Shawn Zhong

Shawn Zhong

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Shawn Zhong

钟万祥
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Home / 2017 / Page 10

Math 375 – 11/29

  • Nov 29, 2017
  • Shawn
  • Math 375
  • No comments yet
Question 1 (from Monday) • Given ○ Let V be a vector space and let T:V→V be a linear map ○ Suppose x,y∈V are eigenvectors of T with eigenvalues λ and μ. • Prove ○ If ax+by (a≠0, b≠0) is an eigenvector of T, then λ=μ • Proof ○ Tx=λx, Ty=μy ○ ⇒T(ax+by)=aλx+bμy ○ Denote the eigenvalue for ax+by to be k ○ ⇒T(ax+by)=k(ax+by) ○ ⇒aλx+bμy=akx+bky ○ ⇒a(λ−k)x−b(μ−k)y=0 ○ If x,y are linearly independet § a(λ−k)=b(μ−k)=0 § Because a≠0, b≠0 § ⇒λ=μ=k ○ If x,y are linearly dependet § x=cy for some c § Tx=cTy=cμy=μ(cy)=μx § ⇒λ=μ Question 2 • Given ○ Let A be a real n×n matrix such that A^2=−I • Note ○ [■8(0&a@−1/a&0)]^2=−I, (a≠0) • Proof: A is invertibe ○ A(−A)=−A^2=−(−I)=I ○ ⇒A^(−1)=−A ○ ⇒A is invertibe • Proof: n is even ○ Suppose n is odd ○ det⁡〖A^2 〗=(det⁡A )^2≥0 ○ det⁡(−I)=−1<0 ○ Which makes a contradiction ○ Therefore n is even • Proof: A has no real eigenvalues ○ Suppose ∃λ∈R, x∈Rn, s.t. Ax=λx ○ A^2 x=−Ix=−x=λ^2 x ○ So λ^2=−1⇒λ=±i ○ Which makes a contradiction ○ Therefore A has no real eigenvalues • Proof: det⁡A=1 (when n=2) ○ A=[■8(a&b@c&d)] ○ A^2=[■8(a^2+bc&ab+bd@ac+cd&d^2+bc)] ○ {█(a^2+bc=d^2+bc=−1@ab+bd=ac+cd=0)┤ ○ ⇒ad−bc=1 • Proof: det⁡A=1 (general case) ○ (det⁡A )^2=det⁡〖A^2 〗=det⁡(−I)=(−1)^n=1 ○ ⇒det⁡A=±1 ○ Ax=λx⇒(Ax) ̅=(λx) ̅⇒Ax ̅=λ ̅x ̅ ○ Therefore the eigenvalues come in complex conjugate pairs ○ det⁡A=(λ_1 (λ_1 ) ̅ )(λ_2 (λ_2 ) ̅ )⋯(λ_k (λ_k ) ̅ )≥0 ○ Therefore det⁡A=1 Question 3 • Given ○ Let T:V→V be a finite-dimensional real linear transformation ○ T has no real eigenvalues • Proof: n is even ○ Suppose n is odd ○ f(λ)=−λ^n+a_(n−1) λ^(n−1)+…+a_1 λ+a_0 ○ As λ→∞, f(λ)⇒−∞ ○ As λ→−∞, f(λ)⇒∞ ○ By the Intermediate Value Theorem ○ f(λ) must have a real root ○ Which makes a contradiction ○ Therefore n is even • Proof: n=dim⁡V
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Math 375 – 11/28

