# Question for a theorem proof

1. Aug 6, 2009

### Ylle

1. The problem statement, all variables and given/known data
Let pA(x) and pB(x) be the characteristic polynomial for A and B. If B is similar to A, then there exists a non-singular matrix S so B = S-1AS. Thus there:

pB(x) = det(B-$$\lambda$$I)
= det(S-1AS-$$\lambda$$I)
= det(S-1(A-$$\lambda$$I)S)
= det(S-1)det(A-$$\lambda$$I)det(S)
= pA(x)

2. Relevant equations

3. The attempt at a solution

My question is, that I really don't know why the 3rd step is legal ? Can I just put the S's anywhere I want ?

Regards

2. Aug 6, 2009

### Dick

No, you can't 'put S's anywhere you want'. That would be silly. What you can do is use that I=S^(-1)*I*S and use the rules of algebra to move the S's around by factoring.