Ch07 3
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The document summarizes key concepts related to systems of linear equations and linear algebra, including: 1) A system of n linear equations can be expressed in matrix form as Ax = b, where A is the coefficient matrix, x is the vector of unknowns, and b is the constant vector. If b = 0, the system is homogeneous, otherwise it is nonhomogeneous. 2) If the coefficient matrix A is nonsingular, the system Ax = b has a unique solution that can be found by computing x = A^-1b. If A is singular, the system may have no solution or infinitely many solutions. 3) A set of vectors is linearly dependent if there exist scalar multiples of
Ch07 3
- 1. Ch 7.3: Systems of Linear Equations, Linear Independence, Eigenvalues A system of n linear equations in n variables, can be expressed as a matrix equation Ax = b: If b = 0, then system is homogeneous; otherwise it is nonhomogeneous. = nnnnnn n n b b b x x x aaa aaa aaa 2 1 2 1 ,2,1, ,22,21,2 ,12,11,1 ,,22,11, 2,222,211,2 1,122,111,1 nnnnnn nn nn bxaxaxa bxaxaxa bxaxaxa =+++ =+++ =+++
- 2. Nonsingular Case If the coefficient matrix A is nonsingular, then it is invertible and we can solve Ax = b as follows: This solution is therefore unique. Also, if b = 0, it follows that the unique solution to Ax = 0 is x = A-1 0 = 0. Thus if A is nonsingular, then the only solution to Ax = 0 is the trivial solution x = 0. bAxbAIxbAAxAbAx 1111 −−−− =⇒=⇒=⇒=
- 3. Example 1: Nonsingular Case (1 of 3) From a previous example, we know that the matrix A below is nonsingular with inverse as given. Using the definition of matrix multiplication, it follows that the only solution of Ax = 0 is x = 0: − −− −− = − = − 2/122/3 142 2/372/9 , 834 301 210 1 AA = − −− −− == − 0 0 0 0 0 0 2/122/3 142 2/372/9 1 0Ax
- 4. Example 1: Nonsingular Case (2 of 3) Now let’s solve the nonhomogeneous linear system Ax = b below using A-1 : This system of equations can be written as Ax = b, where Then 0834 2301 220 321 321 321 =+− −=++ =++ xxx xxx xxx − − = − − −− −− == − 7 12 23 0 2 2 2/122/3 142 2/372/9 1 bAx −= = − = 0 2 2 ,, 834 301 210 3 2 1 bxA x x x
- 5. Example 1: Nonsingular Case (3 of 3) Alternatively, we could solve the nonhomogeneous linear system Ax = b below using row reduction. To do so, form the augmented matrix (A|b) and reduce, using elementary row operations. ( ) − − =→ = =+ −=+ → − → − → −− − → − − → − −= 7 12 23 7 22 23 7100 2210 2301 14200 2210 2301 8430 2210 2301 0834 2210 2301 0834 2301 2210 3 32 31 x bA x xx xx 0834 2301 220 321 321 321 =+− −=++ =++ xxx xxx xxx
- 6. Singular Case If the coefficient matrix A is singular, then A-1 does not exist, and either a solution to Ax = b does not exist, or there is more than one solution (not unique). Further, the homogeneous system Ax = 0 has more than one solution. That is, in addition to the trivial solution x = 0, there are infinitely many nontrivial solutions. The nonhomogeneous case Ax = b has no solution unless (b, y) = 0, for all vectors y satisfying A* y = 0, where A* is the adjoint of A. In this case, Ax = b has solutions (infinitely many), each of the form x = x(0) + ξ, where x(0) is a particular solution of Ax = b, and ξ is any solution of Ax = 0.
