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 - $$\ell$$E, we get a polynomial algebraic equation in $$\ell$$ which has solutions $$\ell$$$$^{1}$$,...,$$\ell$$$$^{p}$$, i.e. one can write:

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

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-$$\ell$$E) 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.