Chapter 3 Determinants and Eigenvectors 大葉大學 資訊工程系 黃鈴玲 Linear Algebra

Ch03_2 3.1 Introduction to Determinants Definition The determinant ( 行列式 ) of a 2  2 matrix A is denoted |A| and is given by Observe that the determinant of a 2  2 matrix is given by the different of the products of the two diagonals of the matrix. The notation det(A) is also used for the determinant of A. Example 1

Ch03_3 Definition Let A be a square matrix. The minor ( 子式 ) of the element a ij is denoted M ij and is the determinant of the matrix that remains after deleting row i and column j of A. The cofactor ( 餘因子 ) of a ij is denoted C ij and is given by C ij = (–1) i+j M ij Note that C ij = M ij or  M ij.

Ch03_4 Example 2 Solution Determine the minors and cofactors of the elements a 11 and a 32 of the following matrix A. 隨堂作業： 3(c)

Ch03_5 Definition The determinant of a square matrix is the sum of the products of the elements of the first row and their cofactors. These equations are called cofactor expansions ( 餘因子展開式 ) of |A|.

Ch03_6 Example 3 Evaluate the determinant of the following matrix A. Solution

Ch03_7 Theorem 3.1 The determinant of a square matrix is the sum of the products of the elements of any row or column and their cofactors. ith row expansion: jth column expansion: Example 4 Find the determinant of the following matrix using the second row. Solution 隨堂作業： 9(d)

Ch03_8 Example 5 Evaluate the determinant of the following 4  4 matrix. Solution 隨堂作業： 11(b)

Ch03_9 Example 6 Solve the following equation for the variable x. Solution Expand the determinant to get the equation Proceed to simplify this equation and solve for x. There are two solutions to this equation, x = – 2 or 3. 隨堂作業： 14

Ch03_10 Computing Determinants of 2  2 and 3  3 Matrices Note ：此法不可用在 4  4 及更大的矩陣 !

Ch03_11 Homework Exercise 3.1: 3, 9, 11, 14

Ch03_12 Let A be an n  n matrix and k be a nonzero scalar. (a)If or, then |B| = k|A|. (b)If or, then |B| = –|A|. (c)If or, then |B| = |A|. 3.2 Properties of Determinants Theorem 3.2 Proof (a) |A| = a j1 C j1 + a j2 C j2 + … + a jn C jn |B| = ka j1 C j1 + ka j2 C j2 + … + ka jn C jn  |B| = k|A|.

Ch03_13 Example 1 Solution Evaluate the determinant 隨堂作業： 4(a)(b)(c)

Ch03_14 Example 2 If and |A| = 12 is known. Evaluate the determinants of the following matrices. Solution (a) Thus |B 1 | = 3|A| = 36. (b) Thus |B 2 | = – |A| = –12. (c) Thus |B 3 | = |A| = 12. 隨堂作業： 10(b,d)

Ch03_15 Theorem 3.3 Let A be a square matrix. A is singular if (a)all the elements of a row (column) are zero. (b)two rows (columns) are equal. (c)two rows (columns) are proportional ( 成比例的 ). (i.e., Ri=cRj) Proof (a) Let all elements of the kth row of A be zero. (c) If Ri=cRj, then, row i of B is [0 0 … 0].  |A|=|B|=0 Definition A square matrix A is said to be singular ( 奇異 ) if |A|=0. A is nonsingular if |A|  0.

Ch03_16 Example 3 Show that the following matrices are singular. Solution (a)All the elements in column 2 of A are zero. Thus |A| = 0. (b)Row 2 and row 3 are proportional. Thus |B| = 0.

Ch03_17 Theorem 3.4 Let A and B be n  n matrices and c be a nonzero scalar. (a)|cA| = c n |A|. (b)|AB| = |A||B|. (c)|A t | = |A|. (d) (assuming A –1 exists) Proof (a) (d)

Ch03_18 Example 4 If A is a 2  2 matrix with |A| = 4, use Theorem 3.4 to compute the following determinants. (a) |3A| (b) |A 2 | (c) |5A t A –1 |, assuming A –1 exists Solution (a)|3A| = (3 2 )|A| = 9  4 = 36. (b)|A 2 | = |AA| =|A| |A|= 4  4 = 16. (c)|5A t A –1 | = (5 2 )|A t A –1 | = 25|A t ||A –1 | Example 5 Prove that |A –1 A t A| = |A| Solution 隨堂作業： 9(a,b,e)

Ch03_19 Example 6 Prove that if A and B are square matrices of the same size, with A being singular, then AB is also singular. Is the converse true? Solution (  ) |A| = 0  |AB| = |A||B| = 0 Thus the matrix AB is singular. (  ) |AB| = 0  |A||B| = 0  |A| = 0 or |B| = 0 Thus AB being singular implies that either A or B is singular. The inverse is not true.

