align} A_k &\to A_l + A_k \qquad &A_k &= L + K
    \\ A_l &\to - A_l \qquad &A_l &= -L
    \\ A_l &\to A_k + A_l \qquad &A_l &= L + K - L = K
    \\ A_k &\to -A_l + A_k \qquad &A_k &= K+ L - K = L
    \end{align}##

    Sorry I did not understand completely what you were saying.
     
  8. Jun 7, 2017 #7

    andrewkirk

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    Is that 4th step allowed: ##A_k\to -A_l+A_k##?

    It seems implied by your OP that the 'other two types' of allowed operations to change a row are
    (1) add another row to it; or
    (2) multiply it by a scalar
    That does not allow setting a row equal to a linear combination of rows, which is what your 4th step does. If that is allowed, then the whole thing can be done in three steps.

    Nevertheless, if operations like your 4th step are included in 'the other two types', your solution is valid.
     
  9. Jun 7, 2017 #8
    Sorry I did that unconsciously. It would require two more steps.
     
  10. Jun 7, 2017 #9

    PeroK

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    In a problem such as this, for your own and everyone else's benefit, I would do it like this:

    ##A_k = K; A_l = L##
    ##A_k = K + L; A_l = L## (add ##A_l## to ##A_k##)
    ##A_k = K + L; A_l = 2L## (multiply ##A_l## by ##2##)

    Etc. Then all is clear
     
  11. Jun 7, 2017 #10

    mathwonk

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    here are my 6 steps, imitating those of the OP, in which the first, 3rd and 5th are adding a row to another, and the 2nd, 4th and 6th are multiplying a row by a scalar: ( the rows are written side by side)

    [ A : B] --> [A+B : B] --> [A+B : -B] --> [A+B : A] --> [A+B : -A] --> [B : -A] --> [B : A].
     
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