Linear Algebra

362 Chapter Five. Similarity2.4 Lemma A transformation t is diagonalizable if and only if there is a basisB = 〈⃗β 1 , . . . , ⃗β n 〉 and scalars λ 1 , . . . , λ n such that t(⃗β i ) = λ i⃗β i for each i.Proof Consider a diagonal representation matrix.⎛⎞ ⎛⎞..λ 1 0Rep B,B (t) = ⎜⎝Rep B (t(⃗β 1 )) · · · Rep B (t(⃗β n )) ⎟⎠ = ⎜⎝.. ..⎟. ⎠..0 λ nConsider the representation of a member of this basis with respect to the basisRep B (⃗β i ). The product of the diagonal matrix and the representation vector⎛ ⎞ ⎛ ⎞0 0⎛⎞λ 1 0..⎜Rep B (t(⃗β i )) = ⎝.. ..⎟. ⎠1=λ i0 λ⎜ ⎟ ⎜ ⎟n ⎝.⎠ ⎝ . ⎠0 0has the stated action.2.5 Example To diagonalize( )3 2T =0 1QEDwe take it as the representation of a transformation with respect to the standardbasis T = Rep E2 ,E 2(t) and we look for a basis B = 〈⃗β 1 , ⃗β 2 〉 such that( )λ 1 0Rep B,B (t) =0 λ 2that is, such that t(⃗β 1 ) = λ 1⃗β 1 and t(⃗β 2 ) = λ 2⃗β 2 .( ) ( )3 23 2⃗β 1 = λ 1 · ⃗β 1⃗β 2 = λ 2 · ⃗β 20 10 1We are looking for scalars x such that this equation( ) ( ) ( )3 2 b 1 b 1= x ·0 1 b 2 b 2has solutions b 1 and b 2 , which are not both zero (the zero vector is not themember of any basis). That’s a linear system.(3 − x) · b 1 + 2 · b 2 = 0(1 − x) · b 2 = 0(∗)Focus first on the bottom equation. The two numbers multiply to give zero onlyif at least one of them is zero so there are two cases, b 2 = 0 or x = 1. In the