- 7. Example 2: Singular Case (1 of 3) Solve the nonhomogeneous linear system Ax = b below using row reduction. To do so, form the augmented matrix (A|b) and reduce, using elementary row operations. ( ) solnno 10 153 12 1000 1530 1121 4000 1530 1121 3530 1530 1121 61060 1530 1121 1545 0651 1121 3 32 321 → = =+ =−− → −− → − −− → − −− → − −− → −− − −− = x xx xxx bA 1545 0651 1121 321 321 321 −=+− =++− =−− xxx xxx xxx
- 8. Example 2: Singular Case (2 of 3) Solve the nonhomogeneous linear system Ax = b below using row reduction. Reduce the augmented matrix (A|b) as follows: ( ) 072 2 7 2 1 000 530 121 2 5 2 1 530 530 121 51060 530 121 545 651 121 123 123 12 1 13 12 1 13 12 1 3 2 1 =−−→ −− + −− → − + −− → − + −− → − − −− = bbb bbb bb b bb bb b bb bb b b b b bA 3321 2321 1321 545 651 121 bxxx bxxx bxxx =+− =++− =−−
- 9. Example 2: Singular Case (3 of 3) From the previous slide, we require Suppose Then the reduced augmented matrix (A|b) becomes: ξxxxx += − + ==→ − − + =→ − − =→ = =+ =−− → −− + −− )0( 3 3 3 3 3 32 321 123 12 1 3 5 7 0 0 1 1 3/5 3/7 0 0 1 3/5 3/71 00 053 112 2 7 2 1 000 530 121 cx x x x x xx xxx bbb bb b 072 123 =−− bbb 5,1,1 321 =−== bbb
- 10. Linear Dependence and Independence A set of vectors x(1) , x(2) ,…, x(n) is linearly dependent if there exists scalars c1, c2,…, cn, not all zero, such that If the only solution of is c1= c2 = …= cn = 0, then x(1) , x(2) ,…, x(n) is linearly independent. 0xxx =+++ )()2( 2 )1( 1 n nccc 0xxx =+++ )()2( 2 )1( 1 n nccc
- 11. Example 3: Linear Independence (1 of 2) Determine whether the following vectors are linear dependent or linearly independent. We need to solve or = − ⇔ = + − + 0 0 0 834 301 210 0 0 0 8 3 2 3 0 1 4 1 0 3 2 1 21 c c c ccc 0xxx =++ )3( 3 )2( 2 )1( 1 ccc = − = = 8 3 2 , 3 0 1 , 4 1 0 )3()2()1( xxx
- 12. Example 3: Linear Independence (2 of 2) We thus reduce the augmented matrix (A|b), as before. Thus the only solution is c1= c2 = …= cn = 0, and therefore the original vectors are linearly independent. ( ) =→ = =+ =+ → → − = 0 0 0 0 02 03 0100 0210 0301 0834 0301 0210 3 32 31 c bA c cc cc
- 13. Example 4: Linear Dependence (1 of 2) Determine whether the following vectors are linear dependent or linearly independent. We need to solve or − = − − = −= 5 6 1 , 4 5 2 , 5 1 1 )3()2()1( xxx 0xxx =++ )3( 3 )2( 2 )1( 1 ccc = − − −− ⇔ = − + − − + − 0 0 0 545 651 121 0 0 0 5 6 1 4 5 2 5 1 1 3 2 1 321 c c c ccc
- 14. Example 4: Linear Dependence (2 of 2) We thus reduce the augmented matrix (A|b), as before. Thus the original vectors are linearly dependent, with ( ) − =→ − − =→ = =+ =−− → −− → − − −− = 3 5 7 3/5 3/7 00 053 012 0000 0530 0121 0545 0651 0121 3 3 3 3 32 321 k c c c c cc ccc cc bA = − − − − + − 0 0 0 5 6 1 3 4 5 2 5 5 1 1 7
- 15. Linear Independence and Invertibility Consider the previous two examples: The first matrix was known to be nonsingular, and its column vectors were linearly independent. The second matrix was known to be singular, and its column vectors were linearly dependent. This is true in general: the columns (or rows) of A are linearly independent iff A is nonsingular iff A-1 exists. Also, A is nonsingular iff detA ≠ 0, hence columns (or rows) of A are linearly independent iff detA ≠ 0. Further, if A = BC, then det(C) = det(A)det(B). Thus if the columns (or rows) of A and B are linearly independent, then the columns (or rows) of C are also.
- 16. Linear Dependence & Vector Functions Now consider vector functions x(1) (t), x(2) (t),…, x(n) (t), where As before, x(1) (t), x(2) (t),…, x(n) (t) is linearly dependent on I if there exists scalars c1, c2,…, cn, not all zero, such that Otherwise x(1) (t), x(2) (t),…, x(n) (t) is linearly independent on I See text for more discussion on this. ( ) ( )βα,,,,2,1, )( )( )( )( )( )( 2 )( 1 =∈= = Itnk tx tx tx t k m k k k x Ittctctc n n ∈=+++ allfor,)()()( )()2( 2 )1( 1 0xxx
- 17. Eigenvalues and Eigenvectors The eqn. Ax = y can be viewed as a linear transformation that maps (or transforms) x into a new vector y. Nonzero vectors x that transform into multiples of themselves are important in many applications. Thus we solve Ax = λx or equivalently, (A-λI)x = 0. This equation has a nonzero solution if we choose λ such that det(A-λI) = 0. Such values of λ are called eigenvalues of A, and the nonzero solutions x are called eigenvectors.
- 18. Example 5: Eigenvalues (1 of 3) Find the eigenvalues and eigenvectors of the matrix A. Solution: Choose λ such that det(A-λI) = 0, as follows. − = 63 32 A ( ) ( )( ) ( )( ) ( )( ) 7,3 73214 3362 63 32 det 00 01 63 32 detdet 2 −==⇒ +−=−+= −−−−= −− − = − − =− λλ λλλλ λλ λ λ λλIA
- 19. Example 5: First Eigenvector (2 of 3) To find the eigenvectors of the matrix A, we need to solve (A-λI)x = 0 for λ = 3 and λ = -7. Eigenvector for λ = 3: Solve by row reducing the augmented matrix: ( ) = − − ⇔ = −− − ⇔=− 0 0 93 31 0 0 363 332 2 1 2 1 x x x x 0xIA λ =→ = =→ = =− → − → − − → − − 1 3 choosearbitrary, 1 33 00 031 000 031 093 031 093 031 )1( 2 2)1( 2 21 xx cc x x x xx
- 20. Example 5: Second Eigenvector (3 of 3) Eigenvector for λ = -7: Solve by row reducing the augmented matrix: ( ) = ⇔ = +− + ⇔=− 0 0 13 39 0 0 763 372 2 1 2 1 x x x x 0xIA λ − =→ − = − =→ = =+ → → → 3 1 choosearbitrary, 1 3/13/1 00 03/11 000 03/11 013 03/11 013 039 )2( 2 2)2( 2 21 xx cc x x x xx
- 21. Normalized Eigenvectors From the previous example, we see that eigenvectors are determined up to a nonzero multiplicative constant. If this constant is specified in some particular way, then the eigenvector is said to be normalized. For example, eigenvectors are sometimes normalized by choosing the constant so that ||x|| = (x, x)½ = 1.
