Determinant: question on proof
$begingroup$
For every matrix $A in M_n(K)$: $det(A) = sum_{sigma in S} operatorname{sgn}(sigma)a_{sigma(1),1} cdots a_{sigma(n),n}$.
Proof: Consider $B = (b_{ij}) in M_n(K)$. Then $ C := AB in M_n(K) $ with $C = (C_1,dots,C_n) $ and $C_k = b_{1k}A_1 + cdots + b_{nk}A_n$. Using the linearity of the determinant we get: $underline{det(C_1,dots,C_n) = sum_{i_1,i_2,dots,i_n} b_{i_1,1}b_{i_2,2}cdots b_{i_n,n} det(A_{i_1},dots,A_{i_n}) }$, where all $i_j$ are independently range from $1$ to $n$.
If $i_k = i_{ell}$, matrix $(A_{i_1},dots,A_{i_n})$ will have two identical columns and therefore $det(A_{i_1},dots,A_{i_n}) = 0$. Only the terms with $sigma = (i_1,dots,i_n)$ a permutation of ${1,dots,n}$ appear in the sum. Using $det(A_{i_1},dots,A_{i_n}) = operatorname{sgn}(sigma)det(A)$, we get:
$underline{det(C_1,dots,C_n) = det(A_{1},dots,A_{n}) sum_{sigmain S} operatorname{sgn}(sigma)b_{sigma(1),1}b_{sigma(2),2}cdots b_{sigma(n),n} }$. With $S$ the set of all permutations of ${1,dots,n}$.
The given statement can be proven by setting $A = I_n$ in the above.
I'm having troubles understanding where the underlined parts come from, especially the first expression for $det(C)$. Hopefully, someone could help me.
linear-algebra
asked Jan 12 at 6:58
ZacharyZachary
3229
$endgroup$
add a comment |
$begingroup$
For every matrix $A in M_n(K)$: $det(A) = sum_{sigma in S} operatorname{sgn}(sigma)a_{sigma(1),1} cdots a_{sigma(n),n}$.
Proof: Consider $B = (b_{ij}) in M_n(K)$. Then $ C := AB in M_n(K) $ with $C = (C_1,dots,C_n) $ and $C_k = b_{1k}A_1 + cdots + b_{nk}A_n$. Using the linearity of the determinant we get: $underline{det(C_1,dots,C_n) = sum_{i_1,i_2,dots,i_n} b_{i_1,1}b_{i_2,2}cdots b_{i_n,n} det(A_{i_1},dots,A_{i_n}) }$, where all $i_j$ are independently range from $1$ to $n$.
If $i_k = i_{ell}$, matrix $(A_{i_1},dots,A_{i_n})$ will have two identical columns and therefore $det(A_{i_1},dots,A_{i_n}) = 0$. Only the terms with $sigma = (i_1,dots,i_n)$ a permutation of ${1,dots,n}$ appear in the sum. Using $det(A_{i_1},dots,A_{i_n}) = operatorname{sgn}(sigma)det(A)$, we get:
$underline{det(C_1,dots,C_n) = det(A_{1},dots,A_{n}) sum_{sigmain S} operatorname{sgn}(sigma)b_{sigma(1),1}b_{sigma(2),2}cdots b_{sigma(n),n} }$. With $S$ the set of all permutations of ${1,dots,n}$.
The given statement can be proven by setting $A = I_n$ in the above.
I'm having troubles understanding where the underlined parts come from, especially the first expression for $det(C)$. Hopefully, someone could help me.
linear-algebra
asked Jan 12 at 6:58
ZacharyZachary
3229
$endgroup$
1
$begingroup$
First expression is just multilinearity of the determinant.
$endgroup$
– Dietrich Burde
Jan 12 at 9:18
add a comment |
$begingroup$
For every matrix $A in M_n(K)$: $det(A) = sum_{sigma in S} operatorname{sgn}(sigma)a_{sigma(1),1} cdots a_{sigma(n),n}$.
