Need help with an Eigenvalue

1. Dec 29, 2004

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

2. Dec 29, 2004

matt grime

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

Last edited: Dec 29, 2004
3. Dec 29, 2004

dextercioby

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.

4. Dec 29, 2004

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. For a square,symmetrical matrix it has to be true.For other cases (meaning square form and nonsymetry),probably not.

Daniel.

5. Dec 29, 2004

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.

6. Dec 30, 2004

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 $\lambda$)

7. Dec 30, 2004

matt grime

let lambda = 0

8. Dec 30, 2004

Niels

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