Linear-2 3

Motivation behind diagonalization

A∼DA=P−1DPAn⏟Hard=(P−1DP)n=P−1Dn⏟EasyP

Diagonalizable matrix

A is called diagonalizable iff ∃D:A∼D

Diagonalizable matrix and eigenvectors

Let A∈Fn×nA is diagonalizable⟺∃B basis of Fn:∀i∈[1,n]:Avi=λivi
A=(111021023)∈R3×3Determine whether A is diagonalizableSolution:det(λI−A)=|λ−1−1−10λ−2−10−2λ−3|=(λ−1)((λ−2)(λ−3)−2)=(λ−1)2(λ−4)λ=1⟹(0−1−100−1−100−2−20)→(011000000000)⟹E1=sp{(100),(01−1)}λ=4⟹(3−1−100−2−100−2−10)→(3−1−1002100000)⟹E4=sp{(112)}P=(11001102−1)⟹D=(100040001)
T:R2→R2B={(10),(11)}T(v1)=2v1T(v2)=v1+v2Determine whether T is diagonalizableSolution:[T]BB=(2101)PT(λ)=(λ−2)(λ−1)⟹T is diagonalizableE2=sp{(10)}E1=sp{(1−1)}CB={(10),(1−1)}={[d1]B,[d2]B}⟹D={(10),(0−1)}[T]CBCB=(2101)−Not diagonal![T]DD=(2001)−Diagonal!
T:R2[x]→R2[x]B={1+x,x,x2}T(1+x)=−1+x2T(x)=1+x2T(x2)=1+2x−x2Determine whether T is diagonalizableSolution:[T]BB=(−1111−1111−1)PT(λ)=|λ+1−1−1−1λ+1−1−1−1λ+1|=(λ−1)|111−1λ+1−1−1−1λ+1|==(λ−1)|1110λ+2000λ+2|=(λ−1)(λ+2)2λ=1⟹(0000−12−10−1−120)⟹E1=sp{(111)}λ=−2⟹(0000−1−1−10−1−1−10)⟹E−2=sp{(10−1),(1−10)}⟹D={1+2x+x2,1+x−x2,1}⟹[T]DD=(1000−2000−2)
A=(1000π200−1e30−123534)B=(444401π−e00230003)Determine whether A∼BSolution:PA(λ)=(λ−1)(λ−2)(λ−3)(λ−4)PB(λ)=(λ−1)(λ−2)(λ−3)(λ−4)A∼(1000020000300004)∼B⟹A∼B
A=(3000120000200a0b)b=2⟹[A is diagonalizable⟺a=0]b=3⟹[A is diagonalizable⟺a=0]b∉{2,3}⟹A is diagonalizable
A∈C3×3PA(λ)=λ3A=(010001000)⟹dim⁡(N(0I−A))=dim⁡(N(−A))=dim⁡(N(A))=1⟹A is not diagonalizable