Proof of one of the properties of a matrix

[mess1n]

Hey, I've come across a part of my notes which states:

Statement:The determinant of a matrix is equal to the product of all its eigenvalues:

Proof: We know that, after opening up the determinant of A - \ellE, we get a polynomial algebraic equation in \ell which has solutions \ell^{1},...,\ell^{p}, i.e. one can write:

det(A-\ellE) = (\ell^{1} - \ell)...(\ell^{p} - \ell)

By putting \ell = 0 in the above equation, we can obtain the desired result.

I understand that if \ell = 0, then what's left is detA = \ell^{1}\ell^{2}...\ell^{p}. Or at least, I assume that's what my lecturer is getting at.

What I don't understand is why the \ell in the characteristic equation (A-\ellE) is the same as the \ell's in the (\ell^{p} - \ell), but not the same as the \ell^{p}'s in (\ell^{p} - \ell).

Why doesn't detA = 0 ?

I'd really appreciate any help.

[note: I couldn't find a lambda symbol so I had to use \ell instead]

Cheers,
Andrew

 

[jbunniii]

I think the notation is poor. What you have written as

\lambda^k

is meant to be the k'th eigenvalue, NOT \lambda to the k'th power. It would be more customary to use subscripts instead of superscripts to avoid this ambiguity.

The characteristic equation is by definition

\mathrm{det}(A - \lambda I) = 0

(I'm using the more standard I for the identity matrix, instead of E.)

The left-hand side of this equation is a polynomial in \lambda of degree exactly n (the characteristic polynomial). Since its degree is n, it can be factored into n linear factors:

\mathrm{det}(A - \lambda I) = (\lambda - \lambda_1)(\lambda - \lambda_2)\cdots(\lambda - \lambda_n)

where the \lambda_k's are the n roots of the characteristic polynomial, which are by definition the eigenvalues of A. In particular, with respect to the VARIABLE \lambda, the eigenvalues \lambda_1,\ldots,\lambda_n are constants, so plugging in \lambda = 0 does not affect them.