Proof of Determinant property via definition in linear algebra












1












$begingroup$


Let A=$(a_{ij})_n$ be a square matrix & B=$(b_{ij})_n$ be another square matrix obtained after interchanging two rows of A, say $p^{th}$ and $q^{th}$. Then |B|=-|A|.



Proof Given:-
We have,



$b_{pj}$ = $a_{qj}$



$b_{qj} = $$a_{pj}$ and



$a_{ij}$ = $b_{ij}$ $forall$ j,i($neq$p,q)



Consider the transposition $tau$=(p q). Note that $S_n$={$sigmacirctau$ | $sigma$ $epsilon$ $S_n$}



Hence,



$|B|=sum_{sigma epsilon S_n} = operatorname{sign}(sigmacirctau$) $b_{1,sigma circ tau(1)} b_{2,sigma circ tau(2)}b_{p,sigma circ tau(p)}dots b_{q,sigma circtau(q)} b_{n,sigma circ tau(n)}$



$|B|=sum_{sigma epsilon S_n} = operatorname{sign}(sigma) operatorname{sign}( tau ) b_{1sigma(1)}$$b_{2sigma(2)}$....$b_{psigma(q)}$.....$b_{qsigma(p)}$....$b_{nsigma(n)}$



|B| = sign($tau$) $sum_{sigma epsilon S_n} sign(sigma) a_{1sigma(1)} a_{2sigma(2)}....a_{qsigma(q)}.....a_{psigma(p)}....a_{nsigma(n)}$



|B| = -|A|



My doubts are:-




  1. Upon interchanging rows, the order will not change and hence, won't change the sign($sigma$). Then why sign($sigma$) is being replaced by sign($sigma$ o $tau$)?


  2. How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?


  3. How come $S_n$={$sigma circtau$ | $sigma$ $epsilon$ $S_n$}? How to explain this equality?



See link: Proof of Part 1



Thanks in advance!!






linear-algebra determinant permutation-cycles







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let A=$(a_{ij})_n$ be a square matrix & B=$(b_{ij})_n$ be another square matrix obtained after interchanging two rows of A, say $p^{th}$ and $q^{th}$. Then |B|=-|A|.



    Proof Given:-
    We have,



    $b_{pj}$ = $a_{qj}$



    $b_{qj} = $$a_{pj}$ and



    $a_{ij}$ = $b_{ij}$ $forall$ j,i($neq$p,q)



    Consider the transposition $tau$=(p q). Note that $S_n$={$sigmacirctau$ | $sigma$ $epsilon$ $S_n$}



    Hence,



    $|B|=sum_{sigma epsilon S_n} = operatorname{sign}(sigmacirctau$) $b_{1,sigma circ tau(1)} b_{2,sigma circ tau(2)}b_{p,sigma circ tau(p)}dots b_{q,sigma circtau(q)} b_{n,sigma circ tau(n)}$



    $|B|=sum_{sigma epsilon S_n} = operatorname{sign}(sigma) operatorname{sign}( tau ) b_{1sigma(1)}$$b_{2sigma(2)}$....$b_{psigma(q)}$.....$b_{qsigma(p)}$....$b_{nsigma(n)}$



    |B| = sign($tau$) $sum_{sigma epsilon S_n} sign(sigma) a_{1sigma(1)} a_{2sigma(2)}....a_{qsigma(q)}.....a_{psigma(p)}....a_{nsigma(n)}$



    |B| = -|A|



    My doubts are:-




    1. Upon interchanging rows, the order will not change and hence, won't change the sign($sigma$). Then why sign($sigma$) is being replaced by sign($sigma$ o $tau$)?


    2. How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?


    3. How come $S_n$={$sigma circtau$ | $sigma$ $epsilon$ $S_n$}? How to explain this equality?



    See link: Proof of Part 1



    Thanks in advance!!






    linear-algebra determinant permutation-cycles







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let A=$(a_{ij})_n$ be a square matrix & B=$(b_{ij})_n$ be another square matrix obtained after interchanging two rows of A, say $p^{th}$ and $q^{th}$. Then |B|=-|A|.



