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![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)
![Expansion by Rows Theorem • Cofactor Matrix ○ C_kl=(−1)^(k+l) {█((n−1)×(n−1) determinant obtained@by deleting row k and column l @from the original determinant)} • Determinant and Cofactor Matrix ○ det(A)=|■8(a_11&a_12&⋯&a_1n@⋮&⋮&⋮&⋮@a_k1&a_k2&…&a_kn@⋮&⋮&⋮&⋮@a_n1&a_n2&⋯&a_nn )|=a_k1 C_k1+a_k2 C_k2+…+a_kn C_kn ○ [■8(a_11&a_12&⋯&a_1n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )] ⏟([■8(C_11&C_21&⋯&C_n1@C_12&C_22&…&C_n2@⋮&⋮&⋱&⋮@C_1n&C_2n&⋯&C_nn )] )┬(adjugate matrix of A: adj(A))=detA⋅[■(1&&&@&1&&@&&⋱&@&&&1)] • Expansion by Rows ○ |■8(a_11&a_12&⋯&a_1n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )|=a_11 C_11+a_12 C_12+…+a_1n C_1n ○ |■8(x_1&x_2&⋯&x_n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )|=x_1 C_11+x_2 C_12+…+x_n C_1n • Calculating A⋅adj(A) ○ Expanding A⋅adj(A) § [■8(a_11&a_12&⋯&a_1n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )][■8(C_11&C_21&⋯&C_n1@C_12&C_22&…&C_n2@⋮&⋮&⋱&⋮@C_1n&C_2n&⋯&C_nn )] § =[■8(∑_(k=1)^n▒〖a_1k C_1k 〗&∑_(k=1)^n▒〖a_1k C_2k 〗&⋯&∑_(k=1)^n▒〖a_1k C_nk 〗@∑_(k=1)^n▒〖a_2k C_1k 〗&∑_(k=1)^n▒〖a_2k C_2k 〗&…&∑_(k=1)^n▒〖a_2k C_nk 〗@⋮&⋮&⋱&⋮@∑_(k=1)^n▒〖a_nk C_1k 〗&∑_(k=1)^n▒〖a_nk C_2k 〗&⋯&∑_(k=1)^n▒〖a_nk C_nk 〗)] ○ Where § ∑_(k=1)^n▒〖a_1k C_1k 〗=|■8(a_11&a_12&⋯&a_1n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )|=detA § ∑_(k=1)^n▒〖a_1k C_2k 〗=|■8(a_21&a_22&⋯&a_2n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )|=0 § ⋮ ○ Conclusion § A⋅adj(A)=[■(detA&&&@&detA&&@&&⋱&@&&&detA )]=detA [■(1&&&@&1&&@&&⋱&@&&&1)] • Theorem ○ det〖(A)≠0〗⟺A is invertible and A^(−1)=1/detA ⋅adj(A) ○ det(A)=0⟺A is not invertible • Example ○ Let A=[■8(a&b@c&d)] ○ Cofactor Matrix § C=[■8(C_11&C_12@C_21&C_22 )]=[■8(d&−c@−b&a)] ○ Adjugate Matrix § adj(A)=C^T=[■8(d&−b@−c&a)] ○ Determinant § detA=|■8(a&b@c&d)|=ad−bc ○ Inverse Matrix § A^(−1)=[■8(a&b@c&d)]^(−1)=1/detA ⋅adj(A)=1/(ad−bc) [■8(d&−b@−c&a)] Cramer](https://shawnzhong.com/wp-content/uploads/2017/11/img_5a049b6a19df1.png)
![Determinant and Area • |■8(a_1&a_2@b_1&b_2 )| = area of parallelogram with sides a=(█(a_1@a_2 )), b=(█(b_1@b_2 )) • Proof by graph • Proof ○ Area(A_1,A_2 )=signed area of parallelogram spanned by A_1, A_2 ○ If A_1→A_2 is counter-clockwise = area ○ If A_1→A_2 is clockwise =−area ○ Then Area(A_1,A_2 )=det(A_1,A_2 ), because ○ Alternating § Area(A_1,A_2 )=−Area(A_2,A_1 ) § (by definition, same area, but different orientation) ○ Linearity(Homogeneous) § Area(t⋅A_1,A_2 )=t⋅Area(A_1,A_2 ) § (Easy to prove from picture) ○ Linearity(Additive) § Area(A+B,C)=Area(A,C)+Area(B,C) § If A,C is parallel, then □ Area(A,C)=0 § If A,C is independent , then □ Area(A+sC,C)=Area(A,C), ∀A,C § Let B=t⋅A+s⋅C, then □ Area(A+B,C) □ =Area(A+t⋅A+s⋅C,C) □ =Area(A+t⋅A,C) □ =(1+t)Area(A,C) □ =Area(A,C)+t⋅Area(A,C) □ =Area(A,C)+Area(t⋅A,C) □ =Area(A,C)+Area(t⋅A+s⋅C,C) □ =Area(A,C)+Area(B,C) § Therefore Area(A+B,C)=Area(A,C)+Area(B,C) ○ Uniqueness Theorem § Area(A,B) § =det〖(A,B)⋅Area(I_1,I_2 )〗 § =det〖(A,B)⋅Area(unit square)〗 § =det(A,B) Determinant and Volume • det〖(A,B,C)=signed volume〗 of parallelepiped spanned by A,B,C Inverse of a Matrix • Setup ○ T:Rn→Rn linear ○ T has a matrix m(T)=[■8(T_11&⋯&T_1n@⋮&⋱&⋮@T_n1&⋯&T_nn )] • The following statements are equivalent ○ N(T)={0} ○ T is injective ○ T is one\-to\-one ○ T is bijective § because T:Rn→Rn § dimN(T)+dimrange(T)=dim〖Rn 〗 § ⇒dimrange(T)=n § ⇒R(T)=Rn ○ There is a map S:Rn→Rn with ST=TS=I • Find the inverse of 2×2 matrix ○ T=[■8(1&3@2&5)] ○ Find T^(−1), i.e. solve Tx=y ○ Note: Tx=y⟺x=T^(−1) y ○ Normal version § [■8(1&3@2&5)][█(x_1@x_2 )]=[█(y_1@y_2 )] § {█(x_1+3x_2=1⋅y_1+0⋅y_2@2x_2+5x_2=0⋅y_1+1⋅y_2 )┤ § ⇒{█(x_1=−5y_1+3y_2@x_2=2y_1−y_2 )┤ § ⇒x=T^(−1) y § where T^(−1)=[■8(−5&3@2&−1)] ○ Shorthand § [T│I] § ~[■8(1&3@2&5) │ ■8(1&0@0&1)] § ~[■8(1&3@0&−1) │ ■8(1&0@−2&1)] § ~[■8(1&0@0&−1) │ ■8(−5&3@−2&1)] § ~[■8(1&0@0&1) │ ■8(−5&3@2&−1)] § ~[I│T^(−1) ] § Therefore T^(−1)=[■8(−5&3@2&−1)] Minors and Cofactors • Theorem ○ |■8(a_11&a_12&⋯&a_1n@⋮&⋮&⋮&⋮@a_k1&a_k2&…&a_kn@⋮&⋮&⋮&⋮@a_n1&a_n2&⋯&a_nn )|=a_k1 C_k1+a_k2 C_k2+…+a_kn C_kn ○ C_kl=cofactor matrix • Cofactor Matrix C_kl=(−1)^(k+l) {█((n−1)×(n−1) determinant obtained@by deleting row k and column l @from the original determinant)} • Example ○ |■8(1&7&2@4&π&−1@3&ln2&2)| ○ =3×|■8(7&2@π&−1)|−ln2 |■8(1&2@4&−1)|+2|■8(1&7@4&π)| ○ =3×(−7−2π)−ln2×(−9)+2×(π−28) ○ =−77−4π+9ln2 • Matrix Multiplication ○ Let P=[■8(a_11&a_12&⋯&a_1n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )][■8(C_11&C_21&⋯&C_n1@C_12&C_22&…&C_n2@⋮&⋮&⋱&⋮@C_1n&C_2n&⋯&C_nn )] ○ P_11=a_11 C_11+a_12 C_12+…+a_1n C_1n=detA ○ P_21=a_21 C_11+a_21 C_12+…++a_21 C_1n=0 ○ Because we have two equal row ○ Therefore P=detA [■(1&&@&⋱&@&&1)] •](https://shawnzhong.com/wp-content/uploads/2017/11/img_5a020d4a91900.png)
