2.4 Linear Independence (線性獨立) and Linear Dependence(線性相依) A linear combination(線性組合) of the vectors in V is any vector of the form c1v1 + c2v2 + … + ckvk where c1, c2, …, ck are arbitrary scalars. A set of V of m-dimensional vectors is linearly independent(線性獨立) if the only linear combination of vectors in V that equals 0 is the trivial linear combination. A set of V of m-dimensional vectors is linearly dependent (線性相依) if there is a nontrivial linear combination of vectors in V that adds up to 0.

Example 10: LD Set of Vectors (p.33) Show that V = {[ 1 , 2 ] , [ 2 , 4 ]} is a linearly dependent set of vectors. Solution Since 2([ 1 , 2 ]) – 1([ 2 , 4 ]) = (0 0), there is a nontrivial linear combination with c1 =2 and c2 = -1 that yields 0. Thus V is a linear dependent set of vectors.

Linear dependent 線性相依 (p.33) What does it mean for a set of vectors to linearly dependent? A set of vectors is linearly dependent only if some vector in V can be written as a nontrivial linear combination of other vectors in V. If a set of vectors in V are linearly dependent, the vectors in V are, in some way, NOT all “different” vectors. By “different” we mean that the direction specified by any vector in V cannot be expressed by adding together multiples of other vectors in V. For example, in two dimensions, two linearly dependent vectors lie on the same line.

The Rank of a Matrix (p.34) Let A be any m x n matrix, and denote the rows of A by r1, r2, …, rm. Define R = {r1, r2, …, rm}. The rank(秩) of A is the number of vectors in the largest linearly independent subset of R. To find the rank of matrix A, apply the Gauss-Jordan method to matrix A. (Example 14, p.35) Let A’ be the final result. It can be shown that the rank of A’ = rank of A. The rank of A’ = the number of nonzero rows in A’. Therefore, the rank A = rank A’ = number of nonzero rows in A’.

Whether a Set of Vectors Is Linear Independent A method of determining whether a set of vectors V = {v1, v2, …, vm} is linearly dependent is to form a matrix A whose ith row is vi. (p.35) - If the rank A = m, then V is a linearly independent set of vectors. - If the rank A < m, then V is a linearly dependent set of vectors.

2.5 The Inverse of a Matrix (反矩陣) (p.36) A square matrix(方陣) is any matrix that has an equal number of rows and columns. The diagonal elements (對角元素)of a square matrix are those elements aij such that i=j. A square matrix for which all diagonal elements are equal to 1 and all non-diagonal elements are equal to 0 is called an identity matrix(單位矩陣). An identity matrix is written as Im.

The Inverse of a Matrix (continued) For any given m x m matrix A, the m x m matrix B is the inverse of A if BA=AB=Im. Some square matrices do not have inverses. If there does exist an m x m matrix B that satisfies BA=AB=Im, then we write B= . (p.39) The Gauss-Jordan Method for inverting an m x m Matrix A is Step 1 Write down the m x 2m matrix A|Im Step 2 Use EROs to transform A|Im into Im|B. This will be possible only if rank A=m. If rank A<m, then A has no inverse. (p.40)

Matrix inverses can be used to solve linear systems. (example, p.40) The Excel command =MINVERSE makes it easy to invert a matrix. Enter the matrix into cells B1:D3 and select the output range (B5:D7 was chosen) where you want A-1 computed. In the upper left-hand corner of the output range (cell B5), enter the formula = MINVERSE(B1:D3) Press Control-Shift-Enter and A-1 is computed in the output range

Inverse Matrix (反矩陣)

反矩陣之性質  p.41 #8a  p.42 #9  p.42 #10

 p.41 problems group 8a

Example : 例題：

Example (continued) Exercise： problem 2, p.41, 5min

Orthogonal matrix (正規矩陣) A-1=AT Det(A)=1 or Det(A)=-1 Determine matrix A is an orthogonal matrix or not . p.42 #11

