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eigenvector formula

what is the difference between these two formula

[br]p(λ)=det(AλI)=[br]p(λ)=det(λIA)=[br][br][br]p(\lambda) = \det( A - \lambda I) =[br]p(\lambda) = \det( \lambda I - A) =[br][br]

ive just noticed my teacher uses the latter but everywhere on the web uses a different formula

did he make a mistake as surely the result would be different

thanks
elli:smile:
Reply 1
One of them is minus the other. But it doesn't really matter as what you're most concerned about is when p(lambda)=0. Since one is minus the other then this will happen for the same values of lambda regardless of which formula you're using.
Reply 2
Original post by ttoby
One of them is minus the other. But it doesn't really matter as what you're most concerned about is when p(lambda)=0. Since one is minus the other then this will happen for the same values of lambda regardless of which formula you're using.


Ahh great thanks, was a right puzzler, but now explained as usual its painfully obvious :smile:
thankyou for your time to respond
Reply 3
For completeness, note that if A and I are n x n matrices, then det(AλI)=(1)ndet(λIA)\det(A-\lambda I) = (-1)^n \det (\lambda I - A). In other words one is only the minus of the other when n is odd.
Reply 4
Original post by DFranklin

Original post by DFranklin
For completeness, note that if A and I are n x n matrices, then det(AλI)=(1)ndet(λIA)\det(A-\lambda I) = (-1)^n \det (\lambda I - A). In other words one is only the minus of the other when n is odd.


Good point, I didn't realise that.
Reply 5
Original post by DFranklin
For completeness, note that if A and I are n x n matrices, then det(AλI)=(1)ndet(λIA)\det(A-\lambda I) = (-1)^n \det (\lambda I - A). In other words one is only the minus of the other when n is odd.


Thank you for this added bit of info, greatly appreciated. :smile: Elli

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