== Intro to Determinants == Below, N ranges over the natural numbers, and B denotes an NxN matrix over field F. 1a: The Det_N fnc is a degree-N homogeneous polynomial in N^2 variables, namely, in its N^2 many entries. This polynomial is a sum of N! (i.e, N factorial) many monomials, each with coeff either +1 or -1. 1b: Fix an N >= 2. Then exactly half of the N! many monomials has coeff +1, and the other half have coeff -1. 2: Viewed as a fnc of its N many columns, Det_N is linear in each column separately. So Det_N() is "N-multi-linear" in its N columns. Also: Det_N is "N-multi-linear" in its N rows. 3: Viewing a matrix as an element of an N^2-dimensional vectorspace, fnc Det_N() is /not/ linear, except when N=1. 4a: Det_N is "anti-commutative" as a function of its columns; exchanging two columns negates the determinant (i.e, multiplies it by -1). Also, Det_N is anti-commutative in its rows. 4b: More strongly (in the presence of multi-linearity), Det_N is "alternating" as a function of its columns, in that: If some two columns are equal, then the determinant is zero. Similarly, Det_N is alternating in its rows. 5: Det_N(transpose of B) equals Det_N(B). 6: Det_N(B) is non-zero IFF B has full rank, i.e its set of rows is lin-indep; equivalently, its column-set is lin-independent. 7: For two NxN matrices B and C, it turns out that Det(C * B) = Det(C)·Det(B). The lefthand "*" is matrix-mult, whereas the righthand "·" is mult in F, the scalar field. Hence Det_N is a group-homomorphism from GL_N(F) to the multiplicative-group (F - {0} , * , 1) . Note that "GL" stands for "general linear" group, and comprises the matrices with non-zero determinant. Aside: The group SL_N(F) is called the "special linear group". It is the set of NxN matrices over F that have determinant equal to 1. ================================================================ on Mon Oct 11 09:23:09 EDT 2010 |\^/| Maple 10 (SUN SPARC SOLARIS) ._|\| |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2005 \ MAPLE / All rights reserved. Maple is a trademark of <____ ____> Waterloo Maple Inc. | Type ? for help. > M0 := Matrix(0,0,[[]]); M1 := <>; M2 := <,>; M0 := [] M1 := [a] [a b] M2 := [ ] [c d] > M3 := <,,>; [a b c] [ ] M3 := [d e f] [ ] [g h i] > M4 := <,,,>; [a b c d] [ ] [e f g h] M4 := [ ] [i j k l] [ ] [m n o p] > M5 := <,,,,>; [a b c d e] [ ] [f g h i j] [ ] M5 := [k l m n o] [ ] [p q r s t] [ ] [u v w x y] > Det(M0):=Det(M0); Det(M1):=Det(M1); Det(M2):=Det(M2); Det(M3):=Det(M3); Det(M4):=Det(M4); Det(M5):=Det(M5); Det([]) := 1 Det([a]) := a [a b] Det([ ]) := a d - b c [c d] [a b c] [ ] Det([d e f]) := a e i - a f h + d h c - d b i + g b f - g e c [ ] [g h i] [a b c d] [ ] [e f g h] Det([ ]) := [i j k l] [ ] [m n o p] a f k p - a f l o + a j o h - a j g p + a n g l - a n k h - e b k p + e b l o - e j o d + e j c p - e n c l + e n k d + i b g p - i b o h + i f o d - i f c p + i n h c - i n g d - m b g l + m b k h - m f k d + m f c l - m j h c + m j g d [a b c d e] [ ] [f g h i j] [ ] Det([k l m n o]) := [ ] [p q r s t] [ ] [u v w x y] -f b m s y + f b m t x - f b r x o + f b r n y - f b w n t + f b w s o + f l c s y - f l c t x + f l r x e - f l r d y + f l w d t - f l w s e - f q c n y + f q c x o - f q m x e + f q m d y - f q w o d + f q w n e + f v c n t - f v c s o + f v m s e - f v m d t + k v h d t - k v r d j + k v r e i + f v r o d - f v r n e + k b h s y - k b h t x + k b r x j - k b r i y + k b w i t - k b w s j - k g c s y + k g c t x - k g r x e + k g r d y - k g w d t + k g w s e + k q c i y - k q c x j + k q h x e - k q h d y + k q w d j - k q w e i - k