Proving det(A) = lambda_1 * lambda_2 * ... * lambda_n for Eigenvalue A

[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

 

[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

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 going to 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.

 

[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 :)