Let a be matrix such that , then sum of diagonal elements of is:
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For the first,
A=P−1MPA=P−1MP
Where M is a (upper triangular) matrix with eigenvalues of A as diagonal elements. This is what it means to say that A is always similar to its Jordan form.
Use Tr(AB)=Tr(BA)Tr(AB)=Tr(BA)
Tr(A)=Tr(P−1MP)=Tr(MPP−1)=Tr(M)=∑nλnTr(A)=Tr(P−1MP)=Tr(MPP−1)=Tr(M)=∑nλn
b) Let B=A−kIB=A−kI with eigenvalues be χnχn
Eigenvalues are determined by solutions of
|B−χI|=0|B−χI|=0
or,
|A−(χ+k)I|=0|A−(χ+k)I|=0
but since you know
|A−λI|=0|A−λI|=0
you get χn=λn−kχn=λn−k
Let YY be an eigenvector of BB. So BY=χYBY=χY. Now plug stuff in for BB and χχ and see what you'd get.
it may correct
A=P−1MPA=P−1MP
Where M is a (upper triangular) matrix with eigenvalues of A as diagonal elements. This is what it means to say that A is always similar to its Jordan form.
Use Tr(AB)=Tr(BA)Tr(AB)=Tr(BA)
Tr(A)=Tr(P−1MP)=Tr(MPP−1)=Tr(M)=∑nλnTr(A)=Tr(P−1MP)=Tr(MPP−1)=Tr(M)=∑nλn
b) Let B=A−kIB=A−kI with eigenvalues be χnχn
Eigenvalues are determined by solutions of
|B−χI|=0|B−χI|=0
or,
|A−(χ+k)I|=0|A−(χ+k)I|=0
but since you know
|A−λI|=0|A−λI|=0
you get χn=λn−kχn=λn−k
Let YY be an eigenvector of BB. So BY=χYBY=χY. Now plug stuff in for BB and χχ and see what you'd get.
it may correct
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