v c i t + k v c s j - k v h s e + a g m s y - a g m t x + a g r x o - a g r n y + a g w n t - a g w s o - a l h s y + a l h t x - a l r x j + a l r i y - a l w i t + a l w s j + a q h n y - a q h x o + a q m x j - a q m i y + a q w i o - a q w n j - a v h n t + a v h s o - a v m s j + a v m i t - a v r i o + a v r n j - p b h n y + p b h x o - p b m x j + p b m i y - p b w i o + p b w n j + p g c n y - p g c x o + p g m x e - p g m d y + p g w o d - p g w n e - p l c i y + p l c x j - p l h x e + p l h d y - p l w d j + p l w e i + p v c i o - p v c n j + p v h n e - p v h o d + p v m d j - p v m e i + u b h n t - u b h s o + u b m s j - u b m i t + u b r i o - u b r n j - u g c n t + u g c s o - u g m s e + u g m d t - u g r o d + u g r n e + u l c i t - u l c s j + u l h s e - u l h d t + u l r d j - u l r e i - u q c i o + u q c n j - u q h n e + u q h o d - u q m d j + u q m e i ================================================== == A trick for computing Det() of a 3x3 matrix == ================================================== (This trick ONLY works for 3x3; nothing smaller, nothing larger.) Recall that [a b c] [ ] Det([d e f]) := a e i - a f h + d h c - d b i + g b f - g e c . [ ] [g h i] We repeat the first two columns, to get a 3x5 matrix: [a b c a b] [ ] [d e f d e] [ ] [g h i g h] Compute the SUM of the three diagonal products shown: a b c a b \ \ \ d e f d e \ \ \ g h i g h From that sum, SUBTRACT the sum of these three opposite-slope diagonals: a b c a b / / / d e f d e / / / g h i g h The result is Det of the given 3x3. ============================ == Computing a Wronkskian == ============================ Example of linear-Dependence: Consider the follow functions/expressions: C := cosh(x), S := sinh(x), E := exp(x), G := exp(-x). [The "G" is for neGative.] Notice that function-triple (C, E, G) is linearly-DEPENDENT, since C is an element of Span(E,G), seeing as C = [E + G]/2. Thus, we expect the Wronskian of list (C, E, G) to be the zero-function. Computing, the Wronskian-matrix of the list is [C E G] M := [S E -G] . [C E G] Applying the trick yields 3x5 matrix C E G C E S E -G S E C E G C E Using its diagonals, then, Det(M) = [ C*E*G + E*[-G]*C + G*S*E ] - [ G*E*C + C*[-G]*E + E*S*G ]. The terms cancel pairwise, yielding Det=0. Also note that since E*G = 1, we could first simplify to Det(M) = [ C - C + S ] - [ C - C + S ] = 0. ================ Example of linear-INdependence: As functions of x, define Z := 1, (zeroth power) F := x, (first power) S := x^2, (second power) T := x^3. (third power) The Wronskian-matrix of (Z,F,S) is [1 x S] [0 1 2x] . [0 0 2] The Det() of this matrix, since the matrix is upper-triangular, is 1*1*2, which is /not/ the zero-function. Thus we see that function-triple (1, x, x^2) is linearly-independent, as we expect. Second example of linear-Independence: Now consider function-triple (x, x^2, x^3). We expect that this list is linearly-independent. Its Wronskian-matrix is [x S T] [1 2x 3S] [0 2 6x] Use the trick to compute its Wronskian-function. Is its Wronskian the zero-fnc? ================================================== == Characteristic polynomial of a square matrix == ================================================== CharMatB := BDetOne - (x * eye(3)); [-x 1 1 ] [ ] CharMatB := [1 5 - x -1 ] [ ] [0 5 4 - x] > CharPolyB := Det(CharMatB) ; 2 3 CharPolyB := -24 x + 9 x - x + 1 > M := <|> ; [a b] M := [ ] [c d] > CharMatM := M - (x * eye(2)); [a - x b ] CharMatM := [ ] [ c d - x] > CharPolyM := Det(CharMatM) ; 2 CharPolyM := a d - a x - x d + x - b c > quit; END: "determinant-as-poly.txt"