Math 375 - 12/11

Quiz Question • Given ○ A:R3→R3 ○ det⁡〖(A−2I)=0〗 ○ tr (A)=−2 ○ det⁡(A)=6 • Question: Eigenvalue of A ○ det⁡〖(A−2I)=0〗⇒2 is an eigenvalue ○ tr (A)=−2⇒∑_(i=1)^n▒λ_i =λ_1+λ_2+λ_3=−2 ○ det⁡(A)=6⇒∏_(i=1)^n▒λ_i =λ_1 λ_2 λ_3=6 ○ {█(█(λ_1=2@λ_1+λ_2+λ_3=−2)@λ_1 λ_2 λ_3=6)┤⇒{█(λ_2=−3@λ_3=−1)┤ ○ Therefore eigenvalues are 2, −1, −3 • Question: Characteristic Polynomial ○ f(λ)=(λ−λ_1 )(λ−λ_2 )(λ−λ_3 )=(λ−2)(λ+1)(λ+3) • Note: f(λ)=det⁡(λI−A) Question 1 • Given ○ f:R2\0}→R ○ f(x,y)=xy/(x^2+y2 ), ∀(x,y)∈R2 • Question: Find the direction of steepest decedent at (1,3) ○ ∇f(x,y)=[█(∂f/∂x@∂f/∂y)]=[█(y(y^2−x2 )/(x^2+y2 )^2 @x(x^2−y2 )/(x^2+y2 )^2 )] ○ ∇f(1,3)=[█(3(3²⁻¹2 )/(1²⁺³2 )^2 @1(1²⁻³2 )/(1²⁺³2 )^2 )]=[█(6/25@−2/25)] • Question: Find the line best approximate the level set at (1,3) ○ ∇f(1,3)⋅n ⃗=0⇒n ⃗=[█(1@3)] ○ x+3y+c=0 ○ 1+3⋅3+c=0 ○ ⇒c=−10 ○ l: x+3y−10=0 ○ Alternative: ∇f(1,3)⋅[█(x−1@y−3)]=0 • Question: Estimate f(0.8,3.05) ○ f(0.8,3.05) ○ =f(1−0.2,3+0.05) ○ ≈f(1,3)+∇f(1,3)[█(−0.2@0.05)] ○ =3/(1²⁺³3 )+[█(6/25@−2/25)][█(−0.2@0.05)] ○ =0.248 Question 2 • Find a basis in which the matrix (■8(3&0@3&−2)) becomes diagonalized • Let A=(■8(3&0@3&−2)) • det⁡〖(A−λI)=λ^2−λ−6=0〗 • ⇒λ_1=3, λ_2=−2 • When λ_1=3 ○ A−λI=(■8(0&0@3&−5)) ○ ⇒v_1=(5,3) • When λ_2=−2 ○ A−λI=(■8(5&0@3&0)) ○ ⇒v_2=(0,1) • The basis is (5,3), (0,1)