SPRING OF 2008 D. DETERMINANTS

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1 18024 SPRING OF 2008 D DETERMINANTS In many applications of linear algebra to calculus and geometry, the concept of a determinant plays an important role This chapter studies the basic properties of determinants Determinants of order two or three may have been introduced to you in high school calculus as a useful notation to express certain formulae in a compact form Our point of view here is to treat the determinant as a function which assigns a number to a square matrix of order n for any positive integer n It is possible to define this function by an explicit formula generalizing those of order two and three, but the formula is unwieldy for n large (and is hardly used in practice) Thus, we take another approach and characterize determinants by their essential properties, called axioms for a determinant function Axioms for a determinant functions If A = (a ij ) is an n n matrix, we denote its rows by A 1,, A n Thus, the ith row of A is a vector in V n given by A i = (a i1, a i2,, a in ) A function d, defined for each ordered n-tuple of vectors A 1,, A n in V n, is called a determinant function if it satisfies the following axioms: AXIOM1 (homogeneity in each row) If the kth row A k is multiplied by a scalar t then the determinant is also multiplied by t: d(, ta k, ) = td(, A k, ) AXIOM2 (additivity in each row) For each k we have d(, A k + C, ) = d(, A k, ) + d(, C, ) AXIOM3 The determinant vanishes if any two rows are equal: d(a 1,, A n ) = 0 if A i = A j for some i j AXIOM4 The determinant of the identity matrix is 1: d(e 1,, E n ) = d(i n ) = 1, where E 1,, E n are unit coordinate vectors Sometimes a weaker version of AXIOM3 is used AXIOM3 The determinant vanishes if two adjacent rows are equal: d(a 1,, A n ) = 0 if A k = A k+1 for some k It is rather surprising that there is only one function satisfying AXIOMS1,2,3 and 4 The proof uses in an essential way the expansion formula (??) The next theorem establishes some properties of determinants deduce from AX- IOMS 1, 2 and 3 only (AXIOM4 is not used) 1

2 2 DETERMINANTS Theorem 1 If d satisfies AXIOMS 1,2,3, then (a) the determinant vanishes if some row is zero; (b) the determinant changes sign if any two rows A i and A j, i j are interchanged: d(, A i,, A j, ) = d(, A i,, A j, ); (c) the determinant vanishes if its two rows are linearly dependent The proofs are straightforward and thus left as exercises Computations of determinants It is instructive to compute determinants for some special cases, using only the axioms and properties in Theorem 1 (assuming that determinant functions exist) Example 2 (Determinant of a 2 2 matrix) We claim that a11 a det 12 = a a 21 a 11 a 22 a 12 a Writing A 1 = (a 11, a 12 ) = a 11 i + a 12 j and A 1 = (a 21, a 22 ) = a 21 i + a 22 j, where i = (1, 0) and j = (0, 1), det(a 1, A 2 ) = det(a 11 i + a 12 j, a 21 i + a 22 j) = a 11 det(i, a 21 i + a 22 j) + a 12 det(j, a 21 i + a 22 j) (by AXIOM1) = a 11 a 21 det(i, i) + a 11 a 22 det(i, j) + a 12 a 21 det(j, i) + a 12 a 22 det(j, j) = a 11 a 22 det(i, j) a 12 a 21 det(i, j) (by AXIOM3 and Theorem 1(b)) = a 11 a 22 a 12 a 21 This proves the claim Example 3 (Determinant of a diagonal matrix) A square matrix of the form a a 22 0 A = 0 0 a nn is called a diagonal matrix Each entry a ij off the main diagonal is zero We claim that det A = a 11 a 22 a nn Indeed, by AXIOM1 follows that det A = det(a 11 E 1, a 22 E 2,, a nn E n ) = a 11 a 22 a nn The formula can be written in the form det A = a 11 a nn det I n Example 4 (Determinant of an upper triangular matrix) A square matrix of the form a 11 a 12 a 1n 0 a 22 a 2n A = 0 0 a nn

