Thread: similar matrices

hi, this is my first time using this forum so if i have posted in the wrong place i do apologise!
I'm having trouble with a question on my matrices worksheet:

"Show directly from the definition of an eigenvalue that two similar matrices A and B have the same eigenvalues"

I've already done this by showing that they have the same characteristic polynomial but this is the second part of the question and you have to show it from the definition. I started by trying to show if Av=xv for an eigenvalue x and eigenvector v then P-1BPv=xv but I can't seem to get rid of the P next to the v!

someone suggested using basis(?) or something but we haven't learnt that yet so I'm not sure about that!

any help would be appreciated, thankyou.

Hi Claire

I think we cannot do it diresctly from the definition of Av=xv. There is somethingin my mind which seems to be so trivial! I think the only thing is using the characteristic equations as we did in part (i).

How about using diagonalization?

I think that would be the solution:

using characteristic equation we will have that B and P-1BP have the same eigenvalue - because det(xI-B)=det(xI-(p^-1BP)).

Now if x be an eigenvalue for A, then according to the definition we have:

Av=xV but A and B are similar then Av=(P^1BP)v

But we know that (P^1BP)v=Bv=xv

Therefore if x be an eigenvalue for B that would be an eigenvalue for A and viceversa.