Are matrices with the same eigenvalues always similar?

Question

While watching Strang's lecture on similar matrices he stated that if any two matrices,$A$ and $B$, have the same eigenvalues then they can be put in the form $A=P B P^{-1}$ . This is very easy to see when $B$ is the diagonal matrix housing $A's$ eigenvalues. In this example $A$ and $B$ are not diagonal matrices.

But if this is not the case and say we are in $R^2$ and if $A$ has eigenvectors $v_1$ and $v_2$ with eigenvalues $\lambda_1$ and $\lambda_2$ and a angle of $\theta$(say 30 degrees) from each other. And $B$ has the same eigenvalues but they make some angle $\theta_2$(say 120 degrees) then how can I intuitively see that they are similar?

The intuition for similar matrices is often said as "they are the same transformation just in different basis". This is easy to see if the $\theta's$ are same as then $P$ is just a rotation matrix but I want to understand why its true for all matrices of equal eigenvalues. And how can this $P$ then be found?

Edit: Please dont state the example of a matrix having all equal eigenvalues and not being similar to identity. This is stated and we are assuming distinct eigenvalues for this question.

Answer 1

No. Edit: Unless the eigenvalues are distinct and then yes.

Let $A\in\mathcal{M}_2(\mathbb{R})$ be the Jordan block

$$A=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$

The eigenvalues of $A$ are $\lambda_1=\lambda_2 = 1$.

On the other hand the identity matrix $I$ also have two eigenvalues $\lambda_1=\lambda_2=1$.

But clearly $A,I$ are not similar as $I$ is the only matrix that similar to $I$ indeed $PIP^{-1} = I$.

Edit: If $A$ has $n$ distinct eigenvalues then $A$ is diagonalizable (because it has a basis of eigenvalues). Two diagonal matrices with the same eigenvalues are similar and so $A$ and $B$ are similar.

Answer 2

If matrix A has distinct eigenvalues then there exist a matrix P such that $A= PDP^{-1}$ where D is the diagonal matrix with the eigenvalues on its diagonal. If matrix B has the same eigenvalues then there exist a matrix Q such that $B= QDQ^{-1}$. From that $Q^{-1}BQ= D$. Replace D in the first equation with that: $A= P(QBQ^{-1})P^{-1}= (PQ)B(PQ)^{-1}$.