# Definition of the determinant i = 1

definition of the determinant i = 1"

Lemma:Let B be an element in $$M_n_x_n(F)$$, where n >= 2. If row i of B equals $$e_k$$ for some k (1<= k <= n ), then det(B) = (-1)$$^i^+^kdet(B_i_k).$$

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 $$e_k$$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)$$^i^+^kdet(B_i_k).$$? Why isn't there a coefficient in that equation $$B_i_,_k$$? 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) = $$B_i_,_k$$(-1)$$^i^+^kdet(B_i_k).$$

thanks,

JL

tiny-tim
Homework Helper
Hi jeff1evesque!

(try using the X2 tag just above the Reply box )
… I always thought that in calculating a determinant there would be a scalar coefficient- thus: det(B) = $$B_i_,_k$$(-1)$$^i^+^kdet(B_i_k).$$

Yes, that's right …

but in this case, Bik is 1.

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 $$e_k = 1$$. 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 $$F^n$$.

Thanks so much,

JL

tiny-tim
That would imply that every $$e_k = 1$$. 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 $$F^n$$.