LECTURE 4: DETERMINANT (CHAPTER 2 IN THE BOOK)

LECTURE 4: DETERMINANT (CHAPTER 2 IN THE BOOK) Everything with is not required by the course syllabus. Idea Idea: for each n n matrix A we will assign a real number called det(a). Properties: det(a) 0 A is nonsingular; det(i n ) = ; det(a T ) = A; det(αi n ) = α n ; det(a)det(b) = det(ab). Now we begin to define determinant. 2. Permutations and defintion of determinant Definition 2.. Let T = {,, n } be the set of integers form to n. A permutation of T is a bijection map from T to itself. For a pertumtation τ is usually represened 2 by ( 2 3 n τ() τ(2) τ(3) τ(n) ) Note that τ(),, τ(n) are distinct, since τ is bijective. Permutation can be compose with each other. Example 2.2. (i) id: the identity map (ii) τ = ( 2 3 2 3 ) 2 3 (iii) σ = ( 2 3 3 2 ) 2 3 (iv) στ = ( 2 3 2 3 ) 2 3 (v) Clearly (γσ)τ = γ(στ). Facts: ). Every permutation, say τ, is invertiable with the invererse τ() τ(2) τ(3) τ(n) ( 2 3 n ) 2). Denote S n = { τ τis a permutation of T }. 3). S n = n!. 4). In fact, S n form a mathematical object called group which is called symmetric group or permutation group. τ, γ and σ are Greek letter called tau, gamma and sigma respectively. 2 This is NOT a matrix but a notation. τ σ

2 LECTURE 4: DETERMINANT (CHAPTER 2 IN THE BOOK) 5). Transposition: a permutation exchange two numbers and kepp all other number fixed. Example 2.3 (continued). τ and σ are transpositions, but στ is not a transposition. 6). For any τ S n, there exits a finite sequence s,, s k of transposition such that τ = s s k. This factorization of τ is not unique. But the parity of k is fixed. 7). Define the map sgn S n {, } s s k = τ { if k is even if k is odd Example 2.4 (continued). We have (a) sgn(σ) = sgn(τ) = and sgn(στ) = sgn(id) =. (b) For any permutations τ, σ S n, we have sgn(τσ) = sgn(τ)sgn(σ) Definition 2.5. Let A = (a ij ) be an n n matrix. The determinant of A is defined to be det(a) = τ S n sgn(τ)a τ() a 2 τ(2) a n τ(n). Conventionally, people also use A or A to denote the determinate det(a). Example 2.6. (i) det((a) ) = a. (ii) Note that S 2 = { id, ( 2 2 ) } det (a a 2 ) = a a 2 a a 22 a 2 a 2. 22 (iii) Note that S 3 = 6. a a 2 a 3 det a 2 a 22 a 23 = a 3 a 32 a 33 a a 22 a 33 + a 2 a 23 a 3 + a 3 a 2 a 32 a a 23 a 32 a 2 a 2 a 33 a 3 a 22 a 3. (iv) When n 4 it is unpractical to write down the formula. Lemma 2.7. We have det(a T ) = det(a). Proof. Let τ be any permutation, then τ is also a permutation. Moreover, sgn(τ)sgn(τ ) = sgn(id) =, therefore sgn(τ) = sgn(τ ). Set σ = τ and B = (b ij ) = A T. Then we have sgn(τ)a τ() a 2τ(2) a nτ(n) =sgn(τ)a τ ()a τ (2)2 a τ (n)n =sgn(σ)b σ() b 2σ(2) b nσ(n). But when τ running over S n, σ = τ is also running over S n. Therefore, we get det(a) = det(b) by the definition of determinant. 3. Calculate determinant 3.. Minor, cofacter, Laplace expansion.

LECTURE 4: DETERMINANT (CHAPTER 2 IN THE BOOK) 3 Definition 3.. Let A = (a ij ) be an n n matrix, and let M ij denote the (n ) (n ) matrix obtained from A by deleting the row and column containing a ij. The determinant of M ij is called the minor of a ij. We define the cofactor A ij of a ij by A ij = ( ) i+j det(m ij ). There is a way to reduce the calculation of the determinant of an n n-matrix to determinants of (n ) (n )-matrices. Theorem 3.2 (Laplace expansion). Let A be an n n matrix with n 2, then det(a) can be expressed as a cofactor expansion using any row or columns of A, that is: for i =,, n and j =,, n. Example 3.3. (i) det(a) =a i A i + a i2 A i2 + + a in A in =a j A j + a 2j A 2j + + a nj A nj a a 2 a 3 a 2 a 22 a 23 = a ( a 22 a 23 ) a a 3 a 32 a 33 a 32 a 2 ( a 2 a 23 ) + a 33 a 3 a 3 ( a 2 a 22 ) 33 a 3 a 32 (ii) Cross Product For two vectors x and y in R 3. define the Cross Product If C is a matrix of the form i j k x y = x x 2 x 3 y y 2 y 3 w w 2 w 3 C = x x 2 x 3 y y 2 y 3 Hence det(c) = w T (x y). (iii) Consider an upper triangular matrix with diagnals a,, a nn, then a a 2 a n a 22 a 2n a 22 a 2n = a = = a a nn a nn 0 a nn A same result also holds for lower triangular matrix. In particular, we have det(i) = and det(αi) = α n. Lemma 3.4. If A has a row or column consisting entirely of zero, then det(a) = 0. Proof. Expaned the determiant with respect to the row or column with all zeros. We see all terms in the expansion are zero. Hence det(a) = 0. 3.2. The elementry row and column operations. We have following theorem on the effact of elementary row operation to the determinant. The similar statement for column operations also holds, could be obtained by transpose. lem:el Lemma 3.5. Let A be an n n matrix. Elementray Row Operation I: If B = E ij A is the matrix obtained from A by interchanging two rows (say i-th and j-th row) of A, then det(b) = det(a).

