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Lecture 6 Eigenvalue and Vector Space. Lat Time - Evaluation of Determinants - Geometric Interpretations - Properties of Determinants. Elementary Linear Algebra R. Larsen et al. (6th Edition) TKUEE 翁慶昌 -NTUEE SCC_10_2008. Lecture 6: Eigenvalue and Vectors. Today
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Lecture 6 Eigenvalue and Vector Space Lat Time - Evaluation of Determinants - Geometric Interpretations - Properties of Determinants Elementary Linear Algebra R. Larsen et al. (6th Edition) TKUEE翁慶昌-NTUEE SCC_10_2008
Lecture 6: Eigenvalue and Vectors Today • Properties of Determinants • Introduction to Eigenvalues • Applications of Determinants • Vectors in Rn • Vector Spaces Reading Assignment: Secs 3.4 – 3.5, 4.1-4.2 of Textbook Homework #3 due Next Time • Subspaces of Vector Spaces • Spanning Sets and Linear Independence • Basis and Dimension • Rank of a Matrix and Systems of Linear Equations Reading Assignment: Secs 4.3- 4.6 Homework #4 due
Lecture 6: Elementary Matrices & Determinants Today • Properties of Determinants (Cont.) • Introduction to Eigenvalues • Applications of Determinants • Vectors in Rn
3.3 Properties of Determinants • Notes: • Thm 3.5: (Determinant of a matrix product) det (AB) = det (A) det (B) (1) det (EA) = det (E) det (A) (2) (3)
Ex 1: (The determinant of a matrix product) Find |A|, |B|, and |AB| Sol:
Check: |AB| = |A| |B|
Ex 2: • Thm 3.6: (Determinant of a scalar multiple of a matrix) If A is an n × n matrix and c is a scalar, then det (cA) = cn det (A) Find |A|. Sol:
Thm 3.7: (Determinant of an invertible matrix) • Ex 3: (Classifying square matrices as singular or nonsingular) A square matrix A is invertible (nonsingular) if and only if det (A) 0 Sol: A has no inverse (it is singular). B has inverse (it is nonsingular).
Ex 4: • Thm 3.8: (Determinant of an inverse matrix) • Thm 3.9: (Determinant of a transpose) (a) (b) Sol:
If A is an n × n matrix, then the following statements are equivalent. • Equivalent conditions for a nonsingular matrix: (1) A is invertible. (2) Ax = b has a unique solution for every n × 1 matrix b. (3) Ax = 0 has only the trivial solution. (4) A is row-equivalent to In (5) A can be written as the product of elementary matrices. (6) det (A) 0
Ex 5: Which of the following system has a unique solution? (a) (b)
Sol: (a) This system does not have a unique solution. (b) This system has a unique solution.
Lecture 6: Elementary Matrices & Determinants Today Properties of Determinants (Cont.) Introduction to Eigenvalues Applications of Determinants Vectors in Rn Vector Spaces 6 - 15
Eigenvalue and eigenvector: A:an nn matrix :a scalar x: a n1nonzero column matrix Eigenvalue Eigenvector (The fundamental equation for the eigenvalue problem) 3.4 Introduction to Eigenvalues • Eigenvalue problem: If A is an nn matrix, do there exist n1 nonzero matrices x such that Ax is a scalar multiple of x?
Eigenvalue Eigenvalue Eigenvector Eigenvector • Ex 1: (Verifying eigenvalues and eigenvectors)
Note: (homogeneous system) If has nonzero solutions iff . • Characteristic equation of AMnn: • Question: Given an nn matrix A, how can you find the eigenvalues and corresponding eigenvectors?
Sol: Characteristic equation: Eigenvalue: • Ex 2: (Finding eigenvalues and eigenvectors)
Ex 3: (Finding eigenvalues and eigenvectors) Sol: Characteristic equation:
3.5 Applications of Determinants • Matrix of cofactors of A: • Adjoint matrix of A:
Thm 3.10: (The inverse of a matrix given by its adjoint) If A is an n × n invertible matrix, then • Ex:
Ex 2: (a) Find the adjoint of A. (b) Use the adjoint of A to find Sol:
adjoint matrix of A inverse matrix of A • Check: cofactor matrix of A
Thm 3.11: (Cramer’s Rule) (this system has a unique solution)
( i.e. )
Pf: A x = b,
Ex 6: Use Cramer’s rule to solve the system of linear equations. Sol:
Keywords in Section 3.5: • matrix of cofactors : 餘因子矩陣 • adjoint matrix : 伴隨矩陣 • Cramer’s rule : Cramer 法則
Lecture 6: Elementary Matrices & Determinants Today Properties of Determinants (Cont.) Introduction to Eigenvalues Applications of Determinants Vectors in Rn Vector Spaces 6 - 36
Chapter 4 Vector Spaces 4.1 Vectors in Rn 4.2 Vector Spaces 4.3 Subspaces of Vector Spaces 4.4 Spanning Sets and Linear Independence 4.5 Basis and Dimension 4.6 Rank of a Matrix and Systems of Linear Equations 4.7 Coordinates and Change of Basis
n-space: Rn the set of all ordered n-tuple 4.1 Vectors in Rn • An ordered n-tuple: a sequence of n real number
n = 1 R1 = 1-space = set of all real number n = 2 R2 = 2-space = set of all ordered pair of real numbers n = 3 R3 = 3-space = set of all ordered triple of real numbers n = 4 R4 = 4-space = set of all ordered quadruple of real numbers • Ex:
a point a vector • Notes: (1) An n-tuple can be viewed as a point in Rn with the xi’s as its coordinates. (2) An n-tuple can be viewed as a vector in Rnwith the xi’s as its components. • Ex:
(two vectors in Rn) • Equal: • if and only if • Vector addition (the sum of u and v): • Scalar multiplication (the scalar multiple of u by c): • Notes: The sum of two vectors and the scalar multiple of a vector in Rn are called the standard operations in Rn.
Difference: • Zero vector: • Notes: (1) The zero vector 0 in Rn is called the additive identity in Rn. (2) The vector –v is called the additive inverse of v. • Negative:
(1) u+v is a vector inRn (2) u+v = v+u (3) (u+v)+w = u+(v+w) (4) u+0 = u (5) u+(–u) = 0 (6) cu is a vector inRn (7) c(u+v) = cu+cv (8) (c+d)u = cu+du (9) c(du) = (cd)u (10) 1(u) = u • Thm 4.2: (Properties of vector addition and scalar multiplication) Let u, v, and w be vectors inRn , and let c and d be scalars.
Ex 5: (Vector operations in R4) Let u=(2, – 1, 5, 0), v=(4, 3, 1, – 1), and w=(– 6, 2, 0, 3) be vectors in R4. Solve x for x in each of the following. (a) x = 2u– (v + 3w) (b) 3(x+w) = 2u –v+x Sol: (a)
(1) The additive identity is unique. That is, if u+v=v, then u = 0 (2) The additive inverse of v is unique. That is, if v+u=0, then u = –v (3) 0v=0 (4) c0=0 (5) If cv=0, then c=0 or v=0 (6) –(–v) = v • Thm 4.3: (Properties of additive identity and additive inverse) Let v be a vector inRn and c be a scalar. Then the following is true.
Ex 6: Given x = (– 1, – 2, – 2), u = (0,1,4), v = (– 1,1,2), and w = (3,1,2) in R3, find a, b, and c such that x = au+bv+cw. Sol: • Linear combination: The vector x is called a linear combination of , if it can be expressed in the form