  • Nov 28, 2017
  • Shawn
  • Math 375
  • No comments yet
Open Balls and Open Sets • Open Interval • Closed Interval • Interior Point ○ E⊆Rn is a subset ○ p∈E is an interior point if there is an r 0 ○ such that B_r (p)⊆E ○ where B_r (p) is the open disc of radius centered at p ○ B_r (p)={x∈Rn│‖x−p‖ r} • Koch s Snowflake • Open Sets ○ E⊆Rn is open if all x∈E are interior points in E • Example • Boundary Point ○ A point p∈Rn is a boundary point for E if for every r 0 ○ B_r (p) contains x,y with x∈E and y∉E Limits and Continuity • Limits ○ lim_(x→a)⁡f(x)=L⟺lim_(‖x−a‖→0)⁡‖f(x)−L‖=0 ○ If x→a, then f(x)→L • Properties ○ If f(x)→L∈Rm,g(x)→M∈Rm, when x→a, then ○ f(x)±g(x)→L±M ○ f(x)⋅g(x)→L⋅M ○ ‖f(x)‖→‖L‖ ○ f(x)/g(x) →L/M ○ (only when n=1, f(x),g(x)∈Rn) • Graph ○ Graph of f={(x,y,z)|z=f(x,y)} • Continuity ○ f:Rn→Rm is continuous at a∈Rn ○ if lim_(x→a)⁡f(x)=f(a) • Continuous Function Example ○ f(x_1,…,x_n )=x_k ○ f:Rn→R • Properties ○ If f,g is continuous ○ Then f±g, fg, f/g (g(a)≠0) are continuous • Example ○ f:R2→R ○ f(x,y)={■8(xy/(x^2+y^2 )&(x,y)≠(0,0)@0&x=y=0)┤ ○ f is continuous at all point except (0,0) ○ Let (x,y)→(0,0) along a straight line with angle θ ○ x=rcos⁡θ, y=rsin⁡θ ○ f(x,y)=xy/(x^2+y^2 )=(r^2 sin⁡θ cos⁡θ)/(r^2 cos^2⁡θ+r^2 sin⁡θ )=cos⁡θ sin⁡θ ○ Note that f(x,y) does not depend on r ○ lim_█((x,y)→(0,0)@along line@with angle θ)⁡f(x,y)=sin⁡θ cos⁡θ ○ When θ=π/2⇒f=0, when θ=π/4⇒f=1/2⋯ ○ Therefore we get the counter plot near origin ○ And the graph near 0 Derivative • Directional Derivative ○ D_hf(x)=∇_hf(x)= f^′ (x;h⃗ )=df_x⋅h ○ =lim_(t→0)⁡〖(f(x+th⃗ )−f(x))/t〗 ○ =[d/dt f(x+th⃗ )]_(t=0) • Example ○ f:Rn→R ○ f(x)=‖x‖^2 ○ f^′ (x;h⃗ ) ○ =[d/dt f(x+th]_(t=0) ○ =[d/dt ‖x+th^2 ]_(t=0) ○ =[d/dt (h2 t^2+(2hx)t+x^2 )]_(t=0) ○ =[2h2 t+2hx]_(t=0) ○ =2x⋅h • Partial Derivative • Total Derivative - • •
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Math 375 – 11/27

