Math 375 - Midterm 2 Practice 2 Nov 15, 2017 Shawn Math 375 No comments yet Math 375 - 11/14 Nov 15, 2017 Shawn Math 375 No comments yet
![Eigenvalues and Eigenvectors • Definition ○ If T:V→V is linear and V is a vector space ○ Then v∈V is an eignevector of T with eigenvalue λ if § v≠0 § Tv=λv • Example ○ Suppose you have two eigenvectors § v,w∈V with Tv=λv, Tw=μw ○ Then § T(2v+3w)=2Tv+3Tw=2λv+3μw ○ Find a solution to Tx=v+w § Try x=av+bw § Then Tx=T(av+bw) § =λav+μbw § =v+w § ⇒┴? {█(λa=1@μb=1)┤⇒{█(a=1/λ@b=1/μ)┤ (if λ,μ≠0) § Therefore x=1/λ v+1/μ w ○ Compute T^2017 (2v+3w) § T^2017 (2v+3w) § =T^2016 (2λv+3μw) § =T^2015 (2λ^2 v+3μ^2 w) § ⋮ § =2λ^2017 v+3μ^2017 w • Fibonacci Number ○ f_n={■8(0&n=0@1&n=1@f_(n−1)+f_(n−2)&n≥2)┤ ○ For example § f_0=0 § f_1=1 § f_2=1 § f_3=2 § f_4=3 § ⋮ ○ It could be viewed as a sequence of vectors § [█(1@0)],[█(1@1)],[█(2@1)],[█(3@… ○ Consider § x_n=[█(f_n@f_(n−1) )] § x_(n+1)=[█(f_(n+1)@f_n )]=[█(f_n+f_(n+1)@f_n )]=⏟([■8(1&1@1&0)] )┬T ⏟([█(f_n@f_(n−1) )] )┬(x_n ) ○ Try to compute § x_2017=[█(f_2017@f_2016 )]=T[█(f_2016@f_2015 )]=…=T^2016 [█(1@0)] § If we had two eigenvectors/eigenvalues for T § And [█(1@0)]=av+bw § Then [█(f_2017@f_2016 )]=λ^2016 av+μ^2016 bw • Eigenvector Equation ○ By definition, if T:V→V is linear and V is a vector space ○ Then v∈V is an eignevector of T with eigenvalue if § v≠0, and Tv=λv § ⇒Tv=λIv § ⇒Tv−λIv=0 § ⇒(T−λI)v=0 § ⟺v∈Null(T−λI) ○ Therefore § v is an eigenvector with eigenvalue λ § ⇒0≠v∈Null(T−λI) § ⇒T−λI is not injective • Theorem ○ If T: Rn→Rn is given by matrix multipication ○ Then λ is an eigenvalue of T if and only if ○ det〖(T−λI)=0〗 • Proof ○ V=Rn or ℂ^n ○ Tx=[■8(t_11&⋯&t_1n@⋮&⋱&⋮@t_n1&⋯&t_nn )][█(x_1@⋮@x_n )] ○ T−λI=[■8(t_11−λ&t_12&⋯&t_1n@t_21&t_22−λ&…&t_2n@⋮&⋮&⋱&⋮@t_n1&t_n2&⋯&t_nn−λ)] ○ Fibonacci Example § T=[■8(1&1@1&0) § det〖(T−λI)=|■8(1−λ&1@1&−λ)|=λ^2−λ−1=┴? 0〗 • Solving for eigenvalue and eigenvector ○ For T: Rn→Rn (or ℂ^n→ℂ) ○ det(T−λI) is called the characteristic polynimal of T § det(T−λI) § =|■8(t_11−λ&t_12&⋯&t_1n@t_21&t_22−λ&…&t_2n@⋮&⋮&⋱&⋮@t_n1&t_n2&⋯&t_nn−λ)| § =(−λ)^n+c_1 (−λ)^(n−1)+…+c_(n−1) (−λ)+c_n § Where c_1=tr(T), c_n=detT ○ By Fundamental Theorem of Algebra § det(T−λI) § =(−λ)^n+c_1 (−λ)^(n−1)+…+c_(n−1) (−λ)+c_n § =(−λ)^n (λ−λ_1 )(λ−λ_2 )…(λ−λ_n ) § λ_1,λ_2,…,λ_n∈ℂ is called the eigentvalue of T ○ Given eigenvalues λ_1,…,λ_n § We can find eigenvectors N_1,…,N_n by § N_1∈N(T−λ_1 I) § N_2∈N(T−λ_2 I) § ⋮ § N_n∈N(T−λ_n I) • Theorem ○ T:V→V ○ v_1,…,v_k∈V are eigenvectors ○ with distinct eigenvalues λ_1,…,λ_k ○ then {v_1,…,v_k } is linearly indelendent • Proof ○ By induction on k ○ When k=1 § Given v_1∈V, v_1≠0,Tv_1=λ_1 v_2 § Then {v_1 } is independent because v_1≠0 ○ When k 1 § Assume Theorem true for k−1 § Suppose Tv_1=λ_1 v_1,…,Tv_k=λ_k v_k § λ_i≠λ_j for all i≠j, and all v_i≠0 § Suppose c_1 v_1+c_2 v_2+…+c_k v_k=0 § ⇒{█(λ_k c_1 v_1+λ_k c_2 v_2+…+λ_k c_k v_k=0@λ_1 c_1 v_1+λ_1 c_2 v_2+…+λ_1 c_k v_k=0)┤ § ⇒(λ_k−λ_1 ) c_1 v_1+…+(λ_k−λ_(k−1) ) c_(k−1) v_(k−1)=0 § Since Theorem is true for k−1 § ⇒{v_1,…,v_(k−1) } is linearly independent § ⇒{█(⏟((λ_k−λ_1 ) )┬(≠0) c_1=0@⋮@⏟((λ_k−λ_(k−1) ) )┬(≠0) c_(k−1)=0)┤ § ⇒c_1=c_2=…=c_(k−1)=0 § Therefore c_k v_k=0 § Since v_k≠0, we find c_k=0 § ⇒{v_1,…,v_k } is linearly independet](https://shawnzhong.com/wp-content/uploads/2017/11/img_5a145f1818e33.png)