L8_Eigenvectors - MA2213 Lecture 8 Eigenvectors Application...

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MA2213 Lecture 8 Eigenvectors
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Application of Eigenvectors + = + = - - - + + + - - + + λ 1 1 1 1 1 1 1 0 n n n c c c c Vufoil 18, lecture 7 : The Fibonacci sequence satisfies - + ± - ± ± ± ± - - + + + = = + = c c c s n n n , , 2 5 1 1 = = = = + + + 1 1 1 1 1 0 1 1 1 0 1 1 1 0 2 1 1 2 1 n n n n n n s s s s s s 1 1 5 4 3 2 1 , 5 , 3 , 2 , 1 - + + = = = = = = n n n s s s s s s s s + - - - - + + - - + + + = + + = 1 1 1 lim lim n n n n n n n n c c c c s s
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Fibonacci Ratio Sequence
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Fibonacci Ratio Sequence
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Another Biomathematics Application Leonardo da Pisa, better known as Fibonacci, invented his famous sequence to compute the reproductive success of rabbits* Similar sequences describe frequencies in males, females of a sex-linked gene. For genes (2 alleles) carried in the X chromosome** ,... 2 , 1 , 0 , 1 0 1 2 1 2 1 2 1 = = + + + n u u u u n n n n *page i, ** pages 10-12 in The Theory of Evolution and Dynamical Systems ,J. Hofbauer and K. Sigmund, 1984. The solution has the form n n v u , n n c c u ) ( 2 1 2 1 - + = where ) ( ), 2 ( 0 0 3 2 2 0 0 3 1 1 v u c v u c - = + =
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Eigenvector Problem (pages 333-351) Recall that if v is a square matrix then a nonzero A vector if v Av λ = is an eigenvector corresponding to the Eigenvectors and eigenvalues arise in biomathematics where they describe growth and population genetics eigenvalue They arise in physical problems, especially those that involve vibrations in which eigenvalues are related to vibration frequencies They arise in numerical solution of linear equations because they determine convergence properties
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Example 7.2.1 pages 333-334 For ) 2 ( 2 ) 1 ( 1 2 1 v c v c x x x + = = = 25 . 1 75 . 0 75 . 0 25 . 1 A the eigenvalue-eigenvector pairs are = = 1 1 , 2 ) 1 ( 1 v λ We observe that every (column) vector and - = = 1 1 , 5 . 0 ) 2 ( 2 v where 2 / ) ( 2 1 1 x x c + = 2 / ) ( 1 2 2 x x c - =
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Example 7.2.1 pages 333-334 Therefore, since x Ax is a linear transformation ) 2 ( 2 ) 1 ( 1 ) 2 ( 2 ) 1 ( 1 ) ( Av c Av c v c v c A Ax + = + = and since ) 2 ( ) 1 ( , v v We can repeat this process to obtain ) 2 ( 2 2 ) 1 ( 1 1 ) 2 ( 2 ) 1 ( 1 v c v c Av c Av c λ + = + are eigenvectors ) 2 ( 2 2 2 ) 1 ( 2 1 1 ) 2 ( 2 2 ) 1 ( 1 1 2 ) ( v c v c v c v c A x A + = + = ) 2 ( 2 1 2 ) 1 ( 1 ) 2 ( 2 2 ) 1 ( 1 1 ) ( 2 v c v c v c v c x A n n n n n + = + = Question What happens as n ?
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Example 7.2.1 pages 333-334 General Principle : If a vector v can be expressed as a linear combination of eigenvectors of a matrix A, then it is very easy to compute Av = 5 0 1 5 J It is possible to express every vector as a linear combination of eigenvectors of an n by n matrix A iff either of the following equivalent conditions is satisfied : (i) there exists a basis consisting of eigenvectors of A (ii) the sum of dimensions of eigenspaces of A = n Question Does this condition hold for ?
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L8_Eigenvectors - MA2213 Lecture 8 Eigenvectors Application...

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