  • Nov 27, 2017
  • Shawn
  • Math 375
  • No comments yet
Question 1 • Question ○ Let θ∈R. ○ Find all eigenvalues and eigenvectors of the following matrix ○ A=[■8(cos⁡θ&−sin⁡θ@sin⁡θ&cos⁡θ )] • Answer ○ |A−λI|=|■8(cos⁡θ−λ&−sin⁡θ@sin⁡θ&cos⁡〖θ−λ〗 )|=\ (cos⁡θ−λ)^2+sin^2⁡θ=0 ○ ⇒λ^2−(2 cos⁡θ )λ+1=0 ○ ⇒λ=cos⁡θ±isin⁡θ ○ When λ_1=cos⁡θ−isin⁡θ § [■8(i sin⁡θ&−sin⁡θ@sin⁡θ&i sin⁡θ )][█(x_1@x_2 )]=0 § {█(i sin⁡θ x_1−sin⁡θ x_2=0@sin⁡θ x_1+i sin⁡θ x_2=0)┤⇒ix_1=x_2 § ⇒v_1=t(1,i), t∈ℂ ○ When λ_2=cos⁡θ+isin⁡θ § [■8(−i sin⁡θ&−sin⁡θ@sin⁡θ&−i sin⁡θ )][█(x_1@x_2 )]=0 § {█(−i sin⁡θ x_1−sin⁡θ x_2=0@sin⁡θ x_1−i sin⁡θ x_2=0)┤⇒−ix_1=x_2 § ⇒v_1=t(1,−i), t∈ℂ Question 2 • Question ○ Let V be a vector space and let T:V→V be a linear map ○ Suppose x∈V is an eigenvector for T with eigenvalue λ. ○ Prove that, for each polynomial, ○ the linear map P(T) has eigenvector x with eigenvalue P(λ) • Answer ○ Let P(λ)=c_n λ^n+c_(n−1) λ^(n−1)+…+c_1 λ+c_0 ○ (P(T))(x) ○ =(c_n T^n+c_(n−1) T^(n−1)+…+c_1 T+c_0 )(x) ○ =c_n T^n (x)+c_(n−1) T^(n−1) (x)+…+c_1 T(x)+c_0 x ○ =c_n λ^n x+c_(n−1) λ^(n−1) x+…+c_1 λx+c_0 x ○ =(c_n λ^n+c_(n−1) λ^(n−1)+…+c_1 λ+c_0 )x ○ =(P(λ))x Question 3 • Given ○ Let V be a vector space and let T:V→V be a linear map ○ Let c be a scalar. ○ Suppose T^2 has an eigenvalue c^2 • Prove ○ T has either c or −c as an eigenvalue • Proof ○ ∃x∈V, ≠0 ○ (T^2−c^2 I)x=0 ○ (T+cI)[(T−cI)x]=0 ○ When (T−cI)x≠0 § (T−cI)x is a eigenvector for T with eigenvalue of −c ○ When (T−cI)x=0 § x is a eigenvector for T with eigenvalue of c Question 4 • Given ○ Let V be a vector space and let T:V→V be a linear map ○ Suppose x,y∈V are eigenvectors of T with eigenvalues λ and μ. • Prove ○ If ax+by (a,b∈R) is an eigenvector of T, ○ then a=0 or b=0 or λ=μ • Proof
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Math 375 – 11/22

  • Nov 27, 2017
  • Shawn
  • Math 375
  • No comments yet
Theorem • V has a basis v_1,…,v_n, and another basis w_1,…,w_n • Let T be a linear transformation V→V • Define the following matrices ○ A≔matrix(T,v_i ) ○ B≔matrix(T,w_i ) ○ C≔∀i∈{1,…,n}, C(w_i )=v_i • Then B=C^(−1) AC Question • Given ○ T:R3→R3 ○ f(T)=(2−λ)^2 (3−λ) ○ dim⁡(Null(T−2I))=1 • Find T ○ T=[■8(2&1&0@∗&2&0@0&∗&3) ○ For λ=2 ○ Tv=2v ○ ⇒v=k[█(1@0@0)]
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Math 375 – 11/21

  • Nov 21, 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 • Theorem ○ Linear transformation T:Rn→Rn (or ℂ^n→ℂ^n) ○ matrix(T)=T=[■8(t_11&⋯&t_1n@⋮&⋱&⋮@t_n1&⋯&t_nn )] ○ Then λ is an eigenvalue of T if ○ det⁡(T−λI)=0 • Characteristic Polynomial ○ det⁡(T−λI) is the called characteristic polynomial of T ○ f(λ)=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 • How to Find Eigenvalues ○ Solve det⁡(T−λI)=0 ○ Get roots λ_1,…,λ_n (possibly repeated) • How to Find Eigenvectors ○ Solve (T−λI)v=0 ○ For λ=λ_1,λ=λ_2,…,λ=λ_n ○ (T−λI)v=0 is n equations with n unknowns ○ Typically v=0 is the only solution for some λ=λ_i ○ Then det⁡(T−λI)=0, and there is a solution v≠0 • Coefficients of Characteristic Polynomial ○ By definition § f(λ)=(−λ)^n+c_1 (−λ)^(n−1)+…+c_(n−1) (−λ)+c_n ○ By Fundamental Theorem of Algebra § f(λ)=a(λ_1−λ)(λ_2−λ)⋯(λ_n−λ) ○ Comparing the coefficient of (−λ)^n, we get § a=1 ○ Setting λ=0 to both polynomials we get § c_n=det⁡T=λ_1 λ_2…λ_n ○ By Vieta
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