definition of the determinant i = 1"

jeff1evesque

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

tiny-tim

Hi jeff1evesque!

(try using the X₂ tag just above the Reply box )
Yes, that's right …

but in this case, B_ik is 1.

jeff1evesque

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

Hi jeff1evesque!
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,…)