Linear-2 2

Let ARn×nProve or disprove: A invertibleAAT invertibleProof:det(A)0det(AT)0det(AAT)0
Let ARn×n be symmetricProve or disprove: A invertibleA+AT invertibleProof:A+AT=2Adet(A)0det(2A)=2ndet(A)0det(A+AT)0
Let αRLet ARn×nAij={αi=j1ij|α111α111α|i[2,n]:R1+Ri|α+n1α+n1α+n11α111α|1α+n1R1(α+n1)|1111α111α|i[2,n]:RiR1(α+n1)|1110α1000α1|det(A)=(α+n1)(α1)n1
Let AFn×nLet λFλ is called an eigenvalue of AvFn0:Av=λAv is then called an eigenvector of A in respect to eigenvalue λλ0A(1λv)=vvC(A)λ=0Av=0vN(A)

Characteristic polynomial

Characteristic polynomial PA(λ)=|λIA||λ563λ+4|=(λ5)(λ+4)+18=(λ+1)(λ2)PA(λ)=0λ is an eigenvalue of AProof:v0:Av=λv(λIA)v=0vN(λIA)det(λIA)=0PA(λ)=0
A=(0110)PA(λ)=|λ11λ|=λ2+1ARn×nNo eigenvaluesACn×n±i is an eigenvalueAZ2n×n1 is an eigenvalue
A=(311242113)|λ3112λ4211λ3|R1+R2+R3|λ6λ6λ62λ4211λ3|=(λ6)|1112λ4211λ3|(λ6)|1110λ2200λ2|=(λ6)(λ2)2PA(λ)=0[λ=6λ=2λ=6vλ=N(000222113)=N(000111012)=N(000101012)==sp{(121)}λ=2vλ=N(111222111)=N(111000000)=sp{(110),(101)}
Let ABPA(λ)=PB(λ)Proof:|λIB|=|λIP1AP|=|P1(λIA)P|==|P1||λIA||P|=|λIA|
Let T:VV be a linear transformationB basis of V:λ is an eigenvalue of Tλ is an eigenvalue of [T]BProof:Let v0VTv=λv[Tv]B=[λv]B[T]B[v]B=λ[v]B