Definition of the determinant i = 1

In summary, the determinant is defined as the sum of the products of the elements in each row and column of a matrix, with alternating signs. The lemma states that if row i of the matrix B equals a standard ordered basis, then the determinant can be calculated by taking the determinant of the submatrix B_i_k and multiplying it by (-1)^i^+^k. This follows immediately from the definition of the determinant, as shown in the proof by mathematical induction.
  • #1
jeff1evesque
312
0
definition of the determinant i = 1"

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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  • #2
Hi jeff1evesque! :smile:

(try using the X2 tag just above the Reply box :wink:)
jeff1evesque said:
… 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. :wink:
 
  • #3


tiny-tim said:
Hi jeff1evesque! :smile:

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


Yes, that's right …

but in this case, Bik is 1. :wink:

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
 
  • #4
Hi jeff1evesque! :smile:
jeff1evesque said:
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,…)
 

What is the definition of the determinant i = 1?

The determinant of a matrix with only one element, i = 1, is simply the value of that element. In other words, the determinant of i = 1 is just 1.

How do you find the determinant of i = 1?

To find the determinant of i = 1, you do not need to use any special formulas or methods. Simply take the value of the element, which is 1, and that is your determinant.

Why is the determinant of i = 1 equal to 1?

The determinant of i = 1 is equal to 1 because the definition of a determinant is the product of the main diagonal elements in a square matrix. Since i = 1 is a single element matrix, the only element on the main diagonal is 1, and the product of 1 is 1.

What is the significance of the determinant of i = 1?

The determinant of i = 1 does not have any significant mathematical or practical implications. It is simply a single element in a matrix and does not affect any calculations or operations on the matrix.

Can the determinant of i = 1 be negative or zero?

No, the determinant of i = 1 can only be positive 1. This is because the determinant is the product of the main diagonal elements, and the only element on the main diagonal is 1, which has a positive value. A negative or zero value would require the presence of a negative or zero element on the main diagonal, which is not the case for i = 1.

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