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Definition of the determinant i = 1

by jeff1evesque
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jeff1evesque
#1
May1-09, 05:16 PM
P: 312
Lemma:Let B be an element in [tex]M_n_x_n(F)[/tex], where n >= 2. If row i of B equals [tex]e_k[/tex] for some k (1<= k <= n ), then det(B) = (-1)[tex]^i^+^kdet(B_i_k).[/tex]

Proof: The proof is by mathematical induction on n. The lemma is easily proved for n = 2. Assume that for some integer n >= 3, the lemma is true for (n-1) x (n-1) matrices, and let B be an nxn matrix in which row i of B equals [tex]e_k[/tex]for some k (1<=k<=n). The result follows immediately from the definition of the determinant i = 1.

Questions:
1. Can someone help me understand/prove the italized text above.. "...follows from immediately from the definition of the determinant i = 1"

2. And how does det(B) = (-1)[tex]^i^+^kdet(B_i_k).[/tex]? Why isn't there a coefficient in that equation [tex]B_i_,_k[/tex]? This questions is not necessary I guess- since If i follow and understand the proof in the book the logic will work out- but I always thought that in calculating a determinant there would be a scalar coefficient- thus: det(B) = [tex]B_i_,_k[/tex](-1)[tex]^i^+^kdet(B_i_k).[/tex]

thanks,


JL
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tiny-tim
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May1-09, 06:40 PM
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Hi jeff1evesque!

(try using the X2 tag just above the Reply box )
Quote Quote by jeff1evesque View Post
I always thought that in calculating a determinant there would be a scalar coefficient- thus: det(B) = [tex]B_i_,_k[/tex](-1)[tex]^i^+^kdet(B_i_k).[/tex]
Yes, that's right

but in this case, Bik is 1.
jeff1evesque
#3
May1-09, 10:19 PM
P: 312
Quote Quote by tiny-tim View Post
Hi jeff1evesque!

(try using the X2 tag just above the Reply box )


Yes, that's right

but in this case, Bik is 1.
That would imply that every [tex]e_k = 1[/tex]. The symbol e stands for the standard ordered basis- for our particular row 1 correct? So for row i, are all the elements e (standard ordered basis) equal to 1? But how can you assume the ordered basis for our row is all 1's? The e's simply denote the order of the basis for this particular row [tex]F^n[/tex].

Thanks so much,


JL

tiny-tim
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May2-09, 03:39 AM
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Definition of the determinant i = 1

Hi jeff1evesque!
Quote Quote by jeff1evesque View Post
That would imply that every [tex]e_k = 1[/tex]. The symbol e stands for the standard ordered basis- for our particular row 1 correct? So for row i, are all the elements e (standard ordered basis) equal to 1? But how can you assume the ordered basis for our row is all 1's? The e's simply denote the order of the basis for this particular row [tex]F^n[/tex].
Perhaps I'm misunderstanding the notation, but I think it means that eg if row i of B equals e2, then row i is (0,1,0,0,)


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