# 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 :)