- 22. Algebraic and Geometric Multiplicity In finding the eigenvalues λ of an n x n matrix A, we solve det(A-λI) = 0. Since this involves finding the determinant of an n x n matrix, the problem reduces to finding roots of an nth degree polynomial. Denote these roots, or eigenvalues, by λ1, λ2, …, λn. If an eigenvalue is repeated m times, then its algebraic multiplicity is m. Each eigenvalue has at least one eigenvector, and a eigenvalue of algebraic multiplicity m may have q linearly independent eigevectors, 1 ≤ q ≤ m, and q is called the geometric multiplicity of the eigenvalue.
- 23. Eigenvectors and Linear Independence If an eigenvalue λ has algebraic multiplicity 1, then it is said to be simple, and the geometric multiplicity is 1 also. If each eigenvalue of an n x n matrix A is simple, then A has n distinct eigenvalues. It can be shown that the n eigenvectors corresponding to these eigenvalues are linearly independent. If an eigenvalue has one or more repeated eigenvalues, then there may be fewer than n linearly independent eigenvectors since for each repeated eigenvalue, we may have q < m. This may lead to complications in solving systems of differential equations.
- 24. Example 6: Eigenvalues (1 of 5) Find the eigenvalues and eigenvectors of the matrix A. Solution: Choose λ such that det(A-λI) = 0, as follows. = 011 101 110 A ( ) 1,1,2 )1)(2( 23 11 11 11 detdet 221 2 3 −=−==⇒ +−= ++−= − − − =− λλλ λλ λλ λ λ λ λIA
- 25. Example 6: First Eigenvector (2 of 5) Eigenvector for λ = 2: Solve (A-λI)x = 0, as follows. =→ = =→ = =− =− → − − → − − → − − − → − − − → − − − 1 1 1 choosearbitrary, 1 1 1 00 011 011 0000 0110 0101 0000 0110 0211 0330 0330 0211 0112 0121 0211 0211 0121 0112 )1( 3 3 3 )1( 3 32 31 xx cc x x x x xx xx
- 26. Example 6: 2nd and 3rd Eigenvectors (3 of 5) Eigenvector for λ = -1: Solve (A-λI)x = 0, as follows. − = − =→ − + − = −− =→ = = =++ → → 1 1 0 , 1 0 1 choose arbitrary,where, 1 0 1 0 1 1 00 00 0111 0000 0000 0111 0111 0111 0111 )3()2( 3232 3 2 32 )2( 3 2 321 xx x xxxx x x xx x x xxx
- 27. Example 6: Eigenvectors of A (4 of 5) Thus three eigenvectors of A are where x(2) , x(3) correspond to the double eigenvalue λ = - 1. It can be shown that x(1) , x(2) , x(3) are linearly independent. Hence A is a 3 x 3 symmetric matrix (A = AT ) with 3 real eigenvalues and 3 linearly independent eigenvectors. − = − = = 1 1 0 , 1 0 1 , 1 1 1 )3()2()1( xxx = 011 101 110 A
- 28. Example 6: Eigenvectors of A (5 of 5) Note that we could have we had chosen Then the eigenvectors are orthogonal, since Thus A is a 3 x 3 symmetric matrix with 3 real eigenvalues and 3 linearly independent orthogonal eigenvectors. −= − = = 1 2 1 , 1 0 1 , 1 1 1 )3()2()1( xxx ( ) ( ) ( ) 0,,0,,0, )3()2()3()1()2()1( === xxxxxx
- 29. Hermitian Matrices A self-adjoint, or Hermitian matrix, satisfies A = A* , where we recall that A* = AT . Thus for a Hermitian matrix, aij = aji. Note that if A has real entries and is symmetric (see last example), then A is Hermitian. An n x n Hermitian matrix A has the following properties: All eigenvalues of A are real. There exists a full set of n linearly independent eigenvectors of A. If x(1) and x(2) are eigenvectors that correspond to different eigenvalues of A, then x(1) and x(2) are orthogonal. Corresponding to an eigenvalue of algebraic multiplicity m, it is possible to choose m mutually orthogonal eigenvectors, and hence A has a full set of n linearly independent orthogonal eigenvectors.