# Thread: Proof of two similar matrices have same eigenvalue and a question

1. ## Proof of two similar matrices have same eigenvalue and a question

Theorem:

I've a question regarding the proof of a theorem.

There is a theorem in my linear algebra book that states:

"If $A$ and $B$ are similar $n$ x $n$ matrices, then they have the same eigenvalues."

Proof of this theorem:

Let $A$ and $B$ be similar matrices so there exist an invertible matrix $P$ such that $B = P^{-1} A P$

By the properties of determinant it follows that:

\begin{align*} \lvert{\lambda I - B}\rvert = \lvert \lambda I - P^{-1} A P \rvert =& \lvert P^{-1} \lambda I P - P^{-1} A P \rvert \\ =& \lvert P^{-1} (\lambda I - A) P \rvert \\ =& \lvert P^{-1} \rvert \lvert \lambda I - A \rvert \lvert P \rvert \\ =& \lvert P^{-1} \rvert \lvert P \rvert \lvert \lambda I - A \rvert \\ =& \lvert P^{-1} P \rvert \lvert \lambda I - A \rvert \\ =& \lvert \lambda I - A \rvert \end{align*}

My question:

My question is: why is that

$\lambda I = P^{-1} \lambda I P$

How do you proof that left hand side of this statement is equal to the right hand side?

What property of linear algebra makes this true? I can't figure out the answer for this problem. Is it possible help me finding the answer?

2. ## Re: Proof of two similar matrices have same eigenvalue and a question

$\text{det}(P^{-1}AP-\lambda I)=\text{det}(P^{-1}AP-\lambda P^{-1}IP)$

$=\text{det}(P^{-1}(A-\lambda I)P)=\text{det}(P^{-1})\text{det}(A-\lambda I)\text{det}(P)$

$=\text{det}(P^{-1})\text{det}(P)\text{det}(A-\lambda I)=\text{det}(I)\text{det}(A-\lambda I)=\text{det}(A-\lambda I)$

What is $P^{-1}IP$???

It is just I.

3. ## Re: Proof of two similar matrices have same eigenvalue and a question

Originally Posted by dwsmith
$\text{det}(P^{-1}AP-\lambda I)=\text{det}(P^{-1}AP-\lambda P^{-1}IP)$

$=\text{det}(P^{-1}(A-\lambda I)P)=\text{det}(P^{-1})\text{det}(A-\lambda I)\text{det}(P)$

$=\text{det}(P^{-1})\text{det}(P)\text{det}(A-\lambda I)=\text{det}(I)\text{det}(A-\lambda I)=\text{det}(A-\lambda I)$
Thank you so much. I was thinking $\lambda$ as a matrix(I don't know why I was thinking that). That was the error on my part. Sorry about that. Thanks for clarifying that using constant multiplication property of matrices.