Proof of Determinant property via definition in linear algebra
$begingroup$
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:-
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 $tau$)?
How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?
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
edited Jan 23 at 11:18
Davide Giraudo
128k17156268
asked Jan 17 at 7:07
Onkar SinghOnkar Singh
317
$endgroup$
add a comment |
$begingroup$
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:-
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 $tau$)?
How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?
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
edited Jan 23 at 11:18
Davide Giraudo
128k17156268
asked Jan 17 at 7:07
Onkar SinghOnkar Singh
317
$endgroup$
add a comment |
$begingroup$
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:-
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 $tau$)?
How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?
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
edited Jan 23 at 11:18
Davide Giraudo
128k17156268
asked Jan 17 at 7:07
Onkar SinghOnkar Singh
317
$endgroup$
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:-
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 $tau$)?
How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?
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
edited Jan 23 at 11:18
Davide Giraudo
128k17156268
asked Jan 17 at 7:07
Onkar SinghOnkar Singh
317
edited Jan 23 at 11:18
Davide Giraudo
128k17156268
asked Jan 17 at 7:07
Onkar SinghOnkar Singh
317
edited Jan 23 at 11:18
Davide Giraudo
128k17156268
edited Jan 23 at 11:18
Davide Giraudo
128k17156268
edited Jan 23 at 11:18
Davide Giraudo
128k17156268
128k17156268
asked Jan 17 at 7:07
Onkar SinghOnkar Singh
317
asked Jan 17 at 7:07
Onkar SinghOnkar Singh
317
asked Jan 17 at 7:07
Onkar SinghOnkar Singh
317
317
add a comment |
add a comment |
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