napkin solutions > Ch. 10

Back to Chapter Selection

Table of Contents

  1. Problem 10C
  2. Problem 10D
  3. Problem 10G

Problem 10C #

Problem 10C (Switccharoo). Let 𝑉 be the ℂ-vector space with basis 𝑒1 and 𝑒2. The map 𝑇:𝑉→𝑉 sends𝑇(𝑒1)=𝑒2 and 𝑇(𝑒2)=𝑒1. Determine the eigenspaces of 𝑇.Solution by RanolP∀𝑣∈𝑉, 𝑣 can be expressed as 𝑎𝑒1+𝑏𝑒2.𝑇(𝑣)=𝑇(𝑎𝑒1+𝑏𝑒2)=𝑎𝑇(𝑒1)+𝑏𝑇(𝑒2)=𝑎𝑒2+𝑏𝑒1If 𝑣 is an eigenvector of 𝑇, ∃𝜆𝜆(𝑎𝑒1+𝑏𝑒2)=𝜆𝑣=𝑇(𝑣)=𝑎𝑒2+𝑏𝑒1Multiplication is distributive, so 𝜆𝑎𝑒1+𝜆𝑏𝑒2=𝑎𝑒2+𝑏𝑒1.Since they’re basis, they’re linear independent, so 𝜆𝑎=𝑏 and 𝜆𝑏=𝑎.Thus, we get 𝜆2𝑏=𝑏 and 𝜆2𝑎=𝑎. Since it’s eigenvector, 𝑎 and 𝑏 cannot be both zero.Thus if 𝑎≠0, 𝜆2𝑎𝑎=𝑎𝑎, so 𝜆=±1Otherwise 𝑏≠0, 𝜆2𝑏𝑏=𝑏𝑏, so 𝜆=±1The eigenvalue is determined. Let’s find eigenspaces.•For 1-eigenspace𝑎𝑒1+𝑏𝑒2=𝑎𝑒2+𝑏𝑒1, in other words 𝑎(𝑒1+𝑒2) (where 𝑎≠0∈ℂ).The space is spanned by 𝑒1+𝑒2•For −1-eigenspace−(𝑎𝑒1+𝑏𝑒2)=𝑎𝑒2+𝑏𝑒1, in other words 𝑎(𝑒1−𝑒2) (where 𝑎≠0∈ℂ).The space is spanned by 𝑒1−𝑒2.

Problem 10D #

Problem 10D. Suppose 𝑇:ℂ⊕2→ℂ⊕2 is a linear map of ℂ-vector spaces such that 𝑇2011=id. Must 𝑇 bediagonalizable?Solution by kiwiyouYes.Proof. 𝑇 has 2×2 Jordan form, which can contain (1) two 1×1 Jordan blocks, or (2) one 2×2 Jordan block.(1)Jordan form is 𝐽=(𝜆100𝜆2), which is trivially diagonalizable.(2)Jordan form is 𝐽=(𝜆01𝜆), then𝐽2011=(𝜆201102011𝜆2010𝜆2011)=𝐼Since 2011𝜆2010=0, 𝜆=0 but 0=𝜆2011=1, which is contradiction.Therefore 𝑇 is diagonalizable.∎

Problem 10G #

Problem 10G (Differentiation of functions). Let 𝑉 be the infinite-dimensional real vector space of allinfinitely differentiable functions ℝ→ℝ. Note that 𝑑𝑑𝑥:𝑉→𝑉 is a linear map (for example it sends cos𝑥to −sin𝑥). Which real numbers are eigenvalues of this map?Solution by kiwiyouAll numbers.Proof. For all 𝑘∈ℝ, consider infinitely differentiable function 𝑓:ℝ→ℝ=𝑥↦𝑒𝑘𝑥, then𝑑𝑓𝑑𝑥=𝑘𝑒𝑘𝑥=𝑘𝑓Therefore, every real number 𝑘 is eigenvalue.∎