Linear Algebra

Section II. Similarity 3653.2 Remark This definition requires that the eigenvector be non-⃗0. Some authorsallow ⃗0 as an eigenvector for λ as long as there are also non-⃗0 vectors associatedwith λ. Neither style of definition is clearly better; both involve small tradeoffs.In both styles the key point is to not allow a case where λ is such that t(⃗v) = λ⃗vfor only the single vector ⃗v = ⃗0.Also, note that λ could be 0. The issue is whether ⃗ζ could be ⃗0.3.3 Example The projection map⎛ ⎞ ⎛ ⎞x x⎜ ⎟⎝y⎠↦−→π ⎜ ⎟⎝y⎠z 0x, y, z ∈ Chas an eigenvalue of 1 associated with any eigenvector of the form⎛ ⎞x⎜ ⎟⎝y⎠0where x and y are scalars that are not both zero. On the other hand, 2 is notan eigenvalue of π since no non-⃗0 vector is doubled.3.4 Example The only transformation on the trivial space {⃗0 } is ⃗0 ↦→ ⃗0. Thismap has no eigenvalues because there are no non-⃗0 vectors ⃗v mapped to a scalarmultiple λ · ⃗v of themselves.3.5 Example Consider the homomorphism t: P 1 → P 1 given by c 0 + c 1 x ↦→(c 0 + c 1 ) + (c 0 + c 1 )x. While the codomain P 1 of t is two-dimensional, its rangeis one-dimensional R(t) = {c + cx ∣ ∣ c ∈ C}. Application of t to a vector in thatrange will simply rescale the vector c + cx ↦→ (2c) + (2c)x. That is, t has aneigenvalue of 2 associated with eigenvectors of the form c + cx where c ≠ 0.This map also has an eigenvalue of 0 associated with eigenvectors of the formc − cx where c ≠ 0.3.6 Definition A square matrix T has a scalar eigenvalue λ associated with thenonzero eigenvector ⃗ζ if T⃗ζ = λ · ⃗ζ.Although this extension from maps to matrices is natural, we need to makeone observation. Eigenvalues of a map are also the eigenvalues of matricesrepresenting that map and so similar matrices have the same eigenvalues. But theeigenvectors can different — similar matrices need not have the same eigenvectors.3.7 Example Consider again the transformation t: P 1 → P 1 from Example 3.5given by c 0 + c 1 x ↦→ (c 0 + c 1 ) + (c 0 + c 1 )x. One of its eigenvalues is 2, associatedwith the eigenvectors c + cx where c ≠ 0. If we represent t with respect toB = 〈1 + 1x, 1 − 1x〉T = Rep B,B (t) =(2 00 0)