Why is the 3rd step in this theorem proof legal?

[Ylle]

Homework Statement

Let p_A(x) and p_B(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)

Homework Equations

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

 

[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.