Linear-2 5

Reminder

Let A be diagonalizableLet B={v1,…,vn} a set of eigenvectors of AB is then a basis of FnP−1AP=DP=[I]SB

Triangularizable matrix #definition

Let A∈Fn×nLet T∈Fn×n be a triangular matrixA∼T⟺A is triangularizable

Upper and lower triangularizable matrix #lemma

Let A∈Fn×nLet U∈Fn×n be an upper-triangular matrixLet L∈Fn×n be a lower-triangular matrixA∼U⟺A∼L

Triangularizable matrix #lemma

Let A∈Fn×nA is triangularizable ⟺PA(λ) is factorizable into linear factorsCorollary: every matrix is triangularizable over CProof:⟹ Let U∈Fn×n be a triangular matrixLet A∼ULet ∀i∈[1,n]:αi=UiiPA(λ)=PU(λ)=∏i=1n(λ−αi)⟹PA(λ) is factorized into linear factors⟸ Let PA(λ) be factorized into linear factorsBase case. Let n=1,PA(λ)=λ−A11 and A is upper-triangular, A∼AInduction step.Let ∀n′≤n:A∈Fn′×n′:PA(λ) is factorizable into linear factors⟹A∼ULet A∈Fn+1×n+1,PA(λ)=∏i=1n+1(λ−αi)⟹A definitely has eigenvalues in FLet α∈F be an eigenvalue of ALet Eα=sp{v1,…,vt},t≥1Let B={v1,…,vt}∪{ut+1,…,un+1} be a basis of Fn+1Let P=(v1||…vt||ut+1||…un+1||)P−1AP=P−1(αv1||…αvt||Aut+1||…Aun+1||)==(αItB0C)PP−1AP(λ)=(λ−α)t⋅PC(λ)⟹PA(λ)=(λ−α)t⋅PC(λ)⟹PC(λ) is factorizable into linear factorsn+1−t≤n⟹C∈Fn+1−t×n+1−t is triangularizable⟹P−1^CP^=U^,P^∈Fn+1−t×n+1−tLet Q=P(I00P^)Q−1AQ=(I00P^)−1P−1AP(I00P^)=(I00P^)−1(αIB0C)(I00P^)==(I00P−1^)(αIBP^0CP^)=(αIBP^0U^)=U⟹A∼U

Properties of triangularizable matrix #lemma

Let A∈Fn×n be triangularizable⟹PA(λ) is factorizable into linear factorsLet {α1,…,αt} be eigenvalues of Atr(A)=∑i=1tαi⋅μA(αi)|A|=∏i=1tαiμA(αi)

Matrix polynomial expression #definition

Let A∈Fn×nLet P(x)∈Ft[x]P(A)=atAt+⋯+a1A+a0I is then called a matrix polynomial expression

Existence of polynomial with the given matrix as a root #lemma

A∈Fn×n{I,A,A2,…,An2}⊆Fn×n is a linear dependence∑i=0n2αiAi=0⟹P(A)=∑i=0n2αiAi=0⟹P(x)=∑i=0n2αixi

Adjoint matrix #definition

Let A∈Fn×nAdjoint matrix of A is denoted as: adjA∈Fn×n∀i,j∈[1,n]:(adjA)ij=(−1)i+j|Mji|

Adjoint of a transpose #lemma

Let A∈Fn×nThen (adjA)T=adj(AT)Proof:(adjA)ijT=(adjA)ji=(−1)i+j|Mij(A)|=(−1)i+j|(Mij(A))T|==(−1)i+j|Mji(AT)|=adj(AT)ij

Product of matrix and its adjoint #theorem

Let A∈Fn×nA⋅adjA=adjA⋅A=det(A)IProof:Let i∈[1,n](A⋅adjA)ii=∑k=1nAik⋅(adjA)ki=∑k=1nAik⋅(−1)k+i|Mik|=det(A)Let j≠i∈[1,n]Let A^:Rk(A^)={Rk(A)k≠jRi(A)k=j⟹det(A^)=0(A⋅adjA)ij=∑k=1nAik⋅(adjA)kj=∑k=1nAik⋅(−1)k+j|Mjk(A)|==∑i=1nA^jk⋅(−1)k+j|Mjk(A^)|=det(A^)=0⟹A⋅adjA=det(A)I(AT⋅adj(AT))T=(det(AT)I)T⟹(adj(AT))T⋅A=det(A)I⟹adjA⋅A=det(A)I

Cayley-Hamilton theorem #theorem

Let A∈Fn×nThen PA(A)=0Explanation, not proof:PA(A)=det(AI−A)=det(0)=0Proof:PA(λ)=λn+∑i=0n−1aiλi(λI−A)⋅adj(λI−A)=det(λI−A)I=PA(λ)Iadj(λI−A)∈Fn−1[λ]n×n⟹∃{B0,…,Bn−1}⊆Fn×n:adj(λI−A)=∑i=0n−1λiBi⟹(λI−A)⋅∑i=0n−1λiBi=PA(λ)I⟹(λI−A)⋅∑i=0n−1λiBi=∑i=0nλiaiI⟹λnλn−1λn−2…λ1Left sideBn−1Bn−2−ABn−1Bn−3−ABn−2…B0−AB1−AB0Right sideIan−1Ian−2I…a1Ia0I⟹AnBn−1+An−1(Bn−2−ABn−1)+⋯+A(B0−AB1)−AB0==An+an−1An−1+⋯+a0I=PA(A)⟹0=PA(A)