Homogeneous System Equations AX=b為一線性方程組,若b1=b2=…=bm=0，則稱為齊次方程組(homogeneous) ,以 AX=0表之。 x1=x2=…=xn=0為其中一組解，稱為必然解(trivial solution) 若x1、x2、…xn不全為0，則稱為非必然解(nontrivial solution)

Example:

Example:

Example : Solving linear system by inverse matrix

Exercise : Solving linear system by inverse matrix (8 min) Answer:

2.6 Determinants(行列式) (p.42) Associated with any square matrix A is a number called the determinant of A (often abbreviated as det A or |A|). If A is an m x m matrix, then for any values of i and j, the ijth minor of A (written Aij) is the (m - 1) x (m - 1) submatrix of A obtained by deleting row i and column j of A. Determinants can be used to invert square matrices and to solve linear equation systems.

Determinants(行列式)

Minor(子行列式) & Cofactor(餘因子)

餘因子展開式

行列式之性質(1)

行列式之性質(2)

行列式之性質(3)

行列式之性質(4)

Adjoint matrix (伴隨矩陣)

Exercise : adj A = ？ det(A) = ？ Answer： det(A) = -34

Exercise : 若det(λI3-A)=0, λ=？ λ =-5 或λ=0或λ=3

伴隨矩陣之性質

行列式之性質

Cramer’s Rule

Cramer’s Rule 注意事項 若det(A) ≠0，則n元一次方程組為相容方程組， 其唯一解為 若det(A) =det(A1) =det(A2)= …=det(An) =0， 則n元一次方程組為相依方程組，其有無限多組解 若det(A) =0，而det(A1) ≠0 或det(A2) ≠0 ，…， 或det(An) ≠0，則n元一次方程組為矛盾方程組，其 為無解。

Example : Solving Linear System The system has an infinite number of solutions

Exercise : Solving linear system by Cramer’s rule Answer:

LU-Decompositions Factoring the coefficient matrix into a product of a lower and upper triangular matrices. This method is well suited for computers and is the basis for many practical computer programs.

Solving linear systems by factoring Let Ax = b，and coefficient matrix A can be factored into a product of n × n matrices as A = LU， where L is lower triangular and U is upper triangular, then the system Ax = b can be solved by as follows： Step 1. Rewrite Ax = b as LUx = b (1) Step 2. Define a n × 1 new matrix y by Ux = y (2) Step 3. Use (2) to rewrite (1) as Ly = b and solve this system for y. Step 4. Substitude y in(2) and solve for x

Example : Solving linear system by LU decomposition Step 1. A=LU, Ax=b  LUx=b

Step 2. Ux=y Step 3. Ly=b Solving y by forward-substitution. y1=1, y2=5, y3=2

y1=1, y2=5, y3=2 Step 4. Solving y by backward-substitution. x1=2, x2=-1, x3=2

Doolittle method = Assume that A has a Doolittle factorization A = LU. L：unit lower-△，U：upper-△ = The solution X to the linear system AX = b  , is found in three steps:       1.  Construct the matrices  L and U, if possible.       2.  Solve LY = b  for  Y  using forward substitution.     3.  Solve UX = Y   for  X  using back substitution. 

Crout method Assume that A has a Doolittle factorization A = LU. L：lower-△，U：unit upper-△  = The solution X to the linear system AX = b  , is found in three steps:       1.  Construct the matrices  L and U, if possible.       2.  Solve LY = b  for  Y  using forward substitution.     3.  Solve UX = Y   for  X  using back substitution. 

Solving Linear Equations (see *.doc) Gauss backward-substitution 高斯後代法 -- forward-substitution Gauss-Jordan Elimination 高斯約旦消去法 -- reduced row echelon matrix Inversed matrix ─ solve Ax=b as x=A-1b LU method ─ solve Ax=b as LUx=b Cramer's Rule ─ x1=det(A1)/det(A),….