$A$ is diagonalizable, if $A,B$ have then same eigenvalues, then $B$ is also diagonalizable

Question

Given $A_{n\times n},B_{n\times n} \in \mathbb R$ such that $A$ is diagonalizable then:

1. if $A,B$ have then same eigenvalues, then $B$ is also diagonalizable over $ \mathbb R$.

2. if $A,B$ have then same characteristic polynomial, then $B$ is also diagonalizable over $ \mathbb R$.

Those two state basically the same thing and I read that both of these are false, but how?

If both have the same eigenvalues, then why both of them don't have the same diagonal matrix $D$ such that $A= P^{-1} D P = B$ ?

Answer 1

\begin{bmatrix} 1&1 \\ 0& 1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} consider these two matrices.(Jordan Canonical form is the answer).

Answer 2

Answer to your last question: because there are nontrivial Jordan canonical forms.