Proof: Consider $B = (b_{ij}) in M_n(K)$. Then $ C := AB in M_n(K) $ with $C = (C_1,dots,C_n) $ and $C_k = b_{1k}A_1 + cdots + b_{nk}A_n$. Using the linearity of the determinant we get: $underline{det(C_1,dots,C_n) = sum_{i_1,i_2,dots,i_n} b_{i_1,1}b_{i_2,2}cdots b_{i_n,n} det(A_{i_1},dots,A_{i_n}) }$, where all $i_j$ are independently range from $1$ to $n$.
If $i_k = i_{ell}$, matrix $(A_{i_1},dots,A_{i_n})$ will have two identical columns and therefore $det(A_{i_1},dots,A_{i_n}) = 0$. Only the terms with $sigma = (i_1,dots,i_n)$ a permutation of ${1,dots,n}$ appear in the sum. Using $det(A_{i_1},dots,A_{i_n}) = operatorname{sgn}(sigma)det(A)$, we get:
$underline{det(C_1,dots,C_n) = det(A_{1},dots,A_{n}) sum_{sigmain S} operatorname{sgn}(sigma)b_{sigma(1),1}b_{sigma(2),2}cdots b_{sigma(n),n} }$. With $S$ the set of all permutations of ${1,dots,n}$.
The given statement can be proven by setting $A = I_n$ in the above.
I'm having troubles understanding where the underlined parts come from, especially the first expression for $det(C)$. Hopefully, someone could help me.
linear-algebra
asked Jan 12 at 6:58
ZacharyZachary
3229
$endgroup$
For every matrix $A in M_n(K)$: $det(A) = sum_{sigma in S} operatorname{sgn}(sigma)a_{sigma(1),1} cdots a_{sigma(n),n}$.
Proof: Consider $B = (b_{ij}) in M_n(K)$. Then $ C := AB in M_n(K) $ with $C = (C_1,dots,C_n) $ and $C_k = b_{1k}A_1 + cdots + b_{nk}A_n$. Using the linearity of the determinant we get: $underline{det(C_1,dots,C_n) = sum_{i_1,i_2,dots,i_n} b_{i_1,1}b_{i_2,2}cdots b_{i_n,n} det(A_{i_1},dots,A_{i_n}) }$, where all $i_j$ are independently range from $1$ to $n$.
If $i_k = i_{ell}$, matrix $(A_{i_1},dots,A_{i_n})$ will have two identical columns and therefore $det(A_{i_1},dots,A_{i_n}) = 0$. Only the terms with $sigma = (i_1,dots,i_n)$ a permutation of ${1,dots,n}$ appear in the sum. Using $det(A_{i_1},dots,A_{i_n}) = operatorname{sgn}(sigma)det(A)$, we get:
$underline{det(C_1,dots,C_n) = det(A_{1},dots,A_{n}) sum_{sigmain S} operatorname{sgn}(sigma)b_{sigma(1),1}b_{sigma(2),2}cdots b_{sigma(n),n} }$. With $S$ the set of all permutations of ${1,dots,n}$.
The given statement can be proven by setting $A = I_n$ in the above.
I'm having troubles understanding where the underlined parts come from, especially the first expression for $det(C)$. Hopefully, someone could help me.
linear-algebra
linear-algebra
asked Jan 12 at 6:58
ZacharyZachary
3229
asked Jan 12 at 6:58
ZacharyZachary
3229
edited Jan 12 at 8:01
Zachary
asked Jan 12 at 6:58
ZacharyZachary
3229
asked Jan 12 at 6:58
ZacharyZachary
3229
asked Jan 12 at 6:58
ZacharyZachary
3229
3229
1
$begingroup$
First expression is just multilinearity of the determinant.
$endgroup$
– Dietrich Burde
Jan 12 at 9:18
add a comment |
1
$begingroup$
First expression is just multilinearity of the determinant.
$endgroup$
– Dietrich Burde
Jan 12 at 9:18
1
1
$begingroup$
First expression is just multilinearity of the determinant.
$endgroup$
– Dietrich Burde
Jan 12 at 9:18
$begingroup$
First expression is just multilinearity of the determinant.
$endgroup$
– Dietrich Burde
Jan 12 at 9:18
add a comment |
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$begingroup$
First expression is just multilinearity of the determinant.
$endgroup$
– Dietrich Burde
Jan 12 at 9:18