4 LECTURE 4: DETERMINANT (CHAPTER 2 IN THE BOOK) Here E ij is the corresponding elementry matrix and det(e ij ) =. As an application: If A has two identical rows or two identical columns, then det(a) = 0. Elementray Row Operation II: If B = E α,i A is the matrix obtained from A by multiplying the i-th row of A by a number α 0, then det(b) = α det(a). Here E α,i is the corresponding elementary matrix and det(e α,i ) = α. Elementray Row Operation III: If B = E α,ij A is obtained from A by adding the α-multiple of the i-th row to the j-th row, then det(b) = det(a). We have det(e α,ij ) =. In summation, if E is an elementary matrix, then det(ea) = det(e)det(a). Proof. I. Let s be the transposition interchange i and j. For any τ S n, let σ = τs. Then sgn(τ)a τ() a iτ(i) a jτ(j) a nτ(n) = sgn(τs)a τs() a iτs(j) a jτs(i) a nτs(n) = sgn(σ)a σ() a iσ(j) a jσ(i) a nσ(n) = sgn(σ)b σ() b jσ(j) b iσ(i) b nσ(n) When τ running over S n, σ = τs is also running over S n. Therefore, we get det(a) = det(b) by the definition of determinant. To compute det(e ij ), one observe that the only nonzero term inovlved is the term related to s. Hence det(e ij ) = sgn(s) =. Now suppose the i-th and j-th row of A are identical. Interchanging the i-th and j-th row will give the same matrix. Hence det(a) = det(a), i.e. det(a) = 0. II. Suppose the i-th row of B equal to the α multiple of the i-th row of A, i.e b ij = αa ij for any j n. Now the claim is clear from the defintion of determinant. Since E α,i is a diagnal matrix with α on the (i, i)-th entry and on the other diagnal places. Hence det(e α,i ) = α = α. III. Note that an entry on j-th row of B satisfis b jk = a jk + αa ik. Use Laplace expansion with respect to the j-the row of B. We have det(b) =(a j + αa i )A j + (a j2 + αa i2 )A j2 + + (a jn + αa in )A jn =(a j A j + a j2 A j2 + + a jn A jn ) + α(a i A j + a i2 A j2 + + a in A jn ) =det(a) + det(c) = det(a) where C is a matrix with all rows equal to A, except the i-th row which is equal to the j-th row of A. Since E α,ij is a triangular matrix with on its diagnal, we have det(e α,ij ) =.

LECTURE 4: DETERMINANT (CHAPTER 2 IN THE BOOK) 5 The most important property of determinants Theorem 3.6. (i) An n n matrix A is singular if and only if (ii) If A and B are square matrices, then det(a) = 0. det(ab) = det(a)det(b). Proof. (i) Reduce the matrix A into an row echelon form with row operations. Thus U = E k E k E A. So det(u) =det(e k E k E A) =det(e k )det(e k ) det(e )det(a). Since det(e i ) are all non-zero, we have det(u) 0 det(a) 0. If A is singular, then U has a row consiting entirely of zeros, hence det(u) = 0. If A is nonsingular, U is triangular with s along the diagnoal, and hence det(u) =. (ii) If B is singular, Bx = 0 has non-trivial solution say x 0, hence x 0 is also a non-trivial solution of (AB)x = 0. So AB is singular. Hence, by part (i), det(ab) = 0 = det(a)det(b). If B is nonsingular, then B = E k E such that E i are elementary matrices. Now by above Lemma 3.5, Example 3.7. Calculate We have det(ab) =det(ae k E k E ) =det(a)det(e k ) det(e ) =det(a)det(e k E ) =det(a)det(b). 0 5 3 6 9. 2 6 0 5 3 3 6 9 R 0 5 2 R 3 2R 2 +R 3 0 5 3 2 3 3 2 3 2 6 2 6 0 0 5 C 2 2C +C 2 0 5 C 3 3C +C 3 0 5 3 0 3 3 0 0 0 0 5 0 0 5 R 3 R +R 3 0 5 3 0 0 = 3 5 = 65. 0 0

6 LECTURE 4: DETERMINANT (CHAPTER 2 IN THE BOOK) 4. Adjoint of a matrix Definition 4.. Let A be an n n matrix. Defint the adjoint of A by A A 2 A n adj(a) = (A ij ) T A = 2 A 22 A n2, A n A 2n A nn where A ij is the cofactor of the (i, j)-th entry. Lemma 4.2. Let A be an n n matrix. Then det(a) if i = j {eq:ll} () a i A j + a i2 A j2 + + a in A jn = { 0 if i j Proof. When i = j, this is Laplace expansion. When i j, this is the Laplace expansion of a matrix C is a matrix with all rows equal to A, except the i-th row which is equal to the j-th row of A. So det(c). This proves the lemma. Theorem 4.3. We have A adj(a) = det(a)i n. Proof. Clear, since the (i, j)-th entry of A adj(a) is exactly ( ). eq:ll If A is nonsingular, we have A = det(a) adj(a). Example 4.4. Let A = ( a a 2 ) be a nonsingular 2 2-matrix, then a 2 a 22 A = a a 22 a 2 a 2 ( a 22 a 2 a 2 a ) Cramer s Rule. Not required in the exam, but good to known. Theorem 4.5. Let A be an n n nonsingular matrix and b be a column vector in R n. Let A i be the matrix obtained by replacing the i-th column of A by b. If x is the unique solution of Ax = b, then Proof. Since it follows that x = A b = x i = A i A det(a) (adj(a))b, x i = b A i + b 2 A 2i + + b n A ni det(a) = det(a i) det(a)