Jordan matrix two looks

Question

i have question. I have something like this:

$\begin{bmatrix} -2 & 2 \\ 1 & 3 \\ \end{bmatrix}$

$\lambda_{1} = -1$ $\lambda_{2} = -4$

When jordan matrix looks like this:

$\begin{bmatrix} -1 & 0 \\ 0 & -4\\ \end{bmatrix}$

And when to count it we will use : $J= P^{-1}AP$

EDIT: Okey i will specify it more. In my school taught me that Jordan matrix is created by put eigenvalues on main diagonal and the rest is all zero but few days ago someone gave me Jordan matrix and asked me what the eigenvalues are. And I stand and didn't know what too say. The jordan matrix looks like this:

J = $\begin{bmatrix} -1 & 1\\ 0 & 1\\ \end{bmatrix}$

So the main question is : When we crating Jordan matrix as putting eigenvalues on main diagonal and when by using formula above

Answer 1

The general formula is the following one: $$J = P^{-1} \cdot A \cdot P,$$ where $P$ is the matrix that contains the right hand (generalized) eigenvectors of $A$ as its columns.

If matrix $A$ is diagonalizable over a field, then the Jordan normal form of matrix $A$ (which is the matrix $J$) is a diagonal matrix, with the eigenvalues of matrix $A$ on its main diagonal.

Otherwise, matrix $J$ has the eigenvalues of $A$ on its main diagonal and either zeros or ones on its superdiagonal.

I hope this answers your question.