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Final

MATH 4389 Final: Linear_Algebra(part_5)
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Department
Mathematics
Course
MATH 4389
Professor
Almus
Semester
Spring

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Math 2331 Section 5.1 – Eigenvectors and Eigenvalues 1 5  Example: Let A =   0 2    v =     w ==1   −1   u = 1   Definition: An eigenvector of an n××n matrix A is a nonzero vector x such that Ax == λx for some scalar λ λ . A scalar λ is an eigenvalue of A if there exists a nontrivial solution x of Ax == λx ; such an x is called an eigenvector corresponding to λλ . = λ λλ 1 4 2 Example: A = = 1 3     Is u =   an eigenvector?   1 Is v==  an eigenvector?   6 3  Example: A =   2 1  Show that 7 is an eigenvalue of A. 2 ▯ ▯ Notice that Ax = λx for ▯ . x ≠ 0⇒ Ax − λIx = 0⇒ (A− λI)x = 0 ▯ ▯ That is,x is an eigenvector corresponding to λ if and only if x∈Nul(A−λI) . ▯ Definition: Ifx is an eigenvector corresponding to λ , Nul(A− − λI) is called the eigenspace forλ . Notation: Eλ . KEY FACT: λ is an eigenvalue of A if and only ifdet(A−λI) = 0 . Reason: 3 Definition: Suppose A is an nxn matrix. th The characteristic polynomial of A is an n degree polynomial in λ : p(λ) = det(A−λI) 4 2 Example: Let A = . Find all e-values, e-vectors, e-spaces, characteristic 1 3 polynomial. 4 FACT: The special solutions to (A−λI)x = 0 form a basis for Eλ . This basis consist of e-vectors corresponding to e-value λ . 1 1  Example: Let A =  . Find all e-values, e-vectors, e-spaces, characteristic 1 1  polynomial. 5 0 1  Example: Let A = 1 0 . Find all e-values, e-vectors, e-spaces, characteristic   polynomial. 6  1 −1 0  Example: Let A= −1 2 −1 . Find all e-values, e-vectors, e-spaces,    0 −1 1  characteristic polynomial. 8 2 1 0    Example: Let A= 02 5 . Find all e-values, e-vectors, e-spaces, characteristic 0 0 2  polynomial. 9 2 1 0  Example: Let A= 0 2 0  . Find all e-values, e-vectors, e-spaces, characteristic   0 0 2  polynomial. Definition: Algebraic multiplicity of an eigenvalueλ is the degree of the corresponding factor in the characteristic polynomial. Geometric multiplicity of an eigenvalue is the dimension of the corresponding eigenspace. GM ≤ ≤ AM Fact: 10 IMPORTANT FACTS: 1) An nxn matrix A has at most n distinct eigenvalues. 2) If A is a triangular matrix, then the eigenvalues of A are the diagonal entries. 2 1 −1   A=  5 4 ; 2, 5, 6. 0 0 6    3) The product of the eigenvalues of a matrix A equals det(A). 4) The sum of the eigenvalues of a matrix A equals trace(A) . Here, trace(A) = a11 a 22..+ a nn the sum of the diagonal entries. Example:  4 2 A =  1 3 Prdocut of the evalues: 10 Sum of the eigenvalues: 7 ; 2 ad 5. 11 Example: A is a 2x2 matrix with determinant -20 andtrace 1. What are the eigenvalues of A? 5) 0 is an eigenvalue of A if and only if A is singular. 12 1 6) If A is invertible anλ ≠ 0, then λ is an eigenvalue of A if and only iλ is an eigenvalue of A−1. The evectors for λ and 1 are the same. λ Note: Elimination changes eigenvalues! 13 Theorem: If{v ,v ,...}vare eigenvectors corresponding to distinct 1 2 r eigenvaluesλ1λλ2,..λr, then the se{v1,2 ,..r}vis linearly independent. Question: Ix is an eigenvector of A, whatA x = ? 14 Math 2331 Section 5.2 – The Characteristic Equation A:n×××n matrix The characteristic equation of Adet(A− λI)= 0 A scalarλ is an eigenvalue of A if and onlyλisatisfies the characteristic equation det(A−−λI) = 0. Characteristic polynomial of A p(λ)= det(A− λI) Algebraic multiplicity of an eigenvλlis its multiplicity as a ropλ)o. Example: The characteristic polynomial of a matrix is: 6 5 4 pλ) = λ −9λλ −10λλ Find the eigenvalues and their algebraic multiplicities. 1 The Invertible Matrix Theorem (continued) Let A be an n××n matrix. A is invertible if and only if s. The number 0 is NOT an eigenvalue of A. t. The determinant of A is not 0. 2 Similarity Let A and B be two n× ×n matrices. A is similar to B if there exists an invertible matrix P such that P AP = = B . −1 This is equivalent to saying that A = = PBP . −1 Similarity Transformation: A ֏ P AP Theorem: If A and B are similar matrices, then they have the same characteristic polynomial and the same eigenvalues (with the same multiplicities). 3 Similar matrices have the SAME: • Eigenvalues • Determinant • Trace • Rank • Number of linearly independent eigenvectors • Jordan form (later) DIFFERENT: • 4 subspaces (row space, column space, etc.) • Eigenvectors FACT: If B = P AP and x▯is an eigenvector for A, then P x1▯ is an = eigenvector for B. 4  2 1  2 0 Warning: A =  0 2  and B = 0 2 have the same eigenvalues. However,     they are NOT similar! Similar ▯ same eigenvalues Same eigenvalues ▯ may be similar, may be not. Example : A = 2 3 and B =  4 1  are not similar. 0 1   0 5  5 2 3 Example : A = 0 1  Find a matrix which is similar to A.   6 Math 2331 Section 5.3 – Diagonalization Recall: Diagonal matrices -- 1 Definition: A:n××n matrix A can be diagonalized ( or “A is diagonalizable”) if A is similar to a DIAGONAL matrix D . That is, if there exists an invertible matrix P and a diagonal matrix D such thatA = PDP −.  2 3  1 1  2 0 1 − −1  Example:  =      0 5  0 1  0 5 0 1  2 Theorem: The Diagonalization Theorem An n××n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. −1 In fact,A = PDP , with D a diagonal matrix, if and only if the columns of P are n linearly independent vectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P. P: The eigenvector matrix for A D: The eigenvalue matrix for A. In other words, A is diagonalizable if it has enough eigenvectors to form a n n basis forℝ . Such a basis is called an eigenvector basis oℝ . 3 4 2 Example: Diagonalize A =1 3.   −1    From Section 5.1: e-value 2, e-vector  , e-value: 5, e-vector   1    4 1 −1 0  Example: Diagonalize A= − 2 −1  0 −1 1    From Section 5.1- e-values are 0, 1, 3.   −1  1        e-vector for 0: 1  e-vector for 1: 0 , e-vector for 3: −−2          1  1  5 FACTS: 1) A is diagonalizable if every eigenvalue has enough eigenvectors; GM=AM. (That is, if the sum of the dimensions of eigenspaces equals n ) 2) If an ×n matrix A has n distinct eigenvalues, then A is diagonalizable. (Reason: the set of n correspond
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