Math 375 - 11/29

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(a2+bc&ab+bd@ac+cd&d^2+bc)] ○ {█(a^2+bc=d2+bc=−1@ab+bd=ac+cd=0)┤ ○ ⇒ad−bc=1 • Proof: det⁡A=1 (general case) ○ (det⁡A )^2=det⁡〖A2 〗=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