      Proof Given:-
      We have,



      $b_{pj}$ = $a_{qj}$



      $b_{qj} = $$a_{pj}$ and



      $a_{ij}$ = $b_{ij}$ $forall$ j,i($neq$p,q)



      Consider the transposition $tau$=(p q). Note that $S_n$={$sigmacirctau$ | $sigma$ $epsilon$ $S_n$}



      Hence,



      $|B|=sum_{sigma epsilon S_n} = operatorname{sign}(sigmacirctau$) $b_{1,sigma circ tau(1)} b_{2,sigma circ tau(2)}b_{p,sigma circ tau(p)}dots b_{q,sigma circtau(q)} b_{n,sigma circ tau(n)}$



      $|B|=sum_{sigma epsilon S_n} = operatorname{sign}(sigma) operatorname{sign}( tau ) b_{1sigma(1)}$$b_{2sigma(2)}$....$b_{psigma(q)}$.....$b_{qsigma(p)}$....$b_{nsigma(n)}$



      |B| = sign($tau$) $sum_{sigma epsilon S_n} sign(sigma) a_{1sigma(1)} a_{2sigma(2)}....a_{qsigma(q)}.....a_{psigma(p)}....a_{nsigma(n)}$



      |B| = -|A|



      My doubts are:-




      1. Upon interchanging rows, the order will not change and hence, won't change the sign($sigma$). Then why sign($sigma$) is being replaced by sign($sigma$ o $tau$)?


      2. How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?


      3. How come $S_n$={$sigma circtau$ | $sigma$ $epsilon$ $S_n$}? How to explain this equality?



      See link: Proof of Part 1



      Thanks in advance!!






      linear-algebra determinant permutation-cycles







      share|cite|improve this question











      $endgroup$




      Let A=$(a_{ij})_n$ be a square matrix & B=$(b_{ij})_n$ be another square matrix obtained after interchanging two rows of A, say $p^{th}$ and $q^{th}$. Then |B|=-|A|.



      Proof Given:-
      We have,



      $b_{pj}$ = $a_{qj}$



      $b_{qj} = $$a_{pj}$ and



      $a_{ij}$ = $b_{ij}$ $forall$ j,i($neq$p,q)



      Consider the transposition $tau$=(p q). Note that $S_n$={$sigmacirctau$ | $sigma$ $epsilon$ $S_n$}



      Hence,



      $|B|=sum_{sigma epsilon S_n} = operatorname{sign}(sigmacirctau$) $b_{1,sigma circ tau(1)} b_{2,sigma circ tau(2)}b_{p,sigma circ tau(p)}dots b_{q,sigma circtau(q)} b_{n,sigma circ tau(n)}$



      $|B|=sum_{sigma epsilon S_n} = operatorname{sign}(sigma) operatorname{sign}( tau ) b_{1sigma(1)}$$b_{2sigma(2)}$....$b_{psigma(q)}$.....$b_{qsigma(p)}$....$b_{nsigma(n)}$



      |B| = sign($tau$) $sum_{sigma epsilon S_n} sign(sigma) a_{1sigma(1)} a_{2sigma(2)}....a_{qsigma(q)}.....a_{psigma(p)}....a_{nsigma(n)}$



      |B| = -|A|



      My doubts are:-




      1. Upon interchanging rows, the order will not change and hence, won't change the sign($sigma$). Then why sign($sigma$) is being replaced by sign($sigma$ o $tau$)?


      2. How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?


      3. How come $S_n$={$sigma circtau$ | $sigma$ $epsilon$ $S_n$}? How to explain this equality?



      See link: Proof of Part 1



      Thanks in advance!!






      linear-algebra determinant permutation-cycles




      linear-algebra determinant permutation-cycles






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      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 23 at 11:18









      Davide Giraudo

      127k16151264




      127k16151264










      asked Jan 17 at 7:07









      Onkar SinghOnkar Singh

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