- #1

mess1n

- 24

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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