Ch03_20 Determinant of an Upper Triangular Matrix Definition A square matrix is called an upper triangular matrix ( 上三角矩 陣 ) if all the elements below the main diagonal are zero. It is called a lower triangular matrix ( 下三角矩陣 ) if all the elements above the main diagonal are zero.

Ch03_21 Example Note: The determinant of a triangular matrix is the product of its diagonal elements. 快速求行列式的方法： 利用 elementary row operations 將矩陣三角化 ( 對角線上的 數字可以不等於 1) ，再將對角線上的數字相乘即可

Ch03_22 Numerical Evaluation of a Determinant Example 8 Evaluation the determinant. Solution

Ch03_23 Example 9 Evaluation the determinant. Solution

Ch03_24 Example 10 Evaluation the determinant. Solution diagonal element is zero and all elements below this diagonal element are zero. 隨堂作業： 13(a,b)

Ch03_25 Homework Exercise 3.2: 4, 9, 10, 13, 14

Ch03_ Determinants, Matrix Inverse, and Systems of Linear Equations Definition Let A be an n  n matrix and C ij be the cofactor of a ij. The matrix whose (i, j)th element is C ij is called the matrix of cofactor of A. The transpose of this matrix is called the adjoint of A and is denoted adj(A).

Ch03_27 Example 1 Give the matrix of cofactors and the adjoint matrix of the following matrix A. Solution The cofactors of A are as follows. The matrix of cofactors of A is The adjoint of A is

Ch03_28 Theorem 3.5 Let A be a square matrix with |A|  0. A is invertible with Proof Consider the matrix product A  adj(A). The (i, j)th element of this product is

Ch03_29 Therefore  A  adj(A) = |A|I n Proof of Theorem 3.6 If i = j, If i  j, let Since |A|  0, Similarly,. Thus row i = row j in B Matrices A and B have the same cofactors C j1, C j2, …, C jn.

Ch03_30 Theorem 3.6 A square matrix A is invertible if and only if |A|  0. Proof (  ) Assume that A is invertible.  AA –1 = I n.  |AA –1 | = |I n |.  |A||A –1 | = 1  |A|  0. (  ) Theorem 3.6 tells us that if |A|  0, then A is invertible. A –1 exists if and only if |A|  0.

Ch03_31 Example 2 Use a determinant to find out which of the following matrices are invertible. Solution |A| = 5  0. A is invertible. |B| = 0. B is singular. The inverse does not exist. |C| = 0. C is singular. The inverse does not exist. |D| = 2  0. D is invertible.

Ch03_32 Example 3 Use the formula for the inverse of a matrix to compute the inverse of the matrix Solution |A| = 25, so the inverse of A exists.We found adj(A) in Example 1 隨堂作業： 7(a)

Ch03_33 Determinants and Systems of Linear Equations Proof (1)If |A|  0  A –1 exists (Thm 3.6)  there is then a unique solution given by X = A –1 B (Thm 2.8). (2) If |A| = 0  since A  C implies that if |A|  0 then |C|  0 (Thm 3.2).  the reduced echelon form of A is not I n.  The solution to the system AX = B is not unique.  many or no solutions. Theorem 3.7 Let AX = B be a system of n linear equations in n variables. (1) If |A|  0, there is a unique solution. (2) If |A| = 0, there may be many or no solutions.

Ch03_34 Example 4 Determine whether or not the following system of equations has an unique solution. Solution Since Thus the system does not have an unique solution. 隨堂作業： 14(b)

Ch03_35 Theorem 3.8 Cramer’s Rule Let AX = B be a system of n linear equations in n variables such that |A|  0. The system has a unique solution given by Where A i is the matrix obtained by replacing column i of A with B. Proof |A|  0  the solution to AX = B is unique and is given by

Ch03_36 x i, the ith element of X, is given by Thus Proof of Cramer’s Rule the cofactor expansion of |A i | in terms of the ith column

Ch03_37 Example 5 Solving the following system of equations using Cramer’s rule. Solution The matrix of coefficients A and column matrix of constants B are It is found that |A| = –3  0. Thus Cramer’s rule be applied. We get

