determinant of matrix is equal to product of eigenvalues

J

Julius

New member
Joined
Apr 24, 2009
Messages
8
Hi

I got such great help from my last question that I thought I would get some help with the related question. How do I prove that the determinant of a matrix is equal to the product of it's eigenvalues. ( Hopefully this will be my last question for a considerable time. )


The hint is to use the fact that det( A-LI) = (-1)^n (L-L1)...(L-Ln)

L= lambda. I am having trouble getting through the (-1)^n . I understand the FTA will give the factoring spart and it also seems to be true working examples with 2x2 and 3x3 matirces but I cannot prove the hint! I feel that I don't really understand the materal unless I can first prove the hint and show that the det(A-LI) = (-1)^n(L-L1)...(L-Ln) .

Thank you again so much if you can help. ( I think galactica took my last problem and quickly helped me get it.)
 
Julius said:
Hi

I got such great help from my last question that I thought I would get some help with the related question. How do I prove that the determinant of a matrix is equal to the product of it's eigenvalues. ( Hopefully this will be my last question for a considerable time. )


The hint is to use the fact that det( A-LI) = (-1)^n (L-L1)...(L-Ln)

L= lambda. I am having trouble getting through the (-1)^n . I understand the FTA will give the factoring spart and it also seems to be true working examples with 2x2 and 3x3 matirces but I cannot prove the hint! I feel that I don't really understand the materal unless I can first prove the hint and show that the det(A-LI) = (-1)^n(L-L1)...(L-Ln) .

Thank you again so much if you can help. ( I think galactica took my last problem and quickly helped me get it.)
Click to expand...

For a quick proof go to:

http://www.mathhelpforum.com/math-help/ ... proof.html
 
Thank you SubhotoshKhan for your response.

It is interesting. Maybe you can help clarify two items. I assume det (A-lambdaI ) is the same as det( lambdaI-A). I see one version being used in Poole's book and the other version in others.

The other point of confusion is for me is are we allowed to use 0 in lambda if we don't know for sure if 0 is an eigenvalue.

Thank you once again and any help is appreciated.
Julius
 
Those two matricies are off by a scalar multiple (-1). When taking the determinant you are given a polynomial in \(\displaystyle \lambda\). Multiplying by a scalar will not change the roots of this polynomial.
 
Thank you Daon, for reply. I understand your explanation.

I was confusing a scalar determinant with the roots of the charateristic polynomial. Now I am Ok with this.


Have a great day and thanx again!
Julius
 
You must log in or register to reply here.
Top