Eigenvalues

[Niels]

How do you prove that det(A) = \lambda_1*\lambda_2*...*\lambda_n, where \lambda_i is the eigenvalues of A? I'm stuck :cry:

[matt grime]

It isn't true, so you can't prove it. You should examine the question carefully.

[dextercioby]

How do you prove that det(A) = \lambda_1*\lambda_2*...*\lambda_n, where \lambda_i is the eigenvalues of A? I'm stuck :cry:

For that to happen,u must make certain assumptions on the matrix 'A'.
The most important is that the matrix 'A' is of square form.If it is symmetrical,then:
a)if A has real elements,then exists a nonsingular orthogonal matrix M which can bring A to diagonal form:
\exists M\in O_{n}(R) ,so that MAM^{T}=A_{diag}
Then it's easy to show that det A=det A_{diag}=product of eigenvalues.
b)if A has complex elements,then exists a unitary matrix Z which can bring A to diagonal form
\exists Z\in U_{n}(C) ,so that ZAZ^{\dagger}=A_{diag}
Again,it's easy to show that the eigenvalues are on the diagonal and hence the det.is the product of eigenvalues.

Daniel.

[dextercioby]

Well,Matt,if u're right and i'm wrong,then i'm gonna kill my QFT teacher since he graduated both physics and maths. :mad: For a square,symmetrical matrix it has to be true.For other cases (meaning square form and nonsymetry),probably not.

Daniel.

[matt grime]

No, we're both correct, I said you should be careful, and you showed something inthe special case the matrix is diagonalizable, which is in some sense the notion I meant when I said that you should be careful. This depends upon how we dsitinguish between algebraic and geometric multiplicity.

[Niels]

Ok,sorry here's the whole text:
Let A be an nxn matrix, and suppose A har n real eigenvalues \lambda_1 ... \lambda_n repeated accordingly to multiplicities, so that
det(A - \lambda I) = (\lambda_1 - \lambda)*(\lambda_2 - \lambda)*...*(\lambda_n - \lambda)
Explain why det(A) is the product of the n eigenvalues of A.
(Hint: the equation holds for all \lambda)

[matt grime]

let lambda = 0

[Niels]

Thanks! I know now that I'm stupied :)