trace of a matrix trace(a)=Σaii if λ is an value of a (ax=λx) prove / disprove 1. -λ is an value of -a 2. λ^2 is an value of a^2 3.prodoct valus of a = deta 4. (a^-1 λa)^n=a^-1λ^n a
By definition, if a has an eigenvalue λ , ax=λx [*] for some nonzero (eigen)vector x. [*] then implies -ax=-Iax=-λx, where I is the identity matrix. Thus (-a)x=(-λ)x, for the same nonzero vector x. ==> The statement 1 is correct.
Also from [*], (a^2)x=a(ax)=a(λx)=λ(ax)=(λ^2)x ==> The statement 2 is correct.
The statement 3 is usually incorrect. Take a look at the 2 by 2 matrix a=[2 1; 0 2], it has only one eigenvalue 2 with associated eigenvector x=[1 0]^t { and of course its constant multiples}. In this case det a=4, which is not the same as the product of eigenvalues. A similar but correct statement is the following: Let axi=λix and {xi} form a linearly independent set of vectors. Then the prduct of λi= det of a.
(a^-1 λa)^n=(a^-1 λa)(a^-1 λa)(a^-1 λa)...(a^-1 λa) [product of n pieces]
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By definition, if a has an eigenvalue λ , ax=λx [*] for some nonzero (eigen)vector x. [*] then implies -ax=-Iax=-λx, where I is the identity matrix. Thus (-a)x=(-λ)x, for the same nonzero vector x. ==> The statement 1 is correct.
Also from [*], (a^2)x=a(ax)=a(λx)=λ(ax)=(λ^2)x ==> The statement 2 is correct.
The statement 3 is usually incorrect. Take a look at the 2 by 2 matrix a=[2 1; 0 2], it has only one eigenvalue 2 with associated eigenvector x=[1 0]^t { and of course its constant multiples}. In this case det a=4, which is not the same as the product of eigenvalues. A similar but correct statement is the following: Let axi=λix and {xi} form a linearly independent set of vectors. Then the prduct of λi= det of a.
(a^-1 λa)^n=(a^-1 λa)(a^-1 λa)(a^-1 λa)...(a^-1 λa) [product of n pieces]
=(a^-1) λ[(a)(a^-1)] λ[(a)(a^-1)] λ[(a)...(a^-1)] λ(a) =(a^-1)λλ..λ(a)
=(a^-1)(λ^n) (a) . So the statement 4 is correct.