Linear Algebra

356 Chapter Five. Similarity3.8 Example IfS =( )π 10 3(here π is not a projection map, it is the number 3.14 . . .) then( )∣ π − x 1 ∣∣∣∣= (x − π)(x − 3)0 3 − xso S has eigenvalues of λ 1 = π and λ 2 = 3. To find associated eigenvectors, firstplug in λ 1 for x:( ) ( ) ( ( ) (π − π 1 z1 0 z1 a= =⇒ =0 3 − π z 2 0)z 2 0)for a scalar a ≠ 0, and then plug in λ 2 :( ) ( ) (π − 3 1 z1 0=0 3 − 3 z 2 0)where b ≠ 0.=⇒(z1) ( )−b/(π − 3)=z 2 b3.9 Definition The characteristic polynomial of a square matrix T is thedeterminant of the matrix T − xI, where x is a variable. The characteristicequation is |T − xI| = 0. The characteristic polynomial of a transformation tis the polynomial of any Rep B,B (t).Exercise 30 checks that the characteristic polynomial of a transformation iswell-defined, that is, any choice of basis yields the same polynomial.3.10 Lemma A linear transformation on a nontrivial vector space has at leastone eigenvalue.Proof. Any root of the characteristic polynomial is an eigenvalue. Over thecomplex numbers, any polynomial of degree one or greater has a root. (This isthe reason that in this chapter we’ve gone to scalars that are complex.) QEDNotice the familiar form of the sets of eigenvectors in the above examples.3.11 Definition The eigenspace of a transformation t associated with theeigenvalue λ is V λ = { ζ ⃗ ∣ t( ζ ⃗ ) = λζ ⃗ } ∪ {⃗0 }. The eigenspace of a matrix isdefined analogously.3.12 Lemma An eigenspace is a subspace.Proof. An eigenspace must be nonempty — for one thing it contains the zerovector — and so we need only check closure. Take vectors ⃗ ζ 1 , . . . , ⃗ ζ n from V λ , toshow that any linear combination is in V λt(c 1⃗ ζ1 + c 2⃗ ζ2 + · · · + c n⃗ ζn ) = c 1 t( ⃗ ζ 1 ) + · · · + c n t( ⃗ ζ n )= c 1 λ ⃗ ζ 1 + · · · + c n λ ⃗ ζ n= λ(c 1⃗ ζ1 + · · · + c n⃗ ζn )(the second equality holds even if any ⃗ ζ i is ⃗0 since t(⃗0) = λ · ⃗0 = ⃗0).QED