Ch03_38 Giving Cramer’s rule now gives The unique solution is 隨堂作業： 12(c)

Ch03_39 Homogeneous Systems of Linear Equations Determine values of for which the following system of equations has nontrivial solutions. Find the solutions for each value of. Solution homogeneous system  x 1 = 0, x 2 = 0 is the trivial solution.  nontrivial solutions exist  many solutions      = – 3 or = 2. Example 6

Ch03_40 = – 3 results in the system This system has many solutions, x 1 = r, x 2 = r. = 2 results in the system This system has many solutions, x 1 = – 3r/2, x 2 = r. 隨堂作業： 15

Ch03_41 Homework Exercise 3.3: 7, 14, 15

3.4 Eigenvalues and Eigenvectors Definition Let A be an n  n matrix. A scalar is called an eigenvalue ( 特徵 值, 固有值 ) of A if there exists a nonzero vector x in R n such that Ax = x. The vector x is called an eigenvector corresponding to. Ch03_42 Figure 3.1 >0 <0

Computation of Eigenvalues and Eigenvectors Let A be an n  n matrix with eigenvalue and corresponding eigenvector x.  Ax = x  Ax – x = 0  (A – I n )x = 0  a system of linear equations, and x=0 is a solution.  we need nonzero solutions  many solutions  |A – I n | = 0  Solve |A – I n | = 0 for leads to all the eigenvalues of A. On expending the determinant |A – I n |, we get a polynomial in. This polynomial is called the characteristic polynomial of A. The equation |A – I n | = 0 is called the characteristic equation of A. Ch03_43

Example 1 Find the eigenvalues and eigenvectors of the matrix Find the characteristic polynomial of A: Solution The eigenvalues of A are 2 and –1. The corresponding eigenvectors are found by using these values of in the equation(A – I 2 )x = 0. Ch03_44

(1) = 2 (A – 2I 2 )x = 0 Thus the eigenvectors of A corresponding to = 2 are Ch03_45 

(2) = –1 (A + 1I 2 )x = 0 Thus the eigenvectors of A corresponding to = –1 are 隨堂作業： 9 先不求 eigenspaces Ch03_46  

Eigenspaces Let A be an n  n matrix and an eigenvalue of A. The set V of all eigenvectors corresponding to, together with the zero vector, is a subspace of R n. This subspace is called the eigenspace of. Proof Let x 1, x 2  V and let c be a scalar. Then Ax 1 = x 1 and Ax 2 = x 2. Hence, Thus x 1 +x 2  V. The set is closed under addition. Further, Ch03_47 Theorem 3.9 Therefore cx 1  V. The set is closed under scalar multiplication. Thus V is a subspace of R n.

Example 2 Find the eigenvalues and corresponding eigenspaces of the matrix Solution Ch03_48 =10 or 1

(1) = 10 Thus the eigenspace of = 10 is Ch03_49  The set is a basis, and the dimension is 1.

(2) = 1 Thus the eigenspace of = 1 is Ch03_50  The set is a basis, and the dimension is 2. If an eigenvalue occurs as a k times repeated root of the characteristic equation, we say that it is of multiplicity ( 重根數 ) k. Thus =10 has multiplicity 1, while =1 has multiplicity 2. 隨堂作業： 10

Example 3 Let A be an n  n matrix with eigenvalues 1, …, n, and corresponding eigenvectors x 1, …, x n. Prove that if c  0, then the eigenvalues of cA are c 1, …, c n with corresponding eigenvectors x 1, …, x n. Solution Let i be one of eigenvalues of A with corresponding eigenvectors x i.  Ax i = i x i.  cAx i = c i x i Thus c i is an eigenvalues of cA with corresponding eigenvector x i. Further, since cA is an n  n matrix, the characteristic polynomial of A is of degree n. The characteristic equation has n roots, implying that cA has n eigenvalues. The eigenvalues of cA are therefore c 1, …, c n with corresponding eigenvectors x 1, …, x n. Ch03_51

Homework Exercise 3.4: 9, 10, 13, 24, 26, 32 Ex24: Prove that if A is a diagonal matrix, then its eigenvalues are the diagonal elements. Ex26: Prove that if A and A t have the same eigenvalues. Ex32: Prove that the constant term of the characteristic polynomial of a matrix A is |A|. Ch03_52 (3.5 節跳過 )