Linear-2 5

Triangularizable matrix

A is called triangularizable iff ∃T:T is a triangular matrix:A∼TA∼T⟺PA(λ) is factorizable into linear factors
A=(123402060)Over C− all matrices are triangularizableOver R?PA(λ)=|λ−1−2−3−4λ−20−6λ|=6⋅|λ−1−3−4−2|+λ|λ−1−2−4λ|==12−12λ−72+λ2(λ−1)−8λ==λ3−λ2−20λ−60=−(λ−6)(λ2+5λ+10)⟹A≁T
Let A∈Rn×n be nilpotentDetermine whether A is triangularizableDetermine whether A is diagonalizableSolution:A∼D⟺A=0PA(λ)=λn⟹A∼T

Triangularization

1.Find PA(λ) and eigenvalues2.Find eigenspace for each eigenvalue3.Add vectors to the union of eigenspaces to get a basis of Fn4.Assign each vector from the basis to be a column of PResulting matrix is: P−1AP=(D∗0B)Where D is a diagonal matrix with eigenvalues on the diagonalEach eigenvalue is featured γA(λi) timesAnd B is a matrix with eigenvalues of AFor each eigenvalue: γB(λi)=μA(λi)−γA(λi)
A=(−1−3−4−511−1−325912−1−2−3−3)PA(λ)=|λ+1345−1λ−113−2−5λ−9−12123λ+3|=R1+Ri|λ−1λ−1λ−1λ−1−1λ−113−2−5λ−9−12123λ+3|==(λ−1)|1111−1λ−113−2−5λ−9−12123λ+3|=(λ−1)|1000−1λ24−2−3λ−7−10112λ+2|==(λ−1)|λ24−3λ−7−1012λ+2|=(λ−1)|λ240λ−13λ−412λ+2|==(λ−1)(λ(λ−1)(λ+2)−2λ(3λ−4)+(6λ−8−4λ+4))==(λ−1)(λ(λ+2)(λ−1)−6λ(λ−1)+4(λ−1))==(λ−1)2(λ(λ+2)−6λ+4)=(λ−1)2(λ−2)2λ=1⟹(2345−1013−2−5−8−121234)→(111101240−3−6−100123)→(1111012400010000)⟹E1=sp{(1−210)}…⟹E2=sp{(10−21)}⟹P=(1100−20001−2100101)P−1=(0−12001120023210−1−1201)P−1AP=(10∗∗02∗∗00−12−12003272)PB(λ)=(λ−1)(λ−2)⟹P^−1BP^=(1002)(I00P^)−1=(I00P^−1)⟹(I00P^)−1P−1AP(I00P^)=(D∗0D)