prove linear indep implies detM equals 0

Show that for lin indep vectors, det(M) = 0 PhL 6.10.15 ______________________________________________________________________________ Plan A: Gauss Elimination Approach Theorem. For an n x n square matrix A, if the column rank is n, then det(A) ≠ 0. (A.4.2) Proof #1. If the n columns of A are linearly independent, then each column must contain at least one non-zero number. One cannot have a column of all zeros. We know we can swap rows around without changing the value of |det(A)|. So find a row which has a non-zero number in the left column and swap this with the first row so there is a non-zero number in the upper left corner. Then add multiples of this new first row to all the other rows so as to clear out (to zeros) the rest of the first column. These operations also do not affect det(A). After After doing the above, find a non-zero number in the second column (it must exist WHY?) and swap rows so it is in the A22 matrix position. Repeat this process to get a matrix whose lower left triangle is all zeros and whose diagonal has all non-zero values. This matrix then has upper triangular form. The determinant of such a matrix is the product of the diagonal values and thus we have shown that |det(A)| ≠ 0. This process is usually called Gauss elimination. One might wonder why all the diagonal elements are non-zero for some general case. This is assured at each step of the elimination. We leave it to the reader to make this proof more rigorous. If the n columns of M are linearly independent, then each column must contain at least one non-zero element. One cannot have a column of all zeros. So column 1 must have at least one non-zero element. Pick a row which has such a non-zero element in column 1. Swap this row with the top row without changing |det(M) by Theorem 4. Then add multiples of this new first row to all the other rows so as to clear out (to zeros) the rest of the first column. At this point, what can be said about column 2? It seems that somehow column 2 could end up being 0. How can I show this can't happen? I would have to show that adding multiples of rows within a matrix does not alter the fact that the columns are independent. But I am not allowed to use the det(M) ≠ 0 test since that is what I am trying to show From Theorem 4 we can swap rows around without changing the value of |det(M)|. So find a row which has a non-zero number in the left column and swap this with the first row so there is a non-zero number in the upper left corner. Then add multiples of this new first row to all the other rows so as to clear out (to zeros) the rest of the first column. These operations also do not affect det(M). After doing the above, find a non-zero number in the second column (it must exist) and swap rows so it is in the A22 matrix position. ______________________________________________________________________________ Plan B: Try relating det(M) to the unit vector matrix If the rows ri are linearly independent in En then they form a complete set or basis for En. Thus, one can write any of the unit vectors ei as a linear combination of the ri. ei = Σj Sij rj = Siiri + Σj≠i Sij rj Similarly one could write ri = Σj Qij ej = Qiiei + Σj≠i Qij ej Perhaps r1 = Q11e1 + Σj≠1 Q1j ej Q11 ≠ 0 r2 = Σj≠2 Q2j ej r3 = Q33e3 + Σj≠3 Q3j ej Q33 ≠ 0 Then det(r1, r2, r3) = det ( Q11e1 + Σj≠1 Q1j ej, Σj≠2 Q2j ej, Q33e3+Σj≠3 Q3j ej) = Q11Q33 det ( e1 + Σj≠1 [Q1j/Q11] ej, Σj≠2 Q2j ej, e3 + Σj≠3 [Q3j/Q33] ej) This is a sum of four determinants. Ignore leading factors and write it as det1 = det ( e1, Σj≠2 Q2j ej, e3) = 0 since det (e1,e1,e3) = 0, etc det1 = det ( Σj≠1 [Q1j/Q11] ej, Σj≠2 Q2j ej, e3) = Σj≠1 [Q1j/Q11] Σk≠2 Q2k det (ej, ek, e3) = Σj≠1 [Q1j/Q11] Σk≠2 Q2k εjk3 = Σj≠1 [Q1j/Q11] Q21 εj13 = [Q12/Q11] Q21 ε213 This is too complicated!!! We now apply Theorem 3 twice to eliminate "adding multiples of other rows" to get = Q11Q33 det ( e1, Σj≠2 Q2j ej, e3) = Q11Q33 det ( e1, Q21 e1 + Q23e3, e3) = Q11Q33 [ Q21 det (e1,e1,e3) + Q23 det (e1,e3,e3) ] = 0 + 0 = 0 ! So I have shown exactly the wrong result: det(r1, r2, r3) = 0 if ri are linearly independent! According to Theorem 3 we can ignore mul In general some Sii will vanish and some will not. So here is a typical mixed case: e1 = S11r1 + Σj≠i S1j rj e2 = Σj≠i S2j rj e2 = S33r1 Σj≠i S3j rj First define some new rows ri' ≡ ri + Σj≠i Sij rj According to Theorem 3, det( r'1,r'2....r'n) = det( r1,r2....rn). Now define some new rows r"i, r"i = Now define some new rows ri" ri" = Sii ri' . Then we know that det( ri") = (Sii)n det( ri') = (Sii)n det( ri) But ri" = ei and det(ei) = 1 so 1 = (Sii)n det( ri) and therefore det(ri) ≠ 0 Now suppose Sii= 0. Then we start off with ei = Σj≠i Sij rj ______________________________________________________________________________ Plan C: Linear equations argument If the rows are independent, then we have a set of n linear equations in n unknowns and we know it has a solution so det(M) ≠ 0. But HOW do we know it has a solution if we don't know that det(M) ≠ 0? It could maybe have a solution even if det(M) = 0, ______________________________________________________________________________ Plan D: The volume argument Quote my n-piped thing that if vectors are independent, must have volume ≠ 0 so det(M) ≠ 0. But that volume thing is a real bear. At least this Plan D convinces me the theorem is true. ______________________________________________________________________________ Plan E: What does Shilov say? Have to search high and low to find anything. I searched on "linearly dependent" and could find no theorem. Indep search same thing, there is so much I cannot find anything. So no conclusion here. I need someone who exactly states my theorem. I need to be able to search for linearly near determinant! Witzend can do this. ______________________________________________________________________________ Plan F: What does Lucht Matrix Binder say? As I page through tab 1, I see this theorem supposedly appearing as Phil Theorem 9.2 in Section 2. I move to Section 2 page 9 and there is the theorem. Let's read it. It is the wrong theorem. Still: I say to consider Ax = 0 or ΣkAjkxk = 0 or Σk(ck)j xk = 0 true for all j. This is a vector equation Σkck xk = 0 so lin indep columns {xk} = 0. So this shows that: Ax = 0 x = 0 if columns lin indep This does say something: lin indep nullspace = 0 How about this then. (1) columns linearly independent nullspace is only the trivial nullspace (2) the mapping is 1 to 1 therefore, think I can show that (3) therefore inverse of Ax = y must exist and be unique (4) but A-1 exists implies detA ≠ 0. Maybe that will fly? I think so. ______________________________________________________________________________ Plan G :Here from http://www.math.lsa.umich.edu/~hochster/419/det.html Fact 8. The determinant of an n by n matrix A is 0 if and only if the rows are linearly dependent (and not zero if and only if they are linearly independent). That is, the determinant of A is not zero if and only if A is invertible. Here is why: The issue is not affected by switching rows, adding a multiple of one row to another, or multiplying a row by a nonzero scalar. Thus, we may assume that A is a square matrix in RREF. If the rows are independent, it will then be the identity, while otherwise it has a row of zeros. So at least they are confirming my theorem. RREF means "reduced row echelon form".