Proof of one of the properties of a matrix

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

det(A-[tex]\ell[/tex]E) = ([tex]\ell[/tex][tex]^{1}[/tex] - [tex]\ell[/tex])...([tex]\ell[/tex][tex]^{p}[/tex] - [tex]\ell[/tex])

By putting [tex]\ell[/tex] = 0 in the above equation, we can obtain the desired result.
I understand that if [tex]\ell[/tex] = 0, then what's left is detA = [tex]\ell[/tex][tex]^{1}[/tex][tex]\ell[/tex][tex]^{2}[/tex]...[tex]\ell[/tex][tex]^{p}[/tex]. Or at least, I assume that's what my lecturer is getting at.

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

Why doesn't detA = 0 ?

I'd really appreciate any help.

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

Cheers,
Andrew
 

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  • #2
jbunniii
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I think the notation is poor. What you have written as

[tex]\lambda^k[/tex]

is meant to be the [itex]k'th[/itex] eigenvalue, NOT [itex]\lambda[/itex] 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

[tex]\mathrm{det}(A - \lambda I) = 0[/tex]

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

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

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

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