3 DETERMINANTS 3 is called an upper triangular matrix All entries below the main diagonal are zero We claim that det A = a 11 a 22 a nn First, we prove that det A = 0 if a ii = 0 for some i If a 11 = 0, for example, then each of the n row vectors A 1,, A n has its first component zero, and hence these vectors span a subspace of dimension at most n 1 Therefore, these n rows are dependent, and by Theorem 1(c), the determinant is zero Next, we decompose the first row as A 1 = D 1 + R 1, where B 1 = (a 11, 0,, 0), R 1 = (0, a 12,, a 1n ) By linearity in the first row of the matrix, then, it follows that det A = det(d 1, A 2,, A n ) + det(r 1, A 2,, A n ) On the other hand, det(r 1, A 2,, A n ) = 0 since it is an upper triangular matrix with a diagonal entry being zero But the second matrix is just 0 since the first column is identically 0 By repeating this operation a number of times, we are finally left with the diagonal matrix Therefore, det A = det(d 1, A 2,, A n ) Repeating the argument on each of the succeeding rows on the matrix on the right side, we finally obtain det A = det(d 1, D 2,, D n ), where (D 1, D 2,, D n ) is a diagonal matrix with the same diagonal entries as A Therefore, by the previous example, det A = a 11 a nn In general, by applying the Gauss-Jordan elimination process over and over again in a systematic fashion, we transform any square matrix A to an upper triangular matrix, and we know how to compute the determinant of the upper triangular matrix and how the determinant of A is related to that of the upper triangular matrix Each time we interchange two rows, the determinant changes sign, and each time we multiply a scalar t on a row, the determinant is multiplied by t Therefore, det A = ( 1) p (t 1 t 2 t q ) 1 a 11 a 22 a nn det I n In other words, any function d satisfying AXIOMS 1,2,3 must be of the form (1) d(a 1,, A n ) = cd(e 1,, E n ) for some scalar c (depending on A) The uniqueness theorem We recall that for every n n matrix A, there is a scalar c such that (1) holds true Moreover, this formula is a consequence of AXIOMS 1,2,3 only From this we can prove that there canot be more than one determinant function Theorem 5 Let d and f be functions satisfying AXIOMS 1-4 and 1-3 respectively Then, f(a 1,, A n ) = d(a 1,, A n )f(e 1,, E n ) In particular, if f satisfies, in addition, AXIOM 4, then f = d Proof Let g(a 1,, A n ) = f(a 1,, A n ) d(a 1,, A n )f(e 1,, E n ) We observe that both f and d satisfy AXIOMS 1-3 Therefore, we conclude that g also satisfies axioms 1 3 Thus, we can write g(a 1,, A n ) = cg(e 1,, E n )

4 4 DETERMINANTS for some scalar c Taking A = I n, moreover, and noting that d satisfies AXIOM 4, we have g(e 1,, E n ) = f(e 1,, E n ) d(e 1,, E n )f(e 1,, E n ) = 0, implying that g = 0 and we are done The product formula and nonsingular matrices We use the uniqueness theorem to prove the product formula for determinants and the relation between the determinant of a nonsingular matrix and that of its inverse Theorem 6 (Product formula for determinants) For any n n matrices A and B we have det(ab) = det(a) det(b) This proof of this result uses the proof of the uniqueness theorem with the following observation Lemma 7 For any m n matrix A and any n p matrix B, we have (AB) i = A i B, that is, the ith row of the product AB is equal to the product of row matrix A i with B An application of the product formula is to compute the determinant of an inverse matrix A square matrix is called non-singular if it has a left inverse B such that so that BA = I n Note that if a left inverse exists then it is unique and is also a right inverse, AB = I n The relation between det A and det A 1 is as natural as one could expect Theorem 8 If A is nonsingular, and det(a) 0, then det A 1 = 1/ det A Proof From the product formula, it follow (det A)(det A 1 ) = det(aa 1 ) = det I = 1 Thus, det A 0, and the proof is complete Expansion formulas for determinants Given a square matrix A of order n, where n > 1, the square matrix of order n 1 obtained by deleting the kth row and the jthe column of A is called the k, j minor of A, denoted by A kj For example, if a 11 a 12 a 13 A = a 21 a 22 a 23 a 31 a 32 a 33 then, A 11 = a22 a 23, A a 32 a 12 = 23, A 33 a 31 a 13 = a 31 a 32 Theorem 9 (Exapnsion by kth-row minors) For any matrix n n matrix A, n > 1, (2) det A = n ( 1) k+j a kj det A kj j=1

5 DETERMINANTS 5 For the proof, read the proof of [Apo, Theorem 39] We can use induction on n, the size of a matrix, and the above expansion formula to prove that determinant functions of every order exist Read [Apo, Section 313] for a detailed discussion For example, in the case of a 3 3 matrix, we have det a 11 a 12 a 13 a 21 a 22 a 23 a22 a = a 11 det 23 a a 31 a 32 a 32 a REFERENCES [Apo] T Apostol, Calculus, vol II, Second edition, Wiley, 1967 c 2008 BY VERA MIKYOUNG HUR address: verahur@mathmitedu [ a 12 det 23 a 31 a 33 ] +a 13 det 22 a 31 a 32

II. Determinant Functions

II. Determinant Functions Supplemental Materials for EE203001 Students II Determinant Functions Chung-Chin Lu Department of Electrical Engineering National Tsing Hua University May 22, 2003 1 Three Axioms for a Determinant Function