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Determinants of order n

  1. Apr 29, 2009 #1
    Theorem 4.3:The determinant of an nxn matrix is a linear function of each row when the remaining rows are held fixed. That is, for 1<=r<=n, we have

    det( a1, ..., a_(r-1), u+kv, a_(r+1), ..., a_n ) = det( a1, ..., a_(r-1), u, a_(r+1), ..., a_n )
    + det( a1, ..., a_(r-1), v, a_(r+1), ..., a_n )

    whenever k is a scalar and u, v, and each a_i are row vectors of F^n.

    Proof:The proof is by mathematical induction on n. The result is immediate if n = 1. Assume that for some integer n >= 2 the determinant of an (n-1)x(n-1) matrix is a linear function of each row when the remaining rows are held fixed. Let A be an nxn matrix with rows a1, a2, ..., an, respectively, and suppose that for some r (1<=r<=n), we have a_r = u+kv for some u,v in F^n and some scalar k. Let u = (b1, b2, ..., bn) and v = (c1, c2, ..., cn), and let B and C be the matrices obtained from A by replacing row r of A by u and v, respectively. We must prove that det(A) = det(B) + det(C). We leave the proof of this fact to the reader for the case r = 1. For r > 1 and 1 <= j <= n, the rows of A_(1,j), B_(1,j), C_(1,j) - where these are matrices from taking the determinant coefficient from the first row- are the same except for row r-1. Moreover, row r-1 of A_(1,j) is:
    (b_1 + kC1, ..., b_j-1 + kc_(j-1), b_(j+1) + kc_(j+1), ..., b_n + kc_n), which is the sum of the row r-1 of B_(1,j) and k times row r-1 of C_(1,j).



    Question: Above in italized, I have no idea what they mean by the row are the same (of the matrices left after taking the first determinant to the first row of A) except for row r-1???
     
    Last edited: Apr 29, 2009
  2. jcsd
  3. Apr 29, 2009 #2
    A_(1,j) refers to the n-1xn-1 matrix obtained by crossing out the 1st row and jth column of A. Likewise for B_(1,j) and C_(1,j). Since A, B, C differ at row r only, then the matrices A_(1,j), B_(1,j) and C_(1,j) differ only at row r - 1. Moreover row r-1 of A_(1,j) is a linear combination of the corresponding rows in B_(1,j) and C_(1,j) - this is simple to see.
     
  4. Apr 29, 2009 #3
    Ok I understand that A,B,C differ at row r only, but that should be the same for the matrices A_(1,j), B_(1,j) and C_(1,j)...?
     
  5. Apr 30, 2009 #4
    A_(1,j) is obtained by eliminating the 1st row of A and its jth column. The remaining entries form A_(1,j). The same goes with B_(1,j) with respect to B and C_(1,j) with respect to C.
     
  6. Apr 30, 2009 #5
    I am not asking what the definition of a determinant is. I am asking how in this particular proof, after eliminating the first row and it's corresponding column, the matrices differ at r-1.

    Thanks.
     
  7. Apr 30, 2009 #6
    Come up with three 4x4 matrices A, B and C with identical entries except at one row (say row 3). See what A_(1,j), B_(1,j) and C_(1,j) are and how do they differ from each other. Hopefully, that will illustrate the point clearly.
     
  8. Apr 30, 2009 #7
    I'm claiming they differ at row r, since the entries of these new matrices contain the same elements of the nxn Or 4x4. Therefore row r of A will be u_i+kv_i, row r of B will be u_i, and row r of C will be v_i. This shows that the only row that differs is row r.
     
  9. Apr 30, 2009 #8
    [tex] A = \[ \begin{bmatrix}
    1 & 2 & 3 & 4 \\
    5 & 6 & 7 & 8 \\
    1 & 1 & 1 & 3 \\
    9 & 10 & 11 & 12
    \end{bmatrix}\] [/tex]

    [tex] B = \[ \begin{bmatrix}
    1 & 2 & 3 & 4 \\
    5 & 6 & 7 & 8 \\
    1 & 1 & 1 & 1 \\
    9 & 10 & 11 & 12
    \end{bmatrix}\] [/tex]


    [tex] C = \[ \begin{bmatrix}
    1 & 2 & 3 & 4 \\
    5 & 6 & 7 & 8 \\
    0 & 0 & 0 & 1 \\
    9 & 10 & 11 & 12
    \end{bmatrix}\] [/tex]

    Row 3 of A is a linear combination of row 3 of B and row 3 of C. The matrices differ at row 3. Also, taking j = 1, clearly:


    [tex] A_{1,j} = \[ \begin{bmatrix}
    6 & 7 & 8 \\
    1 & 1 & 3 \\
    10 & 11 & 12
    \end{bmatrix}\] [/tex]

    [tex] B_{1,j} = \[ \begin{bmatrix}
    6 & 7 & 8 \\
    1 & 1 & 1 \\
    10 & 11 & 12
    \end{bmatrix}\] [/tex]


    [tex] C_{1,j} = \[ \begin{bmatrix}
    6 & 7 & 8 \\
    0 & 0 & 1 \\
    10 & 11 & 12
    \end{bmatrix}\] [/tex]

    In which row do the last three matrices differ? Rereading your posts, I'm wondering: maybe you thought "crossing out" meant zero-ing the entries?
     
    Last edited: Apr 30, 2009
  10. May 1, 2009 #9
    No, I know what a determinant is. I was just confused thinking that the new matrices A_(i,j), B_(i,j), and C_(i,j), would have row 1 considered as row 2 (of the original matrix). If the text said they differed at r-1 of the new matrices, I would have understood it right off the bat. For this reason I kept thinking- yes they differ (in your example at row 2) at r-1, but this is really differing at row r, since we transplanted the elements from the original matrix. But I understand the proof now. I don't like the idea of the induction hypothesis, its a big assumption (w/o proof). Thanks a lot.
     
    Last edited: May 1, 2009
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